---
title: "Group Theory"
book: "University Mathematics — Year 3"
subject: math
language: en
chapter: 1
exercises: 12
source: https://one-course.com/books/math/5/en/chapter/1-group-theory
---

# Chapter 1 — Group Theory

The Year 2 volume met groups as bookkeeping devices: Lagrange’s theorem, cyclic groups, the symmetric group and its signature. This chapter turns group theory into a *method*. The engine is the notion of a group *acting* on a set: counting [orbits](#def-b3-groups-action) and fixed points yields the [class equation](#cor-b3-groups-classeq), Cauchy’s theorem and the three Sylow theorems — the fundamental local-to-global principle of finite group theory. We then learn to assemble groups (direct and [semidirect products](#def-b3-groups-semidirect)) and to disassemble them ([composition series](#thm-b3-groups-jordanholder), [solvable groups](#def-b3-groups-derived)), and we prove the theorem that, in [Chapter 4](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#ch-b3-galois), will close a three-century-old question about polynomial equations: the [alternating group](#thm-b3-groups-ansimple) $A_n$ is [simple](#def-b3-groups-simple) for $n
\geq 5$.

## 1.1 Quotient groups and the isomorphism theorems

Throughout, $G$ is a group written multiplicatively, $e$ its identity. Recall from the Year 2 volume: subgroups, cosets $gH$, Lagrange’s theorem ($\abs G = [G:H]\,\abs H$ for finite $G$), the order of an element, cyclic groups, and the symmetric group $S_n$ with its signature morphism $\varepsilon \colon S_n \to \{\pm1\}$.

**Definition 1.1.**

A subgroup $N$ of $G$ is *normal* (written $N \trianglelefteq G$) when $gNg^{-1} = N$ for every $g
\in G$ — equivalently, when left and right cosets coincide: $gN =
Ng$ for all $g$.

**Theorem 1.2 (Quotient group).**

Let $N \trianglelefteq G$. The set $G/N$ of cosets, with the multiplication $(gN)(hN) = ghN$, is a well-defined group, the *quotient group*, and the *canonical projection* $\pi \colon G \to G/N$, $g \mapsto gN$, is a surjective morphism with kernel $N$. Conversely, every kernel of a group morphism is [normal](#def-b3-groups-normal): [normal subgroups](#def-b3-groups-normal) are exactly the kernels.

**Proof.** Well-definedness is the whole point. If $gN = g'N$ and $hN = h'N$, write $g' = gn$, $h' = hm$ with $n, m \in N$. Then $g'h' = gnhm =
gh\,(h^{-1}nh)\,m \in ghN$ since $h^{-1}nh \in N$ by normality: the product of cosets does not depend on the representatives. Associativity, identity $eN = N$ and inverses $(gN)^{-1} = g^{-1}N$ are inherited from $G$. Clearly $\pi$ is a surjective morphism and $\pi(g) = N \iff g \in N$.

If $f \colon G \to H$ is a morphism and $k \in \ker f$, then $f(gkg^{-1}) = f(g)f(k)f(g)^{-1} = e$: kernels are [normal](#def-b3-groups-normal). ∎

**Theorem 1.3 (Universal property; first isomorphism theorem).**

Let $f \colon G \to H$ be a morphism and $N \trianglelefteq G$ with $N \subseteq \ker f$. There is a unique morphism $\bar f \colon G/N
\to H$ with $f = \bar f \circ \pi$. In particular, taking $N = \ker
f$:

$$
G/\ker f \;\xrightarrow{\;\sim\;}\; \operatorname{im} f,
\qquad gN \mapsto f(g).
$$

**Proof.** Uniqueness: $\bar f(gN)$ must be $f(g)$. Existence: if $gN = g'N$ then $g^{-1}g' \in N \subseteq \ker f$, so $f(g) = f(g')$ and $\bar
f(gN) = f(g)$ is well defined; it is a morphism because $f$ is. For $N = \ker f$: $\bar f$ is injective, since $\bar f(gN) = e$ means $g \in \ker f$, i.e. $gN = N$; its image is that of $f$. ∎

**Theorem 1.4 (Second and third isomorphism theorems).**

Let $H \leq G$ and $N \trianglelefteq G$.

1. $HN = \{hn : h \in H,\, n \in N\}$ is a subgroup, $N  \trianglelefteq HN$, $H \cap N \trianglelefteq H$, and $$H/(H \cap N) \;\cong\; HN/N .$$
2. If moreover $N \subseteq K \trianglelefteq G$ , then $K/N  \trianglelefteq G/N$ and $(G/N)\big/(K/N) \cong G/K$ .

**Proof.** (1) $HN$ is a subgroup: $(hn)(h'n') = hh'\,(h'^{-1}nh')n' \in HN$ and $(hn)^{-1} = h^{-1}(hn^{-1}h^{-1}) \in HN$, using normality of $N$. Compose $H \hookrightarrow HN \xrightarrow{\pi} HN/N$: this morphism is surjective ($hnN = hN$) with kernel $\{h \in H : h \in
N\} = H \cap N$; apply [Theorem 1.3](#thm-b3-groups-firstiso).

(2) The projection $G/N \to G/K$, $gN \mapsto gK$, is well defined ($N \subseteq K$), surjective, with kernel $K/N$; apply [Theorem 1.3](#thm-b3-groups-firstiso) again. ∎

**Theorem 1.5 (Correspondence theorem).**

Let $N \trianglelefteq G$. The map $H \mapsto H/N$ is a bijection between subgroups of $G$ containing $N$ and subgroups of $G/N$, preserving inclusions, indices and normality (in both directions).

**Proof.** Its inverse is $\bar H \mapsto \pi^{-1}(\bar H)$. Both maps send subgroups to subgroups, and are mutually inverse: $\pi^{-1}(H/N) =
HN = H$ since $N \subseteq H$, and $\pi(\pi^{-1}(\bar H)) = \bar H$ by surjectivity of $\pi$. Inclusions are clearly preserved; $[G:H] = [G/N : H/N]$ because $gH \mapsto (gN)(H/N)$ is a well-defined bijection between coset spaces; and $gHg^{-1} = H$ for all $g$ iff $(gN)(H/N)(gN)^{-1} = H/N$ for all $gN$, again by surjectivity of $\pi$. ∎

**Example 1.6.**

$\varepsilon \colon S_n \to \{\pm 1\}$ gives $S_n/A_n \cong
\{\pm1\}$; $\det \colon GL_n(K) \to K^\times$ gives $GL_n(K)/SL_n(K) \cong K^\times$; $t \mapsto \eu^{2\iu\pi t}$ gives $\R/\Z \cong \mathbb U$, the circle group. The first isomorphism theorem is how quotients are *computed* in practice: find a surjection with the right kernel.

**Method 1.7.**

To prove $N \trianglelefteq G$, in decreasing order of elegance: exhibit $N$ as the kernel of a morphism defined on $G$; check $gNg^{-1} \subseteq N$ for all $g$ (this suffices: applying it to $g^{-1}$ and conjugating gives the reverse inclusion); verify that $N$ is a union of [conjugacy classes](#ex-b3-groups-actions); or note that $[G:N] = 2$ (then $gN = Ng$ is forced — [Exercise 1.1](#exo-b3-groups-1)).

## 1.2 Group actions

**Definition 1.8.**

An *action* of $G$ on a set $X$ is a morphism $\varphi \colon G \to \mathfrak{S}(X)$ into the group of bijections of $X$; one writes $g \cdot x$ for $\varphi(g)(x)$. Equivalently: a map $G \times X \to X$ with $e \cdot x = x$ and $g
\cdot (h \cdot x) = (gh) \cdot x$. The *orbit* of $x$ is $\mathcal O_x = \{g \cdot x : g \in G\}$, its *stabilizer* is the subgroup $G_x = \{g :
g\cdot x = x\}$, and $X^G = \{x : \forall g,\ g \cdot x = x\}$ is the set of *fixed points*. The action is *transitive* if there is exactly one orbit, *faithful* if $\varphi$ is injective, *free* if all stabilizers are trivial.

**Example 1.9.**

Five [actions](#def-b3-groups-action) run all of finite group theory:

1. $G$ on itself by *left translation* $g \cdot x = gx$ : free and transitive.
2. $G$ on itself by *conjugation* $g \cdot x = gxg^{-1}$ : [orbits](#def-b3-groups-action) are the *conjugacy classes* , [stabilizers](#def-b3-groups-action) the *centralizers* $Z_G(x) = \{g : gx = xg\}$ , fixed points the *center* $Z(G)$ .
3. $G$ on the coset space $G/H$ by $g \cdot xH = gxH$ : transitive, with [stabilizer](#def-b3-groups-action) of the coset $H$ equal to $H$ . Every transitive [action](#def-b3-groups-action) is of this form ( [Exercise 1.8](#exo-b3-groups-8) ).
4. $G$ on its set of subgroups by conjugation: the [stabilizer](#def-b3-groups-action) of $H$ is the *normalizer* $N_G(H) =  \{g : gHg^{-1} = H\}$ , the largest subgroup of $G$ in which $H$ is [normal](#def-b3-groups-normal) .
5. $S_n$ on $\intint{1}{n}$ : the mother of all examples.

**Theorem 1.10 (Orbit–stabilizer).**

The map $gG_x \mapsto g \cdot x$ is a well-defined bijection $G/G_x \to \mathcal O_x$. In particular, for $G$ finite,

$$
\abs{\mathcal O_x} = [G : G_x] \quad\text{divides } \abs G,
$$

and, the [orbits](#def-b3-groups-action) partitioning $X$ (they are the classes of the equivalence $x \sim y \iff y \in \mathcal O_x$),

$$
\abs X = \sum_{i} \,[G : G_{x_i}]
\qquad (x_i\colon \text{one point per orbit}).
$$

**Proof.** Well defined and injective: $gG_x = hG_x \iff h^{-1}g \in G_x \iff
h^{-1}g \cdot x = x \iff g \cdot x = h \cdot x$; read the chain in both directions. Surjectivity is the definition of the [orbit](#def-b3-groups-action). The counting statements follow from Lagrange’s theorem and the partition of $X$ into [orbits](#def-b3-groups-action). ∎

**Corollary 1.11 (Class equation).**

For a finite group $G$, picking one representative $x_i$ in each [conjugacy class](#ex-b3-groups-actions) with more than one element:

$$
\abs G = \abs{Z(G)} + \sum_i \,[G : Z_G(x_i)],
\qquad\text{each } [G : Z_G(x_i)] > 1 \text{ dividing } \abs G.
$$

**Proof.** Apply [Theorem 1.10](#thm-b3-groups-orbitstab) to the conjugation [action](#def-b3-groups-action): singleton [orbits](#def-b3-groups-action) are exactly the elements of $Z(G)$. ∎

**Theorem 1.12 (Fixed points of ppp-groups).**

Let $p$ be prime. A *$p$-group* is a finite group whose order is a power of $p$. If a $p$-group $G$ acts on a finite set $X$, then

$$
\abs{X^G} \equiv \abs X \pmod p .
$$

Consequences: a nontrivial $p$-group has nontrivial [center](#ex-b3-groups-actions), and every group of order $p^2$ is abelian.

**Proof.** Each [orbit](#def-b3-groups-action) has cardinality $[G:G_x]$, a power of $p$; this power is $1$ exactly on fixed points and otherwise divisible by $p$. Summing over [orbits](#def-b3-groups-action) gives the congruence. For the [center](#ex-b3-groups-actions): conjugation [action](#def-b3-groups-action) of $G$ on itself has $X^G = Z(G)$, so $\abs{Z(G)} \equiv
\abs G \equiv 0 \pmod p$, and $Z(G) \ni e$ forces $\abs{Z(G)} \geq
p$. Order $p^2$: if $Z(G) \neq G$ then $\abs{Z(G)} = p$ and $G/Z(G)$ is cyclic of order $p$, which forces $G$ abelian ([Exercise 1.2](#exo-b3-groups-2)) — contradiction. ∎

**Theorem 1.13 (Cauchy).**

If a prime $p$ divides $\abs G$, then $G$ contains an element of order $p$.

**Proof (McKay).** Let $X = \{(g_1, \dots, g_p) \in G^p : g_1 g_2 \cdots g_p = e\}$. Choosing $g_1, \dots, g_{p-1}$ freely determines $g_p$: $\abs X =
\abs G^{p-1}$, divisible by $p$. The cyclic group $\Z/p\Z$ acts on $X$ by cyclic shift $(g_1, \dots, g_p) \mapsto (g_2, \dots, g_p,
g_1)$ — this preserves $X$, since $g_2 \cdots g_p g_1 =
g_1^{-1}(g_1 \cdots g_p)g_1 = e$. By [Theorem 1.12](#thm-b3-groups-pfixed), $\abs{X^{\Z/p\Z}} \equiv \abs X \equiv 0 \pmod p$. Fixed points are the constant tuples $(g, \dots, g)$ with $g^p = e$; the tuple $(e,
\dots, e)$ is one of them, so there are at least $p$ of them, hence at least one $g \neq e$ with $g^p = e$: its order is exactly $p$. ∎

**Theorem 1.14 (Cayley).**

Every group of order $n$ embeds in $S_n$.

**Proof.** Left translation $\varphi \colon G \to \mathfrak S(G) \cong S_n$ is a morphism; $\varphi(g) = \mathrm{id}$ forces $g = ge = e$: it is faithful. ∎

**Method 1.15.**

Fixed-point counting is the universal opening move of finite group theory. To prove that something *exists* (a central element, an element of order $p$, a [normal subgroup](#def-b3-groups-normal), a fixed point), make a well-chosen group act on a well-chosen finite set, then compare $\abs{X^G}$ with $\abs X$ modulo $p$, or let [orbit](#def-b3-groups-action) sizes divide the group order. The proofs of Theorems [1.12](#thm-b3-groups-pfixed) and [1.13](#thm-b3-groups-cauchy) and of all three Sylow theorems below are five variations on this single idea.

## 1.3 The Sylow theorems

Lagrange’s theorem says the order of a subgroup divides $\abs G$; the converse fails ($A_4$, of order $12$, has no subgroup of order $6$ — [Exercise 1.1](#exo-b3-groups-1)). The Sylow theorems salvage the converse for prime powers, and their counting clause is the sharpest general tool we have for producing [normal subgroups](#def-b3-groups-normal).

**Definition 1.16.**

Write $\abs G = p^a m$ with $p \nmid m$. A *Sylow $p$-subgroup* of $G$ is a subgroup of order $p^a$ — a $p$-subgroup of the largest conceivable order. The number of Sylow $p$-subgroups of $G$ is denoted $n_p$.

**Lemma 1.17.**

If $\abs G = p^a m$ with $p \nmid m$, then $\dbinom{p^a m}{p^a}
\equiv m \pmod p$.

**Proof.** In $\mathbb F_p[X]$, the freshman’s dream $(1+X)^p = 1 + X^p$ (the coefficients $\binom pk$, $0<k<p$, are divisible by $p$: $p$ divides the numerator of $\frac{p!}{k!(p-k)!}$ but not the denominator) iterates to $(1+X)^{p^a} = 1 + X^{p^a}$, whence

$$
(1+X)^{p^a m} = \bigl(1 + X^{p^a}\bigr)^m
= \sum_{k=0}^{m} \binom mk X^{k p^a} \quad\text{in } \mathbb
F_p[X].
$$

Identify the coefficient of $X^{p^a}$: on the left $\binom{p^a
m}{p^a} \bmod p$, on the right $\binom m1 = m$. ∎

**Theorem 1.18 (Sylow I: existence).**

For every prime $p$, Sylow $p$-subgroups of $G$ exist.

**Proof (Wielandt).** Let $\Omega$ be the set of *subsets* of $G$ of cardinality $p^a$; $G$ acts on $\Omega$ by left translation $g \cdot S = gS$. By [Lemma 1.17](#lem-b3-groups-binom), $\abs\Omega = \binom{p^a m}{p^a}
\equiv m \not\equiv 0 \pmod p$, so some [orbit](#def-b3-groups-action) $\mathcal O_S$ has size prime to $p$ (if $p$ divided every [orbit](#def-b3-groups-action) size, it would divide $\abs\Omega$). Let $H = G_S$ be the [stabilizer](#def-b3-groups-action) of such an $S$. Since $[G : H] = \abs{\mathcal O_S}$ is prime to $p$ and $p^a \mid
\abs G = [G:H]\,\abs H$, we get $p^a \mid \abs H$. Conversely, fix $s \in S$: the map $H \to S$, $h \mapsto hs$, is injective and lands in $S$ because $hS = S$; hence $\abs H \leq \abs S = p^a$. So $\abs H = p^a$. ∎

**Theorem 1.19 (Sylow II: domination and conjugacy).**

Let $P$ be a Sylow $p$-subgroup and $Q$ any $p$-subgroup of $G$. Then $Q \subseteq gPg^{-1}$ for some $g \in G$. In particular all Sylow $p$-subgroups are conjugate, and $P \trianglelefteq G \iff
n_p = 1$.

**Proof.** Let $Q$ act on the coset space $X = G/P$, of cardinality $m
\not\equiv 0 \pmod p$. By [Theorem 1.12](#thm-b3-groups-pfixed) applied to the $p$-group $Q$, $\abs{X^Q} \equiv m \not\equiv 0 \pmod p$: there is a fixed coset $gP$, i.e. $QgP = gP$, i.e. $g^{-1}Qg \subseteq
P$. If $Q$ is itself a [Sylow subgroup](#def-b3-groups-sylow), equality of orders turns $Q \subseteq gPg^{-1}$ into an equality. Finally $P
\trianglelefteq G$ iff its conjugates $\{gPg^{-1}\}$ — which by the above are *all* the Sylow $p$-subgroups — reduce to $\{P\}$. ∎

**Theorem 1.20 (Sylow III: counting).**

$n_p \equiv 1 \pmod p$, and $n_p = [G : N_G(P)]$, which divides $m$.

**Proof.** Let $\mathrm{Syl}_p$ be the set of Sylow $p$-subgroups; $G$ acts on it transitively by conjugation ([Theorem 1.19](#thm-b3-groups-sylow2)), with [stabilizer](#def-b3-groups-action) of $P$ the [normalizer](#ex-b3-groups-actions) $N_G(P) \supseteq P$: $n_p = [G :
N_G(P)]$, and $m = [G:P] = [G:N_G(P)]\,[N_G(P):P]$ shows $n_p \mid
m$.

Now restrict the [action](#def-b3-groups-action) to $P$ and count fixed points. If $Q \in
\mathrm{Syl}_p$ is fixed by $P$, then $P \subseteq N_G(Q)$; both $P$ and $Q$ are Sylow $p$-subgroups of the group $N_G(Q)$, hence conjugate in it ([Theorem 1.19](#thm-b3-groups-sylow2) applied to $N_G(Q)$); but $Q \trianglelefteq N_G(Q)$, so $Q$ is its only conjugate there: $P = Q$. Thus the only fixed point is $P$ itself, and [Theorem 1.12](#thm-b3-groups-pfixed) gives $n_p = \abs{\mathrm{Syl}_p}
\equiv \abs{\mathrm{Syl}_p^P} = 1 \pmod p$. ∎

**Method 1.21.**

To analyse a group of given order $n = p^a m$: list the divisors of $m$ congruent to $1$ mod $p$ — these are the candidates for $n_p$. If the only candidate is $1$, the Sylow $p$-subgroup is [normal](#def-b3-groups-normal). If $n_p > 1$ is forced to be small, act by conjugation on $\mathrm{Syl}_p$ to obtain a morphism $G \to S_{n_p}$ with small kernel. And count elements: distinct Sylow $p$-subgroups of *prime* order $p$ intersect trivially, so they carry $n_p(p-1)$ elements of order exactly $p$; overlapping counts for different primes often force a contradiction ([Exercise 1.7](#exo-b3-groups-7)).

**Example 1.22.**

Let $\abs G = pq$ with $p < q$ primes and $p \nmid q - 1$. Then $n_q \mid p$ and $n_q \equiv 1 \bmod q$ force $n_q = 1$ (as $p <
q$); $n_p \mid q$ and $n_p \equiv 1 \bmod p$ force $n_p = 1$ (as $q
\not\equiv 1 \bmod p$). Let $P, Q$ be the two [normal](#def-b3-groups-normal) Sylows: $P
\cap Q = \{e\}$ (coprime orders), so $\abs{PQ} = pq$ ([Exercise 1.4](#exo-b3-groups-4)) and $G \cong P \times Q \cong \Z/p\Z
\times \Z/q\Z \cong \Z/pq\Z$ by [Proposition 1.24](#prop-b3-groups-direct) below. Every group of order $15$, $33$, $35$, … is cyclic. The excluded case $p \mid q - 1$ produces exactly one more group, nonabelian — see the weekend problem ([Problem 1.1](#pb-b3-groups-1)).

**Example 1.23 (A complete Sylow census: S4S_4S4​).**

Let us run the method on $G = S_4$, $\abs G = 24 = 2^3\cdot3$. *Sylow $3$:* $n_3 \mid 8$, $n_3 \equiv 1 \bmod 3$, so $n_3
\in \{1, 4\}$; since $\langle(123)\rangle$ and $\langle(124)\rangle$ are distinct, $n_3 = 4$ — the four subgroups $\langle(abc)\rangle$, one for each $3$-element subset $\{a, b, c\}$, accounting for the $8$ three-cycles. By Sylow II they are conjugate, and the conjugation morphism $S_4 \to
S_{\mathrm{Syl}_3} \cong S_4$ is an isomorphism here (its kernel is contained in $N = N_G(\langle(123)\rangle)$ of order $24/4 =
6$, and a [normal subgroup](#def-b3-groups-normal) of $S_4$ inside $S_3$-like $N$ must be trivial: it would consist of even permutations fixing the four Sylows, and only $e$ does). *Sylow $2$:* $n_2 \mid 3$, $n_2
\equiv 1 \bmod 2$: $n_2 \in \{1, 3\}$. The subgroup $D =
\langle(1234), (13)\rangle$ has order $8$ (a dihedral $D_4$: the symmetries of the square with vertices $1, 2, 3, 4$), is not [normal](#def-b3-groups-normal) ($(12)(1234)(12) = (2134)$ generates a different $4$-cycle subgroup), so $n_2 = 3$: the three copies of $D_4$ correspond to the three ways of pairing $4$ points into a “square”. Note the moral of the census: $\abs{S_4} = 24$ leaves room for either Sylow to fail normality, and both do — compare order $12$, where the count forces one of them [normal](#def-b3-groups-normal) (Part IV of [Problem 1.1](#pb-b3-groups-1)).

## 1.4 Products, direct and semidirect

**Proposition 1.24 (Recognizing a direct product).**

Let $H, K \trianglelefteq G$ with $H \cap K = \{e\}$ and $HK = G$. Then $(h,k) \mapsto hk$ is an isomorphism $H \times K \to
G$.

**Proof.** For $h \in H$, $k \in K$, the [commutator](#def-b3-groups-derived) $hkh^{-1}k^{-1}$ lies in $K$ (read it as $(hkh^{-1})k^{-1}$, using normality of $K$) and in $H$ (read it as $h(kh^{-1}k^{-1})$): it is $e$, so $H$ and $K$ commute elementwise and the map is a morphism. It is surjective since $HK = G$, and injective since $hk = e$ gives $h = k^{-1} \in
H \cap K = \{e\}$. ∎

Normality of *both* factors is what fails most often: in $S_3
= \langle (1\,2\,3)\rangle \,\langle(1\,2)\rangle$ both factors intersect trivially and generate, yet $S_3 \not\cong \Z/3\Z \times
\Z/2\Z$. The right notion when only one factor is [normal](#def-b3-groups-normal):

**Definition 1.25.**

Let $H$, $K$ be groups and $\varphi \colon K \to
\operatorname{Aut}(H)$ a morphism. The *semidirect product* $H \rtimes_\varphi K$ is the set $H \times K$ equipped with

$$
(h, k)\,(h', k') = \bigl(h\,\varphi(k)(h'),\; kk'\bigr).
$$

**Proposition 1.26.**

$H \rtimes_\varphi K$ is a group; $H \times \{e\}$ is a [normal subgroup](#def-b3-groups-normal) isomorphic to $H$, $\{e\} \times K$ a subgroup isomorphic to $K$; they intersect trivially and generate. Conversely, if $G =
NK$ with $N \trianglelefteq G$, $K \leq G$ and $N \cap K = \{e\}$, then $G \cong N \rtimes_\varphi K$ for $\varphi(k) = (n \mapsto
knk^{-1})$.

**Proof.** Direct verification: associativity reduces to $\varphi(kk') =
\varphi(k)\circ\varphi(k')$ and each $\varphi(k)$ being a morphism; the identity is $(e,e)$ and $(h,k)^{-1} =
\bigl(\varphi(k^{-1})(h^{-1}), k^{-1}\bigr)$. The projection $(h,k)
\mapsto k$ is a morphism onto $K$ with kernel $H \times \{e\}$, which is therefore [normal](#def-b3-groups-normal). For the converse: every $g \in G$ writes *uniquely* as $nk$ with $n \in N$, $k \in K$ (existence: $G =
NK$; uniqueness: $nk = n'k'$ gives $n'^{-1}n = k'k^{-1} \in N \cap
K$), and

$$
(nk)(n'k') = n\,(kn'k^{-1})\;kk'
$$

shows that $nk \mapsto (n, k)$ transports the law of $G$ to that of $N \rtimes_\varphi K$. ∎

**Example 1.27.**

(a) The *dihedral group* $D_n$ ($n \geq
3$) of the $2n$ symmetries of a regular $n$-gon: the rotations form a [normal subgroup](#def-b3-groups-normal) of index $2$, any reflection generates a complement, and conjugating a rotation by a reflection inverts it: $D_n \cong \Z/n\Z \rtimes_\varphi \Z/2\Z$ with $\varphi(1) = (x
\mapsto -x)$. (b) The affine group of a line, $\{x \mapsto ax + b : a \in
K^\times,\, b \in K\} \cong K \rtimes K^\times$: translations [normal](#def-b3-groups-normal), homotheties a complement. (c) $S_n \cong A_n \rtimes \Z/2\Z$ (complement: any transposition). (d) The [quaternion group](#pb-b3-groups-1) $Q_8$ is *not* a [semidirect product](#def-b3-groups-semidirect) of proper subgroups: every nontrivial subgroup contains $-1$ ([Problem 1.1](#pb-b3-groups-1)), so no two proper subgroups intersect trivially.

## 1.5 Solvable groups; simplicity of $A_n$

**Definition 1.28.**

The *commutator* of $x, y \in G$ is $[x,y]
= xyx^{-1}y^{-1}$; the *derived subgroup* $D(G)$ is the subgroup generated by all commutators. The *derived series* is $D^0(G) = G$, $D^{i+1}(G) = D(D^i(G))$, and $G$ is *solvable* if $D^n(G) =
\{e\}$ for some $n$.

**Proposition 1.29.**

$D(G)$ is [normal](#def-b3-groups-normal) (indeed stable under every automorphism), $G/D(G)$ is abelian, and for $N \trianglelefteq G$: $G/N$ abelian $\iff
D(G) \subseteq N$. Moreover $G$ is [solvable](#def-b3-groups-derived) iff there is a chain $G = G_0 \trianglerighteq G_1 \trianglerighteq \dots
\trianglerighteq G_n = \{e\}$ with each $G_{i+1} \trianglelefteq
G_i$ and each quotient $G_i/G_{i+1}$ abelian. Subgroups and quotients of [solvable groups](#def-b3-groups-derived) are [solvable](#def-b3-groups-derived); conversely, if $N$ and $G/N$ are [solvable](#def-b3-groups-derived), so is $G$.

**Proof.** An automorphism $\alpha$ maps $[x,y]$ to $[\alpha x, \alpha y]$: it permutes the [commutators](#def-b3-groups-derived), so preserves the subgroup they generate; conjugations are automorphisms, whence normality. In $G/D(G)$, $\bar x\bar y \bar x^{-1}\bar y^{-1} = \overline{[x,y]} = \bar e$: the quotient is abelian. If $G/N$ is abelian then every $[x,y] \in
N$, so $D(G) \subseteq N$; conversely if $D(G) \subseteq N$ then $G/N$, a quotient of the abelian $G/D(G)$ by the third isomorphism theorem, is abelian.

If $G$ is [solvable](#def-b3-groups-derived), the [derived series](#def-b3-groups-derived) is such a chain. Conversely, given a chain, $D^i(G) \subseteq G_i$ by induction: $G_i/G_{i+1}$ abelian gives $D(G_i) \subseteq G_{i+1}$, so $D^{i+1}(G) = D(D^i G)
\subseteq D(G_i) \subseteq G_{i+1}$; hence $D^n(G) = \{e\}$.

Heredity: $D^i(H) \subseteq D^i(G)$ for $H \leq G$ (induction), and $D^i(G/N) = \pi(D^i(G))$ since $\pi$ maps [commutators](#def-b3-groups-derived) onto [commutators](#def-b3-groups-derived); this gives the statements for subgroups and quotients. Extension: if $D^m(G/N) = \{e\}$ then $D^m(G) \subseteq N$, and $D^n(N) = \{e\}$ gives $D^{m+n}(G) = D^n(D^m(G)) \subseteq D^n(N)
= \{e\}$. ∎

**Example 1.30.**

Abelian groups are [solvable](#def-b3-groups-derived). $p$-groups are [solvable](#def-b3-groups-derived), by induction on the order: $Z(G) \neq \{e\}$ and $G/Z(G)$ is a smaller $p$-group. $S_3$ and $S_4$ are [solvable](#def-b3-groups-derived): $S_4 \trianglerighteq A_4
\trianglerighteq V \trianglerighteq \{e\}$, where $V = \{e,
(1\,2)(3\,4), (1\,3)(2\,4), (1\,4)(2\,3)\}$ is the Klein group of double transpositions ([normal](#def-b3-groups-normal) in $S_4$: a union of [conjugacy classes](#ex-b3-groups-actions)), with abelian quotients $\Z/2\Z$, $\Z/3\Z$, $V$. In [Chapter 4](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#ch-b3-galois), “the general equation of degree $n$ is [solvable](#def-b3-groups-derived) by radicals” will literally *mean* “$S_n$ is a [solvable group](#def-b3-groups-derived)”. Whence the importance of the next definition.

**Definition 1.31.**

A group $G \neq \{e\}$ is *simple* if its only [normal subgroups](#def-b3-groups-normal) are $\{e\}$ and $G$. A nonabelian simple group is not [solvable](#def-b3-groups-derived): $D(G) \trianglelefteq G$ is not $\{e\}$ (else $G$ abelian), so $D(G) = G$ and the [derived series](#def-b3-groups-derived) is constant. The abelian simple groups are exactly the $\Z/p\Z$, $p$ prime (an abelian group is simple iff it has no proper nontrivial subgroup, iff it is cyclic of prime order by Lagrange).

**Lemma 1.32.**

For $n \geq 3$, $A_n$ is generated by $3$-cycles; for $n \geq 5$, all $3$-cycles are conjugate *in $A_n$*.

**Proof.** An element of $A_n$ is a product of an even number of transpositions; pair them up and use (composition right to left)

$$
(a\,b)(c\,d) = (a\,c\,b)(a\,c\,d), \qquad
(a\,b)(b\,c) = (a\,b\,c), \qquad
(a\,b)(a\,b) = e
$$

for disjoint, overlapping and equal pairs respectively: each pair of transpositions is a product of $3$-cycles.

Conjugacy: $\sigma(a\,b\,c)\sigma^{-1} = (\sigma a\; \sigma b\;
\sigma c)$, so any two $3$-cycles are conjugate by some $\sigma \in
S_n$. If $\sigma$ is odd, replace it by $\sigma' = \sigma (d\,e)$ where $d, e$ are two points outside $\{a, b, c\}$ — they exist since $n \geq 5$; then $\sigma'$ is even and $\sigma'(a\,b\,c)
\sigma'^{-1} = \sigma(a\,b\,c)\sigma^{-1}$, since $(d\,e)$ commutes with $(a\,b\,c)$. ∎

**Theorem 1.33 (Simplicity of the alternating group).**

$A_n$ is [simple](#def-b3-groups-simple) for $n \geq 5$.

**Proof.** Let $N \trianglelefteq A_n$, $N \neq \{e\}$. By [Lemma 1.32](#lem-b3-groups-threecycles) it suffices to show that $N$ contains *one* $3$-cycle: normality and conjugacy of $3$-cycles in $A_n$ then put all $3$-cycles in $N$, so $N = A_n$.

For $\rho \in S_n$ let $F(\rho) = \{x : \rho(x) \neq x\}$ be its *support* and $f(\rho) = \abs{F(\rho)}$. Choose $\sigma \in N
\setminus \{e\}$ with $f(\sigma)$ *minimal*. Note that a nontrivial even permutation has $f \geq 3$, and that $f(\sigma) =
4$ is impossible for $\sigma \in A_n$ unless $\sigma$ is a double transposition (a $4$-cycle is odd). We show $\sigma$ is a $3$-cycle.

*Case A: $\sigma$ is a product of disjoint transpositions*, say $\sigma = (a\,b)(c\,d)\cdots$ with $f(\sigma) \geq 4$. Pick $e'
\notin \{a, b, c, d\}$ (possible: $n \geq 5$), set $\tau =
(c\,d\,e')$ and

$$
\sigma' = \tau\sigma\tau^{-1}\,\sigma^{-1} \in N
\qquad (\tau\sigma\tau^{-1} \in N \text{ by normality}).
$$

Since $\sigma\tau^{-1}\sigma^{-1} = (\sigma c\;\sigma e'\;\sigma
d) = (d\;\sigma e'\;c)$ (using $\sigma c = d$, $\sigma d = c$), we get $\sigma' = (c\,d\,e')(d\;\sigma e'\;c)$.

If $\sigma e' = e'$ (which holds in particular when $f(\sigma) =
4$, i.e. $\sigma = (a\,b)(c\,d)$): then $(d\,e'\,c) =
(c\,d\,e')$ and $\sigma' = (c\,d\,e')^2 = (c\,e'\,d)$, a $3$-cycle lying in $N$, with $f(\sigma') = 3 < 4 \leq f(\sigma)$ — contradicting minimality.

If $\sigma e' \neq e'$: then $\sigma e' \notin \{a, b, c, d,
e'\}$ ($\sigma$ swaps $a,b$ and $c,d$, and $e' \notin \{a,b,c,d\}$ with $\sigma$ injective), so $\sigma$ moves the six points $a, b,
c, d, e', \sigma e'$: $f(\sigma) \geq 6$. On the other hand $\sigma'$, a product of two $3$-cycles with supports in $\{c, d,
e', \sigma e'\}$, satisfies $f(\sigma') \leq 4$; and $\sigma' \neq
e$, since $\sigma'(d) = \tau\sigma\tau^{-1}(c) = \tau\sigma(e') =
\sigma e' \neq d$ ($\tau$ fixes $\sigma e' \notin \{c,d,e'\}$). So $\sigma' \in N \setminus\{e\}$ with $f(\sigma') \leq 4 <
f(\sigma)$: minimality is contradicted.

*Case B: some cycle of $\sigma$ has length $\geq 3$*, say $\sigma(a) = b$, $\sigma(b) = c$ with $a, b, c$ distinct. If $\sigma$ is exactly this $3$-cycle, we are done. Otherwise $f(\sigma) \geq 5$ (the case $f(\sigma) = 4$ with a $\geq3$-cycle is the odd $4$-cycle, excluded), so we may pick $d, e' \in
F(\sigma) \setminus \{a, b, c\}$. Set $\tau = (c\,d\,e')$ and $\sigma' = \tau\sigma\tau^{-1}\sigma^{-1} \in N$. As before $\sigma' = (c\,d\,e')\,(\sigma c\;\sigma e'\;\sigma d)$ moves only points of

$$
M = \{c, d, e'\} \cup \{\sigma c, \sigma d, \sigma e'\}
\subseteq F(\sigma)
$$

(images of moved points are moved: $\sigma(x) \ne x$ implies $\sigma(\sigma x) \neq \sigma x$, $\sigma$ being injective). Also $b \notin M$: the five points $a, b, c, d, e'$ are distinct, so $b
\notin \{c, d, e'\}$; and $b \in \{\sigma c, \sigma d, \sigma
e'\}$ would force $a \in \{c, d, e'\}$ (apply $\sigma^{-1}$, using $\sigma a = b$), which is false. Hence $\sigma'$ fixes $b$, while $\sigma$ moves $b$; and $F(\sigma') \subseteq F(\sigma)$. Finally $\sigma' \neq e$: $\sigma^{-1}(c) = b$, $\tau^{-1}(b) = b$, $\sigma(b) = c$, $\tau(c) = d$, so $\sigma'(c) = d \neq c$. Thus $\sigma' \in N
\setminus \{e\}$ with $f(\sigma') \leq f(\sigma) - 1$, contradicting minimality.

Both cases being impossible, $\sigma$ is a $3$-cycle. ∎

**Corollary 1.34.**

For $n \geq 5$: $A_n$ and $S_n$ are not [solvable](#def-b3-groups-derived), and the only [normal subgroups](#def-b3-groups-normal) of $S_n$ are $\{e\}$, $A_n$ and $S_n$.

**Proof.** $A_n$ is nonabelian [simple](#def-b3-groups-simple), hence not [solvable](#def-b3-groups-derived) ([Definition 1.31](#def-b3-groups-simple)); a group containing a [non-solvable](#def-b3-groups-derived) subgroup is not [solvable](#def-b3-groups-derived) ([Proposition 1.29](#prop-b3-groups-derived)). Let $N
\trianglelefteq S_n$: then $N \cap A_n \trianglelefteq A_n$ equals $\{e\}$ or $A_n$. If $N \cap A_n = A_n$, then $A_n \subseteq N$ and $N \in \{A_n, S_n\}$ by the index. If $N \cap A_n = \{e\}$, the restriction to $N$ of the projection $S_n \to S_n/A_n \cong
\Z/2\Z$ is injective, so $\abs N \leq 2$; if $N = \{e, \sigma\}$, normality makes the [conjugacy class](#ex-b3-groups-actions) of $\sigma$ equal to $\{\sigma\}$, i.e. $\sigma \in Z(S_n)$. But $Z(S_n) = \{e\}$ for $n \geq 3$: if $\sigma \ne e$ moves $a$ to $b \neq a$, pick $c
\notin \{a, b\}$; then $(b\,c)\sigma(b\,c)^{-1}$ sends $a$ to $c
\ne b$, so it differs from $\sigma$. Hence $N = \{e\}$. ∎

**Theorem 1.35 (Jordan–Hölder).**

Every finite group $G \neq \{e\}$ admits a *composition series*

$$
\{e\} = G_0 \trianglelefteq G_1 \trianglelefteq \cdots
\trianglelefteq G_r = G,
\qquad G_{i}/G_{i-1} \text{ simple},
$$

and the multiset of *composition factors* $G_i/G_{i-1}$, up to isomorphism, does not depend on the chosen series. A finite group is [solvable](#def-b3-groups-derived) iff all its composition factors are cyclic of prime order.

**Proof.** *Existence*: induction on $\abs G$. If $G$ is [simple](#def-b3-groups-simple), take $\{e\} \trianglelefteq G$. Otherwise pick a maximal proper [normal subgroup](#def-b3-groups-normal) $N$ (there are finitely many subgroups); $G/N$ is [simple](#def-b3-groups-simple) by the correspondence theorem (a proper nontrivial [normal subgroup](#def-b3-groups-normal) of $G/N$ would lift to a [normal subgroup](#def-b3-groups-normal) of $G$ strictly between $N$ and $G$). Append $N \trianglelefteq G$ to a [composition series](#thm-b3-groups-jordanholder) of $N$.

*Uniqueness*: induction on $\abs G$, the case $G$ [simple](#def-b3-groups-simple) being clear. Take two [composition series](#thm-b3-groups-jordanholder), with penultimate terms $M \trianglelefteq G$ and $N \trianglelefteq G$ (so $G/M$, $G/N$ are [simple](#def-b3-groups-simple)). If $M = N$, conclude by induction applied to $M$. Otherwise $MN$, [normal](#def-b3-groups-normal) in $G$ and strictly containing $M$, equals $G$ ($M$ is maximal [normal](#def-b3-groups-normal): any [normal](#def-b3-groups-normal) $M \subsetneq L
\subsetneq G$ would map to a proper nontrivial [normal subgroup](#def-b3-groups-normal) of the [simple](#def-b3-groups-simple) $G/M$). The second isomorphism theorem gives

$$
G/M = MN/M \cong N/(M \cap N),
\qquad
G/N = MN/N \cong M/(M \cap N).
$$

Set $K = M \cap N$ ($\trianglelefteq G$) and fix a [composition series](#thm-b3-groups-jordanholder) of $K$. Then $M$ carries two [composition series](#thm-b3-groups-jordanholder): its original one, and the series of $K$ followed by $K \trianglelefteq
M$ (the quotient $M/K \cong G/N$ is [simple](#def-b3-groups-simple)). By induction (applied to $M$), the factors of the original series of $M$ are $\{\text{factors of } K\} \cup \{G/N\}$; likewise for $N$. Hence both series of $G$ have factors

$$
\{\text{factors of } K\} \;\cup\; \{\,G/N,\; G/M\,\},
$$

the same multiset.

[Solvability](#def-b3-groups-derived): if all factors are $\Z/p_i\Z$, the series is a chain with abelian quotients, so $G$ is [solvable](#def-b3-groups-derived) ([Proposition 1.29](#prop-b3-groups-derived)). Conversely, a composition factor of a [solvable group](#def-b3-groups-derived) is [solvable](#def-b3-groups-derived) (a quotient of a subgroup) and [simple](#def-b3-groups-simple); a [solvable](#def-b3-groups-derived) [simple group](#def-b3-groups-simple) is abelian ($D(G) \ne G$ forces $D(G) = \{e\}$), hence some $\Z/p\Z$. ∎

**Remark 1.36.**

Jordan–Hölder says every finite group is built from [simple groups](#def-b3-groups-simple), with a well-defined parts list — an arithmetic of groups in which the [simple groups](#def-b3-groups-simple) are the primes, and where *how the parts are glued* (extension data, as in the [semidirect product](#def-b3-groups-semidirect)) replaces mere multiplication. The classification of the finite [simple groups](#def-b3-groups-simple) — the cyclic $\Z/p\Z$, the alternating $A_{n \geq
5}$, sixteen families of Lie type, and $26$ sporadic groups — is one of the monuments of twentieth-century mathematics; its proof, spread over some ten thousand journal pages, is very far beyond this course.

![The ten subgroups of the dihedral group D_4 = r, s r4 = s2 = e,\ srs-1 = r-1. The three subgroups of index 2 (middle row) are normal, as is the center r2 (highlighted); the four reflection subgroups fall into two conjugacy classes of two. Chains from bottom to top give composition series, e.g. \e\ r2 r D_4: factors ℤ/2ℤ, ℤ/2ℤ, ℤ/2ℤ — always the same multiset, as Jordan–Hölder demands.](https://one-course.com/images/onecourse/chapters/math-5/b3-groups/fig-3655fe380ff7.svg)

*The ten subgroups of the [dihedral group](#ex-b3-groups-semidirectexamples) $D_4 = \langle r,
s \mid r^4 = s^2 = e,\ srs^{-1} = r^{-1}\rangle$. The three subgroups of index $2$ (middle row) are [normal](#def-b3-groups-normal), as is the [center](#ex-b3-groups-actions) $\langle r^2\rangle$ (highlighted); the four reflection subgroups fall into two [conjugacy classes](#ex-b3-groups-actions) of two. Chains from bottom to top give [composition series](#thm-b3-groups-jordanholder), e.g. $\{e\} \trianglelefteq \langle
r^2\rangle \trianglelefteq \langle r\rangle \trianglelefteq D_4$: factors $\Z/2\Z, \Z/2\Z, \Z/2\Z$ — always the same multiset, as Jordan–Hölder demands.*

## 1.6 Exercises

**Exercise 1.1 ★.**

(a) Show that every subgroup of index $2$ is [normal](#def-b3-groups-normal). (b) Show that if $[G : H] = 2$, then $x^2 \in H$ for every $x \in
G$. (c) Deduce that $A_4$ has no subgroup of order $6$: Lagrange’s converse fails. *(Count the squares of $3$-cycles.)*

**Solution of Exercise 1.1.**

(a) Let $[G:H] = 2$. For $g \in H$, $gH = H = Hg$. For $g \notin
H$: the two left cosets are $H$ and $gH$, so $gH = G \setminus H$; likewise $Hg = G \setminus H$. Hence $gH = Hg$ for all $g$: $H
\trianglelefteq G$.

(b) By (a), $G/H$ is a group of order $2$; the class $\bar x$ satisfies $\bar x^2 = \bar e$, i.e. $x^2 \in H$.

(c) Suppose $H \leq A_4$ with $\abs H = 6$, hence of index $2$. By (b), $\sigma^2 \in H$ for every $\sigma \in A_4$. Every $3$-cycle is such a square: if $\sigma^3 = e$ then $\sigma =
\sigma^4 = (\sigma^2)^2$. So $H$ contains all eight $3$-cycles of $A_4$: $\abs H \geq 8 > 6$, a contradiction. (Lagrange’s converse fails at the very first opportunity: $6 \mid 12$.)

**Exercise 1.2 ★.**

Show that if $G/Z(G)$ is cyclic then $G$ is abelian. Deduce again that every group of order $p^2$ is abelian, and exhibit, for each prime $p$, a nonabelian group of order $p^3$. *(Think of upper triangular matrices with unit diagonal over $\mathbb F_p$.)*

**Solution of Exercise 1.2.**

Say $G/Z(G) = \langle gZ(G) \rangle$. Every $x \in G$ then writes $x = g^k z$ with $k \in \Z$, $z \in Z(G)$. For $x = g^kz$, $y =
g^l z'$:

$$
xy = g^k z\, g^l z' = g^{k+l} z z' = g^l z'\, g^k z = yx,
$$

central elements commuting with everything: $G$ is abelian.

Order $p^2$: $Z(G) \neq \{e\}$ ([Theorem 1.12](#thm-b3-groups-pfixed)), so $\abs{Z(G)} \in \{p, p^2\}$. If it were $p$, then $G/Z(G)$ would have order $p$, hence be cyclic, forcing $G$ abelian and $Z(G) =
G$ of order $p^2$ — contradiction. So $Z(G) = G$.

Nonabelian of order $p^3$: the *Heisenberg group*

$$
H_p = \left\{ \begin{pmatrix} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1
\end{pmatrix} : a, b, c \in \mathbb F_p \right\}
\leq GL_3(\mathbb F_p),
$$

of order $p^3$ (free choice of $a, b, c$; closure and inverses by direct computation). It is nonabelian: the two elementary matrices $I + E_{12}$ and $I + E_{23}$ have [commutator](#def-b3-groups-derived) $I + E_{13} \neq I$.

**Exercise 1.3 ★.**

(a) Show that $\operatorname{Aut}(\Z/n\Z) \cong (\Z/n\Z)^\times$. (b) Show that the inner automorphisms $\iota_g \colon x \mapsto
gxg^{-1}$ form a [normal subgroup](#def-b3-groups-normal) $\operatorname{Inn}(G)
\trianglelefteq \operatorname{Aut}(G)$, with $\operatorname{Inn}(G) \cong G/Z(G)$.

**Solution of Exercise 1.3.**

(a) A morphism $f \colon \Z/n\Z \to \Z/n\Z$ is determined by $k =
f(\bar 1)$ (then $f(\bar m) = m\bar k$), and every $\bar k$ defines one. It is bijective iff $\bar k$ generates $\Z/n\Z$, iff $\gcd(k,n) = 1$, iff $\bar k \in (\Z/n\Z)^\times$. Composition corresponds to multiplication: $f_k \circ f_l = f_{kl}$. Hence $\operatorname{Aut}(\Z/n\Z) \cong (\Z/n\Z)^\times$.

(b) The map $\iota\colon G \to \operatorname{Aut}(G)$, $g \mapsto
\iota_g$, is a morphism: $\iota_g \circ \iota_h = \iota_{gh}$. Its image is $\operatorname{Inn}(G)$; its kernel is $\{g : gxg^{-1} =
x\ \forall x\} = Z(G)$. The first isomorphism theorem gives $\operatorname{Inn}(G) \cong G/Z(G)$. Normality in $\operatorname{Aut}(G)$: for $\alpha \in \operatorname{Aut}(G)$,

$$
(\alpha \circ \iota_g \circ \alpha^{-1})(x)
= \alpha\bigl(g\,\alpha^{-1}(x)\,g^{-1}\bigr)
= \alpha(g)\, x\, \alpha(g)^{-1} = \iota_{\alpha(g)}(x).
$$

**Exercise 1.4 ★★.**

Let $H, K$ be subgroups of a finite group $G$. (a) Prove the *product formula* $\abs{HK}\,\abs{H\cap K} =
\abs H\, \abs K$, by counting the fibers of the map $H \times K
\to HK$, $(h,k) \mapsto hk$. (b) Show that $HK$ is a subgroup iff $HK = KH$ (automatic when one of the two is [normal](#def-b3-groups-normal)). (c) If $H, K \trianglelefteq G$ and $H \cap K = \{e\}$, show that $hk = kh$ for all $h \in H$, $k \in K$.

**Solution of Exercise 1.4.**

(a) Consider $\mu \colon H \times K \to HK$, $(h,k) \mapsto hk$, surjective by definition. Fix $h_0k_0 \in HK$: then $hk = h_0k_0
\iff h_0^{-1}h = k_0 k^{-1} \in H \cap K$. Writing $u =
h_0^{-1}h$, the fiber of $h_0k_0$ is $\{(h_0u,\, u^{-1}k_0) : u
\in H \cap K\}$, of cardinality $\abs{H \cap K}$. Hence $\abs
H\,\abs K = \abs{H\times K} = \abs{HK}\,\abs{H \cap K}$.

(b) If $HK$ is a subgroup: $KH \subseteq HK$ because $kh =
\bigl(h^{-1}k^{-1}\bigr)^{-1} \in (HK)^{-1} = HK$; and $HK
\subseteq KH$ by taking inverses in $HK = (HK)^{-1} \subseteq
(KH)^{-1}\dots$ more directly, for $hk \in HK$, $(hk)^{-1} =
k^{-1}h^{-1} \in KH$, so $HK = (HK)^{-1} \subseteq KH$; both inclusions give $HK = KH$. Conversely if $HK = KH$: closure, $(hk)(h'k') = h(kh')k' \in h(HK)k' = (hH)(Kk') \subseteq HK$; inverses, $(hk)^{-1} = k^{-1}h^{-1} \in KH = HK$; and $e \in HK$: subgroup. If, say, $K \trianglelefteq G$, then $hK = Kh$ for all $h$, so $HK = KH$ automatically.

(c) For $h \in H$, $k \in K$, the [commutator](#def-b3-groups-derived) $[h,k] = hkh^{-1}
k^{-1}$ equals $(hkh^{-1})k^{-1} \in K$ ($K$ [normal](#def-b3-groups-normal)) and $h(kh^{-1}k^{-1}) \in H$ ($H$ [normal](#def-b3-groups-normal)), hence lies in $H \cap K =
\{e\}$: $hk = kh$.

**Exercise 1.5 ★★.**

(Burnside’s counting lemma) A finite group $G$ acts on a finite set $X$. Show that the number of [orbits](#def-b3-groups-action) is the *average number of fixed points*:

$$
\#\{\text{orbits}\} = \frac{1}{\abs G}\sum_{g \in G}
\abs{\operatorname{Fix}(g)},
\qquad \operatorname{Fix}(g) = \{x \in X : g \cdot x = x\},
$$

by counting the set $\{(g,x) : g\cdot x = x\}$ in two ways. Application: $\Z/p\Z$ ($p$ prime) acts by rotation on necklaces of $p$ beads with $a$ available colors; deduce Fermat’s little theorem $a^p \equiv a \pmod p$.

**Solution of Exercise 1.5.**

Count $E = \{(g,x) \in G \times X : g \cdot x = x\}$ two ways:

$$
\abs E = \sum_{g \in G} \abs{\operatorname{Fix}(g)}
= \sum_{x \in X} \abs{G_x}
= \sum_{x \in X} \frac{\abs G}{\abs{\mathcal O_x}}
= \abs G \sum_{\mathcal O \text{ orbit}} \sum_{x \in \mathcal O}
\frac{1}{\abs{\mathcal O}}
= \abs G \cdot \#\{\text{orbits}\},
$$

using orbit–stabilizer ($\abs{G_x} = \abs G/\abs{\mathcal O_x}$) and the partition into [orbits](#def-b3-groups-action).

Necklaces: let $X$ be the set of maps $\Z/p\Z \to
\{1, \dots, a\}$ (colorings of $p$ positions), $\abs X = a^p$, with $\Z/p\Z$ acting by rotation. The identity fixes all $a^p$ colorings. A rotation $\bar k \neq \bar 0$ generates $\Z/p\Z$ ($p$ prime), so a coloring it fixes is invariant under *all* rotations, hence constant: $a$ fixed colorings. Burnside:

$$
\#\{\text{orbits}\} = \frac{a^p + (p-1)a}{p} \in \N ,
$$

so $p \mid a^p + (p-1)a$, i.e. $p \mid a^p - a$: Fermat’s little theorem, by pure counting.

**Exercise 1.6 ★.**

Using the Sylow theorems, show that every group of order $15$ is cyclic and that every group of order $45$ is abelian.

**Solution of Exercise 1.6.**

Order $15 = 3 \cdot 5$: $n_3 \mid 5$ and $n_3 \equiv 1 \pmod 3$ force $n_3 = 1$; $n_5 \mid 3$ and $n_5 \equiv 1 \pmod 5$ force $n_5 = 1$. The Sylows $P_3, P_5$ are [normal](#def-b3-groups-normal), intersect trivially (coprime orders), and $\abs{P_3P_5} = 15$ ([Exercise 1.4](#exo-b3-groups-4)(a)): by [Proposition 1.24](#prop-b3-groups-direct), $G
\cong \Z/3\Z \times \Z/5\Z \cong \Z/15\Z$ (Chinese remainder).

Order $45 = 3^2 \cdot 5$: $n_3 \mid 5$, $n_3 \equiv 1 \pmod 3$ give $n_3 = 1$; $n_5 \mid 9$, $n_5 \equiv 1 \pmod 5$ give $n_5 =
1$. So $G \cong P_3 \times P_5$ with $\abs{P_3} = 9 = 3^2$ and $\abs{P_5} = 5$: both abelian ([Theorem 1.12](#thm-b3-groups-pfixed) for $p^2$; prime order is cyclic), hence so is $G$.

**Exercise 1.7 ★★.**

Show that no group of order $30$, and none of order $56$, is [simple](#def-b3-groups-simple). *(For $30$: if $n_3 \neq 1$ and $n_5 \neq 1$, count the elements of orders $3$ and $5$. For $56$: count the elements of order $7$.)*

**Solution of Exercise 1.7.**

Order $30 = 2 \cdot 3 \cdot 5$. $n_5 \mid 6$, $n_5 \equiv 1 \pmod
5$: $n_5 \in \{1, 6\}$; $n_3 \mid 10$, $n_3 \equiv 1 \pmod 3$: $n_3 \in \{1, 10\}$. Suppose $G$ [simple](#def-b3-groups-simple), so $n_5 = 6$ and $n_3 =
10$. Two distinct subgroups of prime order $p$ intersect trivially (the intersection is a proper subgroup of $\Z/p\Z$), so the six Sylow $5$-subgroups carry $6 \times 4 = 24$ elements of order $5$, and the ten Sylow $3$-subgroups carry $10 \times 2 = 20$ elements of order $3$: $24 + 20 = 44 > 30 - 1$ non-identity elements — absurd. So $n_5 = 1$ or $n_3 = 1$: a [normal](#def-b3-groups-normal) Sylow exists.

Order $56 = 2^3 \cdot 7$. $n_7 \mid 8$, $n_7 \equiv 1 \pmod 7$: $n_7 \in \{1, 8\}$. If $n_7 = 8$, the Sylow $7$-subgroups carry $8
\times 6 = 48$ elements of order $7$, leaving exactly $56 - 48 =
8$ other elements. A Sylow $2$-subgroup has order $8$ and consists of such elements, so it is *the* set of them: $n_2 = 1$. Either $n_7 = 1$ or $n_2 = 1$: never [simple](#def-b3-groups-simple).

**Exercise 1.8 ★★.**

(a) Let $H \leq G$ of index $n$. Show that the [action](#def-b3-groups-action) of $G$ on $G/H$ yields a morphism $G \to S_n$ whose kernel $\bigcap_{g \in
G} gHg^{-1}$ is the largest [normal subgroup](#def-b3-groups-normal) of $G$ contained in $H$. (b) Deduce: if $G$ is finite and $p$ is the *smallest* prime divisor of $\abs G$, every subgroup of index $p$ is [normal](#def-b3-groups-normal). (c) Show that every transitive [action](#def-b3-groups-action) of $G$ on a set $X$ is isomorphic to the [action](#def-b3-groups-action) on a coset space: there is a bijection $X
\to G/G_x$ commuting with the [actions](#def-b3-groups-action).

**Solution of Exercise 1.8.**

(a) The [action](#def-b3-groups-action) $g \cdot xH = gxH$ gives a morphism $\rho \colon G
\to \mathfrak S(G/H) \cong S_n$. Its kernel is

$$
\ker\rho = \{g : \forall x \in G,\ gxH = xH\}
= \{g : \forall x,\ x^{-1}gx \in H\}
= \bigcap_{x \in G} xHx^{-1},
$$

a [normal subgroup](#def-b3-groups-normal) (a kernel) contained in $H$ (take $x = e$). If $N \trianglelefteq G$ and $N \subseteq H$, then for every $x$: $N
= xNx^{-1} \subseteq xHx^{-1}$, so $N \subseteq \ker\rho$: the kernel is the largest such.

(b) Let $[G:H] = p$, smallest prime dividing $\abs G$, and $K =
\ker\rho \subseteq H$. Then $G/K$ embeds in $S_p$, so $[G:K]$ divides $p!$. Also $[G:K] = [G:H]\,[H:K] = p\,[H:K]$, so $[H:K]$ divides $(p-1)!$. But $[H:K]$ divides $\abs G$, whose prime divisors are all $\geq p$, while the prime divisors of $(p-1)!$ are all $< p$: hence $[H:K] = 1$, i.e. $H = K = \ker \rho$ is [normal](#def-b3-groups-normal).

(c) Let the [action](#def-b3-groups-action) be transitive and $x \in X$. The map $\Phi
\colon G/G_x \to X$, $gG_x \mapsto g \cdot x$, is well defined and bijective (orbit–stabilizer; the [orbit](#def-b3-groups-action) is all of $X$), and it intertwines the [actions](#def-b3-groups-action): $\Phi(h \cdot gG_x) = \Phi(hgG_x) = (hg)
\cdot x = h \cdot \Phi(gG_x)$.

**Exercise 1.9 ★★.**

(a) Show that $D(G)$ is the smallest [normal subgroup](#def-b3-groups-normal) of $G$ with abelian quotient, and that every morphism from $G$ to an abelian group factors uniquely through the *abelianization* $G^{\mathrm{ab}} = G/D(G)$. (b) Compute $D(S_n)$ and $S_n^{\mathrm{ab}}$ for $n \geq 2$, and $D(Q_8)$ and $Q_8^{\mathrm{ab}}$.

**Solution of Exercise 1.9.**

(a) $D(G)$ is [normal](#def-b3-groups-normal) with abelian quotient ([Proposition 1.29](#prop-b3-groups-derived)); and if $N \trianglelefteq G$ has $G/N$ abelian, the same proposition gives $D(G) \subseteq N$: $D(G)$ is the smallest. Universal property: let $f \colon G \to A$ with $A$ abelian. Then $f([x,y]) = [f(x), f(y)] = e$, so $D(G)
\subseteq \ker f$, and [Theorem 1.3](#thm-b3-groups-firstiso) factors $f =
\bar f \circ \pi$ through $G^{\mathrm{ab}}$, uniquely since $\pi$ is surjective.

(b) [Commutators](#def-b3-groups-derived) are even permutations, so $D(S_n) \subseteq A_n$. Conversely every $3$-cycle is a [commutator](#def-b3-groups-derived):

$$
\bigl[(a\,b),\,(a\,c)\bigr] = (a\,b)(a\,c)(a\,b)(a\,c) =
(a\,b\,c),
$$

(direct check on $a, b, c$), and $3$-cycles generate $A_n$ ([Lemma 1.32](#lem-b3-groups-threecycles)): $D(S_n) = A_n$ for $n \geq 3$, and $S_n^{\mathrm{ab}} \cong S_n/A_n \cong \Z/2\Z$. (For $n = 2$: $S_2$ is abelian, $D(S_2) = \{e\}$, $S_2^{\mathrm{ab}} = S_2
\cong \Z/2\Z$ — the formula $S_n^{\mathrm{ab}} \cong \Z/2\Z$ holds for all $n \geq 2$.)

$Q_8$: the quotient $Q_8/\{\pm 1\}$ has order $4$, hence is abelian, so $D(Q_8) \subseteq \{\pm 1\}$; and $[\mathrm i, \mathrm
j] = \mathrm i \mathrm j \mathrm i^{-1}\mathrm j^{-1} = \mathrm
i\mathrm j(-\mathrm i)(-\mathrm j) = (\mathrm i \mathrm j)^2 =
\mathrm k^2 = -1$, so $D(Q_8) = \{\pm 1\}$ and $Q_8^{\mathrm{ab}} \cong (\Z/2\Z)^2$ (order $4$, exponent $2$: the classes of $\mathrm i, \mathrm j$ square to $\bar 1$).

**Exercise 1.10 ★★.**

Let $G$ be a $p$-group and $H \subsetneq G$ a proper subgroup. Show that $H \subsetneq N_G(H)$ (“[normalizers](#ex-b3-groups-actions) grow”), and deduce that every maximal subgroup of a $p$-group is [normal](#def-b3-groups-normal) of index $p$. *(Induction on $\abs G$, using $Z(G) \neq \{e\}$: treat separately $Z(G) \subseteq H$ and $Z(G) \not\subseteq H$.)*

**Solution of Exercise 1.10.**

Induction on $\abs G$; for $\abs G = p$ the only proper subgroup is $H = \{e\}$, and $N_G(\{e\}) = G \supsetneq
\{e\}$. Let $Z = Z(G) \neq \{e\}$ ([Theorem 1.12](#thm-b3-groups-pfixed)).

*If $Z \not\subseteq H$*: pick $z \in Z \setminus H$; $z$ commutes with $H$, so $zHz^{-1} = H$ and $z \in N_G(H) \setminus
H$.

*If $Z \subseteq H$*: pass to $\bar G = G/Z$, a $p$-group of smaller order, and $\bar H = H/Z \subsetneq \bar G$ (correspondence theorem). By induction, $N_{\bar G}(\bar H) \supsetneq \bar H$; pick $\bar g \in N_{\bar G}(\bar H) \setminus \bar H$ and a lift $g$. Then $g \notin H$, and $gHg^{-1} \subseteq HZ = H$: indeed $\overline{ghg^{-1}} = \bar g \bar h \bar g^{-1} \in \bar H$ means $ghg^{-1} \in HZ = H$ (as $Z \subseteq H$). So $g \in
N_G(H)\setminus H$.

Maximal subgroups: if $M$ is maximal, $N_G(M) \supsetneq M$ forces $N_G(M) = G$: $M \trianglelefteq G$. Then $G/M$ is a $p$-group with no proper nontrivial subgroup (correspondence + maximality). Take $\bar x \neq \bar e$ in $G/M$, of order $p^k$; then $\bar x^{p^{k-1}}$ generates a subgroup of order $p$, which must be everything: $\abs{G/M} = p$.

**Exercise 1.11 ★★★.**

(Simplicity of $A_5$, hands on) (a) Show that the [conjugacy classes](#ex-b3-groups-actions) of $A_5$ have cardinalities $1$, $15$, $20$, $12$, $12$. Pay attention to the splitting of the $S_5$-class of $5$-cycles: for a $5$-cycle $\sigma$, compare the [centralizers](#ex-b3-groups-actions) of $\sigma$ in $S_5$ and in $A_5$. (b) Deduce that $A_5$ is [simple](#def-b3-groups-simple): a [normal subgroup](#def-b3-groups-normal) is a union of [conjugacy classes](#ex-b3-groups-actions), contains $e$, and has cardinality dividing $60$. (c) Show that a [simple group](#def-b3-groups-simple) of order $60$ necessarily has $n_5 =
6$.

**Solution of Exercise 1.11.**

(a) $\abs{A_5} = 60$. Cycle types in $A_5$: $e$; double transpositions, $\frac{1}{2}\binom{5}{1}\binom{4}{2}\cdot 1 = 15$ of them ($5 \cdot 3$ ways: choose the fixed point, then pair up); $3$-cycles, $\frac{5 \cdot 4 \cdot 3}{3} = 20$; $5$-cycles, $4! =
24$.

A class of $S_5$ contained in $A_5$ either stays one $A_5$-class or splits in two, according to whether the $S_5$-centralizer of an element contains an odd permutation ($\abs{\text{class in }A_5} =
60/\abs{Z_{A_5}(\sigma)}$ and $Z_{A_5} = Z_{S_5} \cap A_5$). For $\sigma = (1\,2)(3\,4)$: $\abs{Z_{S_5}(\sigma)} = 120/15 = 8$, and $(1\,2) \in Z_{S_5}(\sigma)$ is odd, so $\abs{Z_{A_5}} = 4$ and the class has $60/4 = 15$ elements: no split. For $\sigma =
(1\,2\,3)$: $Z_{S_5}(\sigma) \supseteq \langle \sigma \rangle
\times \langle (4\,5)\rangle$, of order $6 = 120/20$, hence equal; it contains the odd $(4\,5)$: class of $60/3 = 20$: no split. For $\sigma$ a $5$-cycle: $Z_{S_5}(\sigma) = \langle \sigma\rangle$ (order $120/24 = 5$), all even: $Z_{A_5}(\sigma) = \langle
\sigma\rangle$ and the $A_5$-class has $60/5 = 12$ elements — the $24$ five-cycles split into *two* classes of $12$. Class sizes: $1, 15, 20, 12, 12$.

(b) A [normal subgroup](#def-b3-groups-normal) $N$ is a union of [conjugacy classes](#ex-b3-groups-actions) including $\{e\}$, with $\abs N \mid 60$. The possible sums $1 +
(\text{subset of } \{15, 20, 12, 12\})$ are

$$
1,\ 13,\ 13,\ 16,\ 21,\ 25,\ 28,\ 28,\ 33,\ 36,\ 40,\ 40,\ 45,\
48,\ 48,\ 60 ;
$$

the only divisors of $60$ in the list are $1$ and $60$: $N =
\{e\}$ or $A_5$.

(c) Let $G$ be [simple](#def-b3-groups-simple) with $\abs G = 60$. $n_5 \mid 12$, $n_5
\equiv 1 \pmod 5$: $n_5 \in \{1, 6\}$. $n_5 = 1$ would make the Sylow $5$-subgroup [normal](#def-b3-groups-normal), contradicting simplicity ($1 < 5 <
60$). Hence $n_5 = 6$.

**Exercise 1.12 ★★.**

([Normalizers](#ex-b3-groups-actions) of [Sylow subgroups](#def-b3-groups-sylow) are self-normalizing) Let $P$ be a Sylow $p$-subgroup of a finite group $G$ and $H =
N_G(P)$. (a) Show that $P$ is the *unique* Sylow $p$-subgroup of $H$. (b) Deduce $N_G(H) = H$. *(For $g \in N_G(H)$: $gPg^{-1}$ is a Sylow $p$-subgroup of $H$, so $gPg^{-1} = P$.)* (c) Conclude that no Sylow [normalizer](#ex-b3-groups-actions) is contained in a proper [normal subgroup](#def-b3-groups-normal) of $G$, and that a maximal subgroup containing $N_G(P)$ is self-normalizing.

**Solution of Exercise 1.12.**

(a) $P$ is [normal](#def-b3-groups-normal) in $H = N_G(P)$ by definition of the [normalizer](#ex-b3-groups-actions), and it is a Sylow $p$-subgroup of $H$ (its order is already the full $p$-part of $\abs G$, a fortiori of $\abs H$). A [normal](#def-b3-groups-normal) [Sylow subgroup](#def-b3-groups-sylow) is unique: any other would be conjugate to it (Sylow II in $H$), hence equal to it.

(b) Let $g \in N_G(H)$. Then $gPg^{-1} \subseteq gHg^{-1} = H$ is a subgroup of $H$ of the same order as $P$: a Sylow $p$-subgroup of $H$, so $gPg^{-1} = P$ by (a). Thus $g \in
N_G(P) = H$: $N_G(H) \subseteq H$, and the reverse inclusion is trivial.

(c) Suppose $H \subseteq N \trianglelefteq G$ with $N$ proper. $P$ is a Sylow $p$-subgroup of $N$; for any $g \in G$, $gPg^{-1} \subseteq N$ is another, so $gPg^{-1} = nPn^{-1}$ for some $n \in N$ (Sylow II in $N$), giving $n^{-1}g \in N_G(P)
\subseteq N$ and $g \in N$: $N = G$, contradiction (this is the *Frattini argument*). For a maximal subgroup $M \supseteq
N_G(P)$: $N_G(M) \supseteq M$ is either $M$ or $G$; if $G$, then $M \trianglelefteq G$ is a proper [normal subgroup](#def-b3-groups-normal) containing $N_G(P)$ — excluded by the previous point. So $N_G(M) = M$.

## 1.7 Problem: the groups of order at most 15

**Problem 1.1.**

Weekend problem — classification of small groups

The aim is a complete classification, with full proofs, of the groups of order $\leq 15$ up to isomorphism. Orders $1, 2, 3, 5,
7, 11, 13$ are settled by Lagrange (cyclic), and orders $4$ and $9$ by [Theorem 1.12](#thm-b3-groups-pfixed) plus the analysis below of $p^2$: there remain $6, 8, 10, 12, 14, 15$.

**Part I — Tools.**

1. Show that a group in which every element satisfies $x^2 =  e$ is abelian; deduce that such a finite group has order $2^k$ and is isomorphic to $(\Z/2\Z)^k$ . *(View it as a vector space over $\mathbb F_2$.)*
2. Show that a group of order $p^2$ is isomorphic to $\Z/p^2\Z$ or $(\Z/p\Z)^2$ . List the abelian groups of order $8$ up to isomorphism: $\Z/8\Z$ , $\Z/4\Z \times  \Z/2\Z$ , $(\Z/2\Z)^3$ — prove the list is complete and irredundant *without* the structure theorem of [Chapter 3](https://one-course.com/books/math/5/en/chapter/3-modules-over-a-principal-ideal-domain#ch-b3-modules) (discuss by the maximal order of an element).
3. Let $\varphi, \varphi' \colon K \to \operatorname{Aut}(H)$ be two [actions](#def-b3-groups-action) . Show that if $\varphi' = \varphi \circ  \alpha$ with $\alpha \in \operatorname{Aut}(K)$ , then $H  \rtimes_{\varphi} K \cong H \rtimes_{\varphi'} K$ .
4. Determine $\operatorname{Aut}(\Z/n\Z)$ for $n = 3, 4, 5,  7$ explicitly, and show $\operatorname{Aut}\bigl((\Z/2\Z)^2  \bigr) \cong S_3$ .

**Part II — Orders $2p$ ($6$, $10$, $14$) and $pq$.**

5. Let $\abs G = 2p$ with $p$ an odd prime. Show that $G$ has a [normal subgroup](#def-b3-groups-normal) $N = \langle r \rangle$ of order $p$ and an element $s$ of order $2$ outside $N$ .
6. Deduce $G \cong \Z/p\Z \rtimes_\varphi \Z/2\Z$ , where $\varphi(1) \in \operatorname{Aut}(\Z/p\Z)$ is an involution, and conclude: $G \cong \Z/2p\Z$ or $G \cong  D_p$ ; check these two are not isomorphic. This settles orders $6$ , $10$ , $14$ .
7. More generally, let $\abs G = pq$ with $p < q$ primes. Show that if $p \nmid q-1$ then $G$ is cyclic ( [Example 1.22](#ex-b3-groups-pq) ), and that if $p \mid q - 1$ there is, besides $\Z/pq\Z$ , *exactly one* nonabelian group $\Z/q\Z \rtimes \Z/p\Z$ up to isomorphism — use question 3 and the fact that $\operatorname{Aut}(\Z/q\Z) \cong  (\Z/q\Z)^\times$ is cyclic of order $q - 1$ , admitted here and proved in [Chapter 4](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#ch-b3-galois) (cyclicity of $\mathbb F_q^\times$ ). Conclude for order $15$ .

**Part III — Order $8$.** Let $G$ be nonabelian of order $8$.

8. Show that $G$ has an element $r$ of order $4$ (use question 1) and that $N = \langle r\rangle$ is [normal](#def-b3-groups-normal) .
9. Let $s \notin N$ . Show that $s^2 \in N$ ( [Exercise 1.1](#exo-b3-groups-1) (b)), that $srs^{-1} = r^{-1}$ (examine the possible images of $r$ under conjugation, which must have order 4, and exclude $srs^{-1} = r$ ), and that $s^2 \in \{e, r^2\}$ *(what happens if $s^2 = r$ or $r^3$? and why must $s^2$ commute with $s$?)* .
10. In the case $s^2 = e$ , show $G \cong D_4$ .
11. In the case $s^2 = r^2$, show that the multiplication table is entirely determined; the resulting group is the *quaternion group* $Q_8 =  \{\pm 1, \pm \mathrm i, \pm \mathrm j, \pm \mathrm k\}$, $\mathrm i^2 = \mathrm j^2 = \mathrm k^2 = \mathrm i\,  \mathrm j\,\mathrm k = -1$ (set $r = \mathrm i$, $s =  \mathrm j$). Verify that $Q_8$ exists, e.g. inside $GL_2(\C)$ via $$\mathrm i \mapsto \begin{pmatrix} \iu & 0\\ 0 & -\iu  \end{pmatrix},  \qquad  \mathrm j \mapsto \begin{pmatrix} 0 & 1\\ -1 & 0  \end{pmatrix}.$$
12. Show that every nontrivial subgroup of $Q_8$ contains $-1$ ; deduce that every subgroup of $Q_8$ is [normal](#def-b3-groups-normal) , that $D_4 \not\cong Q_8$ (count elements of order $2$ ), and that $Q_8$ is not a [semidirect product](#def-b3-groups-semidirect) of two proper subgroups.

**Part IV — Order $12$.** Let $\abs G = 12$, $P_3
\in \mathrm{Syl}_3(G)$, $P_2 \in \mathrm{Syl}_2(G)$.

13. Show $n_3 \in \{1, 4\}$ , $n_2 \in \{1, 3\}$ , and that $n_3  = 4$ forces $n_2 = 1$ *(count elements of order $3$)* .
14. Suppose $n_3 = 4$ . The conjugation [action](#def-b3-groups-action) on $\mathrm{Syl}_3$ gives $\rho \colon G \to S_4$ . Show that $\ker \rho$ , contained in every $N_G(P_3)$ and hence of order dividing $3$ , is trivial *(why can it not have order $3$?)* ; that the image, a subgroup of order $12$ of $S_4$ , is necessarily $A_4$ *(index 2 subgroups are [normal](#def-b3-groups-normal) and contain all squares — [Exercise 1.1](#exo-b3-groups-1); count the squares in $S_4$)* ; and conclude $G \cong A_4$ .
15. Suppose $n_3 = 1$ , so $G \cong \Z/3\Z \rtimes_\varphi P_2$ with $\varphi \colon P_2 \to \operatorname{Aut}(\Z/3\Z)  \cong \Z/2\Z$ . Enumerate the cases: $\varphi$ trivial yields $\Z/12\Z$ and $\Z/6\Z \times \Z/2\Z$ ; $P_2 =  \Z/4\Z$ with $\varphi$ surjective yields the *dicyclic group* $\mathrm{Dic}_3 = \Z/3\Z \rtimes  \Z/4\Z$ ; $P_2 = (\Z/2\Z)^2$ with $\varphi$ surjective yields, up to the equivalence of question 3, a single group — show it is $D_6$ , e.g. by exhibiting an element of order $6$ and a reflection-like involution.
16. Show that $\Z/12\Z$ , $\Z/6\Z \times \Z/2\Z$ , $D_6$ , $A_4$ , $\mathrm{Dic}_3$ are pairwise non-isomorphic *(count elements of order $2$, or use $n_3$)* . This settles order $12$ .

**Part V — Synthesis.**

17. Assemble the classification table: for each order $n \leq  15$ , the complete list of groups up to isomorphism, with the counts $1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1$ .

**Part VI — Beyond: the groups of order $p^3$, $p$ odd.** The order-$8$ analysis of Part III has a beautiful odd-prime analogue, with one genuinely new phenomenon. Let $p$ be an odd prime and $G$ nonabelian of order $p^3$.

18. Show that $\abs{Z(G)} = p$ , that $G/Z(G) \cong  (\Z/p\Z)^2$ *(a cyclic quotient by the [center](#ex-b3-groups-actions) forces abelianity: [Exercise 1.2](#exo-b3-groups-2))* , and that $D(G) =  Z(G)$ *(for $D(G) \subseteq Z(G)$, use that $G/Z(G)$ is abelian; for equality, $G$ is nonabelian and $D(G) \neq \{e\}$)* . Deduce that every [commutator](#def-b3-groups-derived) $[x, y] = xyx^{-1}y^{-1}$ is central and of order dividing $p$ .
19. (The key identity) Let $x, y \in G$ and $z = [y, x]$, central. Prove by induction on $k$: $$(xy)^k = x^k y^k z^{k(k-1)/2} .$$ *(Move each $y$ past each $x$; every crossing costs one central factor $z$.)*
20. Deduce that for $p$ odd the map $\theta\colon x \mapsto  x^p$ is a group morphism from $G$ to $Z(G)$ *(why is $x^p$ central? why does $z^{p(p-1)/2} = e$ need $p$ odd?)* , and conclude that $G$ has exponent $p$ or $p^2$ , the two cases being distinguished by whether $\theta$ is trivial.
21. (Exponent $p$) Suppose every element satisfies $x^p = e$. Pick $x, y$ whose classes generate $G/Z(G)$ and set $z =  [y, x]$. Show that $z \neq e$, that every element of $G$ is uniquely $x^ay^bz^c$ ($0 \leq a, b, c < p$), and that the multiplication is entirely determined by the relations $x^p = y^p = z^p = e$, $z$ central, $[y, x] =  z$. Verify that the *Heisenberg group* $$H_p = \left\{  \begin{pmatrix} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1  \end{pmatrix} : a, b, c \in \mathbb F_p \right\}  \subseteq GL_3(\mathbb F_p)$$ realizes these relations and has exponent $p$ (compute $(I + N)^p$ with $N$ strictly upper triangular, using $N^3 = 0$ and $p \geq 3$): every exponent-$p$ nonabelian group of order $p^3$ is isomorphic to $H_p$.
22. (Exponent $p^2$ ) Suppose some $r \in G$ has order $p^2$ , and set $N = \langle r\rangle$ , [normal](#def-b3-groups-normal) (index $p$ : [Exercise 1.10](#exo-b3-groups-10) ). Show there is $s \notin N$ with $s^p = e$ *(take any $t \notin N$; using question 20, correct it: $\theta(t) = t^p \in Z(G) \subseteq N$ — justify $Z(G) = \langle r^p\rangle$ — and choose $a$ with $s = tr^{a}$ satisfying $s^p = e$; where is $p$ odd used?)* . Show $srs^{-1} = r^{1+p}$ up to replacing $s$ by a power, and conclude: there is exactly *one* nonabelian group of order $p^3$ and exponent $p^2$ , namely $\Z/p^2\Z \rtimes_\varphi \Z/p\Z$ with $\varphi(1)\colon r \mapsto r^{1+p}$ *(use question 3; $\operatorname{Aut}(\Z/p^2\Z)$ is cyclic of order $p(p-1)$, admitted here, so it has a unique subgroup of order $p$)* .
23. Conclude the count: for odd $p$ there are exactly $5$ groups of order $p^3$ (three abelian, two nonabelian), just as for $p = 2$ — but the two nonabelian ones are no longer $D_4$ and $Q_8$ . Pinpoint exactly where the odd- $p$ argument breaks for $p = 2$ : in the identity of question 19, $z^{k(k-1)/2}$ for $k = p = 2$ is $z^1 \neq  e$ , so squaring is not a morphism — and indeed $Q_8$ has a unique element of order $2$ while exponent- $4$ $D_4$ has five.

**Part VII — Complements.**

24. For $p$ odd, count the elements of order $p$ in each of the two nonabelian groups of order $p^3$ : show that $H_p$ has exactly $p^3 - 1$ of them, while $M_p =  \Z/p^2\Z \rtimes \Z/p\Z$ has exactly $p^2 - 1$ *(use the morphism $\theta$ of question 20: identify its image, then the order of its kernel)* . Verify numerically for $p = 3$ : $26$ against $8$ . Explain why no squaring-morphism argument of this kind can separate $D_4$ from $Q_8$ , and which count does separate them.
25. Call an integer $n \geq 1$ *cyclic* if every group of order $n$ is cyclic. Show that if $p^2 \mid n$ for some prime $p$ , or if $n$ has prime divisors $p < q$ with $p \mid q - 1$ , then $n$ is not cyclic *(in each case exhibit a noncyclic group of order $n$, using Part II for the second)* . Deduce that $n$ cyclic forces $\gcd(n, \varphi(n)) = 1$ , where $\varphi$ is Euler’s totient, and check against the table of question 17: among $n \leq 15$ , the orders carrying a single group are exactly $n \in \{1, 2, 3, 5, 7, 11, 13, 15\}$ , precisely those with $\gcd(n, \varphi(n)) = 1$ .

**Solution of Problem 1.1.**

**1.** For $x, y \in G$: $(xy)^2 = e$ gives $xy = (xy)^{-1} =
y^{-1}x^{-1} = yx$ (each element is its own inverse): abelian. Such a $G$, written additively, is a vector space over $\mathbb
F_2$ ($2x = 0$, and the axioms are the abelian group axioms); if finite, it has a finite basis: $G \cong (\Z/2\Z)^k$, of order $2^k$.

**2.** Order $p^2$: $G$ is abelian ([Theorem 1.12](#thm-b3-groups-pfixed)). If some element has order $p^2$, $G$ is cyclic. Otherwise all $x \neq e$ have order $p$; additively $G$ is then a vector space over $\mathbb F_p$ ($px = 0$), of dimension $2$ ($p^2$ elements): $G \cong (\Z/p\Z)^2$.

Abelian of order $8$, by the maximal order $m$ of an element: $m =
8$: cyclic $\Z/8\Z$. $m = 2$: $(\Z/2\Z)^3$ by question 1. $m = 4$: let $x$ have order $4$ and $y \notin \langle x \rangle$; $y^2 \in
\langle x\rangle$ (index $2$). $y^2 \in \{x, x^3\}$ would give $y$ order $8$; so $y^2 \in \{e, x^2\}$. If $y^2 = x^2$, replace $y$ by $xy$: $(xy)^2 = x^2y^2 = x^4 = e$ ($G$ abelian) and $xy \notin
\langle x\rangle$. So we may assume $y^2 = e$: then $\langle x
\rangle \cap \langle y \rangle = \{e\}$, both [normal](#def-b3-groups-normal) (abelian), $\abs{\langle x\rangle \langle y\rangle} = 8$: [Proposition 1.24](#prop-b3-groups-direct) gives $G \cong \Z/4\Z \times \Z/2\Z$. Irredundant: the numbers of solutions of $x^2 = e$ are $2, 4, 8$ in the three groups.

**3.** Define $\psi \colon H \rtimes_{\varphi'} K \to H
\rtimes_{\varphi} K$ by $\psi(h, k) = (h, \alpha(k))$, a bijection. Morphism:

$$
\psi\bigl((h,k)(h',k')\bigr)
= \bigl(h\,\varphi'(k)(h'),\, \alpha(kk')\bigr)
= \bigl(h\,\varphi(\alpha k)(h'),\, \alpha(k)\alpha(k')\bigr)
= \psi(h,k)\,\psi(h',k').
$$

**4.** $\operatorname{Aut}(\Z/n\Z) \cong (\Z/n\Z)^\times$ ([Exercise 1.3](#exo-b3-groups-3)): for $n = 3$: $\{\pm 1\} \cong \Z/2\Z$; $n = 4$: $\{\bar 1, \bar 3\} \cong \Z/2\Z$; $n = 5$: $\{\bar 1,
\bar 2, \bar 3, \bar 4\}$, cyclic of order $4$ generated by $\bar
2$ ($2, 4, 3, 1$); $n = 7$: cyclic of order $6$ generated by $\bar
3$ ($3, 2, 6, 4, 5, 1$). For $V = (\Z/2\Z)^2$: an automorphism is $\mathbb F_2$-linear (it preserves addition, and scalars are $0,
1$), so $\operatorname{Aut}(V) = GL_2(\mathbb F_2)$, of order $(4-1)(4-2) = 6$; it acts faithfully on the $3$ nonzero vectors, giving an injective morphism to $S_3$ between groups of order $6$: $\operatorname{Aut}(V) \cong S_3$.

**5.** Cauchy provides $r$ of order $p$; $N = \langle r
\rangle$ has index $2$, hence is [normal](#def-b3-groups-normal) ([Exercise 1.1](#exo-b3-groups-1)). Cauchy also provides $s$ of order $2$, and $s \notin N$ (all non-identity elements of $N$ have odd order $p$).

**6.** $N \cap \langle s \rangle = \{e\}$ and $\abs{N\langle
s\rangle} = 2p$ ([Exercise 1.4](#exo-b3-groups-4)(a)): by [Proposition 1.26](#prop-b3-groups-semidirect), $G \cong \Z/p\Z \rtimes_\varphi
\Z/2\Z$ with $\varphi(1) = (x \mapsto sxs^{-1})$ an automorphism of order dividing $2$. In $(\Z/p\Z)^\times$, $k^2 = 1$ has only the solutions $k = \pm 1$ ($X^2 - 1$ has at most two roots in the field $\mathbb F_p$). If $\varphi(1) = \mathrm{id}$: the product is direct, $G \cong \Z/p\Z \times \Z/2\Z \cong \Z/2p\Z$. If $\varphi(1) = -\mathrm{id}$: $G = \langle r, s \mid r^p = s^2 = e,
\ srs^{-1} = r^{-1}\rangle \cong D_p$ ([Example 1.27](#ex-b3-groups-semidirectexamples)). They are not isomorphic: $D_p$ is nonabelian for $p \geq 3$ ($srs^{-1} = r^{-1} \neq r$).

**7.** $n_q \equiv 1 \pmod q$ divides $p < q$: $n_q = 1$, so $N \cong \Z/q\Z$ is [normal](#def-b3-groups-normal). Let $P \cong \Z/p\Z$ be a Sylow $p$-subgroup: $N \cap P = \{e\}$, $NP = G$ (order $pq$), so $G
\cong \Z/q\Z \rtimes_\varphi \Z/p\Z$ with $\varphi \colon \Z/p\Z
\to \operatorname{Aut}(\Z/q\Z) \cong \Z/(q-1)\Z$ (cyclic, admitted). If $p \nmid q - 1$: the image of $\varphi$ has order dividing both $p$ and $q-1$, hence is trivial, and $G \cong
\Z/pq\Z$ ([Example 1.22](#ex-b3-groups-pq)). If $p \mid q - 1$: besides the trivial $\varphi$, any nontrivial $\varphi$ is injective (its kernel, a subgroup of $\Z/p\Z$, is trivial) with image *the* unique subgroup $C$ of order $p$ of the cyclic group $\Z/(q-1)\Z$. Two nontrivial [actions](#def-b3-groups-action) $\varphi, \varphi'$ are then two isomorphisms $\Z/p\Z \to C$, so $\alpha =
\varphi^{-1}\circ\varphi' \in \operatorname{Aut}(\Z/p\Z)$ satisfies $\varphi' = \varphi\circ\alpha$: by question 3 the two [semidirect products](#def-b3-groups-semidirect) are isomorphic. Hence exactly one nonabelian group of order $pq$ (nonabelian since $\varphi \neq
\mathrm{id}$ makes some conjugation nontrivial). Order $15$: $p =
3$, $q = 5$, $3 \nmid 4$: cyclic only.

**8.** Not every element has order $\leq 2$ (else abelian by question 1), and no element has order $8$ (else cyclic, abelian): some $r$ has order $4$, and $N = \langle r\rangle$, of index $2$, is [normal](#def-b3-groups-normal).

**9.** $s^2 \in N$ by [Exercise 1.1](#exo-b3-groups-1)(b). The conjugate $srs^{-1} \in N$ has order $4$, so $srs^{-1} \in \{r,
r^3\}$; if $srs^{-1} = r$ then $r$ and $s$ commute and $G =
\langle r, s\rangle$ is abelian — excluded. So $srs^{-1} =
r^{-1}$. If $s^2 = r$ or $r^3$, then $s$ has order $8$: excluded. (Alternatively: $s^2$ commutes with $s$, but $s r s^{-1} =
r^{-1}$ and $s r^3 s^{-1} = r^{-3} = r$: neither $r$ nor $r^3$ is fixed by conjugation by $s$.) So $s^2 \in \{e, r^2\}$.

**10.** If $s^2 = e$: $G = \langle r, s \mid r^4 = s^2 = e,\
srs^{-1} = r^{-1}\rangle$. The eight elements $r^is^j$ ($0 \leq i
< 4$, $0 \leq j < 2$) are distinct ($s \notin \langle r\rangle$) and the relations determine all products: the assignment $r
\mapsto$ (rotation by $\pi/2$), $s \mapsto$ (a reflection) defines a surjective morphism onto $D_4$, between groups of order $8$: an isomorphism.

**11.** If $s^2 = r^2$: again $G = \{r^i s^j\}$ and the relations $r^4 = e$, $s^2 = r^2$, $srs^{-1} = r^{-1}$ force the whole table. With $\mathrm i = r$, $\mathrm j = s$, $\mathrm k =
rs$, $-1 = r^2$: $\mathrm i^2 = \mathrm j^2 = -1$, $\mathrm k^2 =
rsrs = r r^{-1} s s = s^2 = -1$ (using $sr = r^{-1}s$), and $\mathrm i \mathrm j \mathrm k = r\,s\,rs = r\,r^{-1}s\,s = s^2 =
-1$. Existence: the matrices

$$
A = \begin{pmatrix} \iu & 0\\ 0 & -\iu \end{pmatrix},
\qquad
B = \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}
$$

satisfy $A^4 = I$, $B^2 = -I = A^2$ and $BAB^{-1} = A^{-1}$ — for the last one, check

$$
BA = \begin{pmatrix} 0 & -\iu\\ -\iu & 0 \end{pmatrix} = A^{-1}B .
$$

So $\{\pm I, \pm A, \pm B, \pm AB\}$ is a group of order $8$ realizing the table: $Q_8$ exists.

**12.** Let $H \neq \{e\}$ be a subgroup and $x \in H
\setminus \{e\}$. If $x \neq -1$ then $x \in \{\pm\mathrm i,
\pm\mathrm j, \pm\mathrm k\}$ and $x^2 = -1 \in H$. So $-1 \in H$ always. The subgroups are $\{e\}$, $\{\pm 1\}$ (the [center](#ex-b3-groups-actions)), $\langle \mathrm i\rangle, \langle \mathrm j\rangle, \langle
\mathrm k\rangle$ (index $2$) and $Q_8$: all [normal](#def-b3-groups-normal) ($\{e\}$ and the [center](#ex-b3-groups-actions) trivially, index $2$ by [Exercise 1.1](#exo-b3-groups-1), $Q_8$ itself). $D_4$ has five elements of order $2$ ($r^2$ and the four reflections), $Q_8$ only one ($-1$): not isomorphic. A [semidirect product](#def-b3-groups-semidirect) $H \rtimes K$ with $H, K \neq \{e\}$ requires $H \cap K =
\{e\}$, impossible since both contain $-1$.

**13.** $n_3 \mid 4$, $n_3 \equiv 1 \pmod 3$: $n_3 \in \{1,
4\}$; $n_2 \mid 3$, odd: $n_2 \in \{1, 3\}$. If $n_3 = 4$: the four Sylow $3$-subgroups pairwise intersect trivially (prime order), giving $4 \times 2 = 8$ elements of order $3$; the remaining $4$ elements must constitute the unique Sylow $2$-subgroup: $n_2 = 1$.

**14.** $\ker\rho$ normalizes every Sylow $3$-subgroup, so $\ker \rho \subseteq N_G(P_3)$, which has index $n_3 = 4$, i.e. order $3$: $\abs{\ker\rho} \in \{1, 3\}$. Order $3$ would make $\ker\rho$ a *[normal](#def-b3-groups-normal)* Sylow $3$-subgroup, contradicting $n_3
= 4$. So $\rho$ is injective and its image $H \leq S_4$ has order $12$, index $2$: $H \trianglelefteq S_4$ and $H$ contains all squares ([Exercise 1.1](#exo-b3-groups-1)(b)). The squares of $S_4$ include $e$ and all eight $3$-cycles ($\sigma = (\sigma^2)^2$ for a $3$-cycle), which generate $A_4$ (they lie in $A_4$, and together with their products give all twelve elements; or: [Lemma 1.32](#lem-b3-groups-threecycles) for $n = 4$’s generation part, which only needs $n \geq 3$). So $A_4 \subseteq H$ and $\abs{A_4}
= \abs H$: $G \cong H = A_4$.

**15.** $P_3 \trianglelefteq G$, $P_3 \cap P_2 = \{e\}$, $P_3P_2 = G$: $G \cong \Z/3\Z \rtimes_\varphi P_2$ ([Proposition 1.26](#prop-b3-groups-semidirect)), $\varphi \colon P_2 \to
\operatorname{Aut}(\Z/3\Z) = \{\pm\mathrm{id}\} \cong \Z/2\Z$.

- $\varphi$ trivial: [direct products](#prop-b3-groups-direct) $\Z/3\Z \times \Z/4\Z  \cong \Z/12\Z$ and $\Z/3\Z \times (\Z/2\Z)^2 \cong \Z/6\Z  \times \Z/2\Z$ .
- $P_2 = \Z/4\Z$ , $\varphi$ surjective: necessarily $\varphi(1) = -\mathrm{id}$ (the only nontrivial choice): one group, $\mathrm{Dic}_3 = \Z/3\Z \rtimes \Z/4\Z$ .
- $P_2 = (\Z/2\Z)^2$ , $\varphi$ surjective: $\ker\varphi$ is one of the three subgroups of order $2$ ; the three resulting $\varphi$ differ by automorphisms of $(\Z/2\Z)^2$ permuting these subgroups (question 4: $\operatorname{Aut}  \cong S_3$ acts transitively on the three involutions), so by question 3 they give one isomorphism class. It is $D_6$ : pick $t$ generating $\ker\varphi$ and $x$ generating $\Z/3\Z$ ; the element $\rho = (x, t)$ satisfies $\rho^2 = (2x, 0)$ , $\rho^3 = (0, t)$ , $\rho^6 = e$ and no smaller power is $e$ : order $6$ ; for $s = (0, u)$ with $u  \notin \ker\varphi$ : $s^2 = e$ and $s\rho s^{-1} = (-x, t)  = \rho^{-1}$ . As $\langle \rho, s\rangle$ has order $12$ , $G \cong D_6$ .

**16.** Counting elements of order $2$: $\Z/12\Z$ has $1$; $\Z/6\Z\times\Z/2\Z$ has $3$; $D_6$ has $7$ (six reflections and the half-turn $\rho^3$); $A_4$ has $3$; $\mathrm{Dic}_3$ has $1$ (only $(0, 2)$: an element $(h, k)$ with $k$ of order $4$ in $\Z/4\Z$ has order $4$). This separates all but the pairs $\{\Z/12\Z, \mathrm{Dic}_3\}$ and $\{\Z/6\Z\times\Z/2\Z, A_4\}$: the first members are abelian, the second not ($\mathrm{Dic}_3$: the [action](#def-b3-groups-action) is nontrivial; $A_4$: $(1\,2\,3)$ and $(1\,2)(3\,4)$ do not commute). Five distinct groups; parts II–IV show the list is complete.

**17.** The classification table:

| $n$ | groups of order $n$ | # |
| --- | --- | --- |
| $1$ | $\{e\}$ | $1$ |
| $2$ | $\Z/2\Z$ | $1$ |
| $3$ | $\Z/3\Z$ | $1$ |
| $4$ | $\Z/4\Z$, $(\Z/2\Z)^2$ | $2$ |
| $5$ | $\Z/5\Z$ | $1$ |
| $6$ | $\Z/6\Z$, $S_3 = D_3$ | $2$ |
| $7$ | $\Z/7\Z$ | $1$ |
| $8$ | $\Z/8\Z$, $\Z/4\Z{\times}\Z/2\Z$, $(\Z/2\Z)^3$, $D_4$, $Q_8$ | $5$ |
| $9$ | $\Z/9\Z$, $(\Z/3\Z)^2$ | $2$ |
| $10$ | $\Z/10\Z$, $D_5$ | $2$ |
| $11$ | $\Z/11\Z$ | $1$ |
| $12$ | $\Z/12\Z$, $\Z/6\Z{\times}\Z/2\Z$, $D_6$, $A_4$, $\mathrm{Dic}_3$ | $5$ |
| $13$ | $\Z/13\Z$ | $1$ |
| $14$ | $\Z/14\Z$, $D_7$ | $2$ |
| $15$ | $\Z/15\Z$ | $1$ |

Orders $6, 10, 14$ are Part II with $p = 3, 5, 7$; order $15$ is question 7; order $8$ is Part III together with question 2; order $12$ is Part IV; prime orders are Lagrange; orders $4$ and $9$ are question 2.

**18.** $Z = Z(G)$ is nontrivial ([Theorem 1.12](#thm-b3-groups-pfixed)) and $Z \neq G$ (nonabelian), so $\abs Z \in \{p, p^2\}$. If $\abs Z = p^2$, then $G/Z$ is cyclic of order $p$ and [Exercise 1.2](#exo-b3-groups-2) makes $G$ abelian: excluded, so $\abs Z = p$ and $\abs{G/Z} = p^2$. By question 2, $G/Z$ is $\Z/p^2\Z$ or $(\Z/p\Z)^2$; cyclic is again excluded by [Exercise 1.2](#exo-b3-groups-2): $G/Z \cong (\Z/p\Z)^2$. Since $G/Z$ is abelian, every [commutator](#def-b3-groups-derived) lies in $Z$: $D(G) \subseteq Z$; and $D(G) \neq \{e\}$ ($G$ nonabelian), so $D(G) = Z$ ($\abs Z = p$ leaves no room). [Commutators](#def-b3-groups-derived) are central of order dividing $\abs Z = p$.

**19.** Induction on $k$, the case $k = 1$ being trivial. Using $yx = xyz^{-1}\cdot$ — precisely, $z = [y, x] =
yxy^{-1}x^{-1}$ gives $yx = zxy$, i.e. moving one $y$ leftward past one $x$ produces one factor $z$, which is central and can be parked anywhere. Then

$$
(xy)^{k+1} = (xy)^k\,xy = x^ky^kz^{k(k-1)/2}\,xy
= x^k\,(y^kx)\,y\,z^{k(k-1)/2}
= x^{k+1}y^{k+1}\,z^{k(k-1)/2 + k},
$$

since carrying $x$ past $y^k$ costs $k$ factors of $z$ ($y^kx =
z^kxy^k$, by $k$ applications of $yx = zxy$); and $k(k-1)/2 + k
= k(k+1)/2$.

**20.** With $k = p$: $(xy)^p = x^py^pz^{p(p-1)/2}$. For $p$ odd, $(p-1)/2$ is an integer, so $z^{p(p-1)/2} =
(z^p)^{(p-1)/2} = e$ (question 18: $z$ has order dividing $p$): $\theta(xy) = \theta(x)\theta(y)$, a morphism. Its values are central: the class of $x$ in $G/Z \cong (\Z/p\Z)^2$ has order dividing $p$, so $x^p \in Z$. If $\theta$ is trivial, every element has order dividing $p$: exponent $p$ (not $1$: $G \neq
\{e\}$). Otherwise some $x^p \neq e$, and $x$ has order $p^2$ (order divides $p^3$, and $x$ cannot have order $p^3$: $G$ would be cyclic, hence abelian): exponent $p^2$.

**21.** Classes $\bar x, \bar y$ generating $G/Z$: their [commutator](#def-b3-groups-derived) $z = [y, x]$ is $\neq e$, else $x, y, Z$ would generate an abelian $G$ (their classes generate the quotient and $Z$ is central) — and $z$ generates $Z$ ($\abs Z = p$). Every $g \in G$ has class $\bar x^a\bar y^b$ for unique $0 \leq a, b <
p$, so $g = x^ay^bz^c$ with a unique $0 \leq c < p$: $p^3$ elements, all accounted for. Products of such [normal](#def-b3-groups-normal) forms are computed using only $yx = zxy$, $z$ central, and $x^p = y^p =
z^p = e$: the table is forced, so any two exponent-$p$ nonabelian groups of order $p^3$ are isomorphic (match the generators). The Heisenberg group realizes the relations: with $X = I + E_{12}$, $Y = I + E_{23}$, one computes $[Y, X] = I -
E_{13}$ (central in $H_p$), and for any strictly upper triangular $N$, $N^3 = 0$ gives

$$
(I + N)^p = I + pN + \binom p2N^2 = I
\qquad\text{in characteristic } p,\ p \geq 3,
$$

since $p \mid p$ and $p \mid \binom p2$ for odd $p$: exponent $p$. So the exponent-$p$ group is $H_p$.

**22.** $Z(G) = \langle r^p\rangle$: indeed $r^p$ is central (question 20 argument: the class of $r$ in the exponent-$p$ quotient $G/Z$ gives $r^p \in Z$) and is $\neq e$, so it generates the order-$p$ [center](#ex-b3-groups-actions). Take any $t \notin N$. If $t^p = e$, set $s = t$. Otherwise $\theta(t) = t^p \in Z =
\langle r^p\rangle$, say $t^p = r^{pb}$; set $s = tr^{-b}$: by question 20 ($\theta$ a morphism, $p$ odd), $s^p =
t^pr^{-pb} = e$, and $s \notin N$. Conjugation: $srs^{-1} \in
N$ ($N$ [normal](#def-b3-groups-normal)) has order $p^2$, so $srs^{-1} = r^m$ with $p
\nmid m$; also $s^p = e$ forces $m^p \equiv m \pmod{p^2}$ — conjugating $p$ times returns $r$, so $m^p \equiv 1 \pmod
{p^2}$, and $m \equiv m^p \equiv 1 \pmod p$ (Fermat): $m = 1 +
ap$. Nontriviality ($G$ nonabelian) gives $a \not\equiv 0$; replacing $s$ by the power $s^{a'}$ with $aa' \equiv 1 \pmod p$ turns the [action](#def-b3-groups-action) into $r \mapsto r^{1+p}$. This presents $G$ as $\Z/p^2\Z\rtimes_\varphi\Z/p\Z$ with $\varphi(1)\colon r
\mapsto r^{1+p}$; by question 3, any two nontrivial morphisms $\Z/p\Z \to \operatorname{Aut}(\Z/p^2\Z)$ with the same image — and the image is *the* unique subgroup of order $p$ of the cyclic $\operatorname{Aut}(\Z/p^2\Z)$ — give isomorphic [semidirect products](#def-b3-groups-semidirect): uniqueness.

**23.** Abelian: $\Z/p^3\Z$, $\Z/p^2\Z\times\Z/p\Z$, $(\Z/p\Z)^3$ (question 2’s argument, one degree up: classify by maximal order). Nonabelian: exactly $H_p$ (exponent $p$, question 21) and $\Z/p^2\Z\rtimes\Z/p\Z$ (exponent $p^2$, question 22), distinguished by their exponents. Total: five. For $p = 2$ the morphism argument of question 20 collapses: $z^{p(p-1)/2} = z^{1} = z \neq e$, squaring is not a morphism, and indeed both nonabelian groups of order $8$ have exponent $4$ — the invariant that separates $D_4$ from $Q_8$ is the number of elements of order $2$ (five against one), not the exponent. The odd-$p$ world is, for once, tidier than characteristic $2$.

**24.** In $H_p$ every element $\neq e$ has order $p$ (exponent $p$, question 21): $p^3 - 1$ elements of order $p$. In $M_p$, the map $\theta \colon x \mapsto x^p$ is a morphism $M_p \to Z(M_p) = \langle r^p\rangle$ (question 20, $p$ odd); $\theta(r) = r^p \neq e$, so the image is the whole order-$p$ [center](#ex-b3-groups-actions) and $\ker\theta = \{x : x^p = e\}$ has order $p^3/p =
p^2$. The elements of order $p$ are the nonidentity elements of this kernel: $p^2 - 1$ of them. For $p = 3$: $H_3$ has $27 - 1 = 26$ elements of order $3$, and $M_3 = \Z/9\Z \rtimes
\Z/3\Z$ has $9 - 1 = 8$. For $p = 2$ the argument dies at the start: squaring is not a morphism on a nonabelian group of order $8$ (question 23), and indeed the set $\{x : x^2 = e\}$ has $6$ elements in $D_4$ — not the order of a subgroup of $D_4$. The count that does separate the pair is the number of elements of order $2$: five in $D_4$, one in $Q_8$ (question 11).

**25.** If $p^2 \mid n$, the group $\Z/p\Z \times
\Z/(n/p)\Z$ has order $n$ and is not cyclic: every element’s order divides $\operatorname{lcm}(p, n/p) = n/p < n$, since $p \mid n/p$. If $p < q$ are primes dividing $n$ with $p \mid
q - 1$, question 7 provides a nonabelian group $\Z/q\Z \rtimes
\Z/p\Z$ of order $pq$; then $(\Z/q\Z \rtimes \Z/p\Z) \times
\Z/(n/pq)\Z$ has order $n$ and is nonabelian, hence not cyclic. Now suppose $\gcd(n, \varphi(n)) > 1$ and pick a prime $p$ dividing both. Writing $\varphi(n) = \prod_{q^a \parallel
n} q^{a-1}(q - 1)$, the divisibility $p \mid \varphi(n)$ means either $p^2 \mid n$ (the factor $q^{a-1}$ with $q = p$, $a \geq 2$) or $p \mid q - 1$ for some prime $q \mid n$, $q
\neq p$: in both cases $n$ is not cyclic by the above. By contraposition, $n$ cyclic forces $\gcd(n, \varphi(n)) = 1$. Check for $n \leq 15$: the values $\varphi(n)$ for $n = 1,
\dots, 15$ are $1, 1, 2, 2, 4, 2, 6, 4, 6, 4, 10, 4, 12, 6,
8$, and $\gcd(n, \varphi(n)) = 1$ exactly for $n = 1, 2, 3, 5,
7, 11, 13, 15$ — exactly the entries of the table of question 17 with a single group. The other orders are witnessed noncyclic as above: $4, 8, 9, 12$ by a square factor, $6, 10, 12, 14$ by $2 \mid q - 1$. (The converse — $\gcd(n, \varphi(n)) = 1$ implies $n$ cyclic — is also true; question 7 proves its first nontrivial case, $n = pq$ with $p \nmid q - 1$.)