University Mathematics — Year 3 · Bachelor Year 3

1Group 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 and fixed points yields the class equation, 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) and to disassemble them (composition series, solvable groups), and we prove the theorem that, in Chapter 4, will close a three-century-old question about polynomial equations: the alternating group $A_n$ is 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: normal subgroups 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. ∎

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 $N \trianglelefteq G$ .

$HN = \{hn : h \in H,\, n \in N\}$ is a subgroup, $N \trianglelefteq HN$ , $H \cap N \trianglelefteq H$ , and

$H/(H \cap N) \;\cong\; HN/N .$
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.

(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 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; or note that $[G:N] = 2$ (then $gN = Ng$ is forced — Exercise 1.1).

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 run all of finite group theory:

$G$ on itself by left translation $g \cdot x = gx$ : free and transitive.
$G$ on itself by conjugation $g \cdot x = gxg^{-1}$ : orbits are the conjugacy classes, stabilizers the centralizers $Z_G(x) = \{g : gx = xg\}$ , fixed points the center $Z(G)$ .
$G$ on the coset space $G/H$ by $g \cdot xH = gxH$ : transitive, with stabilizer of the coset $H$ equal to $H$ . Every transitive action is of this form (Exercise 1.8).
$G$ on its set of subgroups by conjugation: the stabilizer of $H$ is the normalizer $N_G(H) = \{g : gHg^{-1} = H\}$ , the largest subgroup of $G$ in which $H$ is normal.
$S_n$ 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 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. The counting statements follow from Lagrange’s theorem and the partition of $X$ into orbits. ∎

Corollary 1.11 (Class equation)

For a finite group $G$ , picking one representative $x_i$ in each conjugacy class 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 to the conjugation action: singleton orbits are exactly the elements of $Z(G)$ . ∎

Theorem 1.12 (Fixed points of $p$ -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, and every group of order $p^2$ is abelian.

Proof. Each orbit 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 gives the congruence. For the center: conjugation 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) — 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, $\abs{X^{\Z/p\Z}} \equiv \abs X \equiv 0 \pmod p$ . 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, 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 sizes divide the group order. The proofs of Theorems 1.12 and 1.13 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). The Sylow theorems salvage the converse for prime powers, and their counting clause is the sharpest general tool we have for producing normal subgroups.

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, $\abs\Omega = \binom{p^a m}{p^a} \equiv m \not\equiv 0 \pmod p$ , so some orbit $\mathcal O_S$ has size prime to $p$ (if $p$ divided every orbit size, it would divide $\abs\Omega$ ). Let $H = G_S$ be the stabilizer 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 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, 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), with stabilizer of $P$ the normalizer $N_G(P) \supseteq P$ : $n_p = [G : N_G(P)]$ , and $m = [G:P] = [G:N_G(P)]\,[N_G(P):P]$ shows $n_p \mid m$ .

Now restrict the action 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 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 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. 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).

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 Sylows: $P \cap Q = \{e\}$ (coprime orders), so $\abs{PQ} = pq$ (Exercise 1.4) and $G \cong P \times Q \cong \Z/p\Z \times \Z/q\Z \cong \Z/pq\Z$ by Proposition 1.24 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).

Example 1.23 (A complete Sylow census: $S_4$ )

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 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 ( $(12)(1234)(12) = (2134)$ 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 (Part IV of Problem 1.1).

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 $hkh^{-1}k^{-1}$ 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:

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 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. 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 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, homotheties a complement. (c) $S_n \cong A_n \rtimes \Z/2\Z$ (complement: any transposition). (d) The quaternion group $Q_8$ is not a semidirect product of proper subgroups: every nontrivial subgroup contains $-1$ (Problem 1.1), 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 (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 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 are solvable; conversely, if $N$ and $G/N$ are solvable, so is $G$ .

Proof. An automorphism $\alpha$ maps $[x,y]$ to $[\alpha x, \alpha y]$ : it permutes the commutators, 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, the derived series 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 onto commutators; 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. $p$ -groups are solvable, 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: $S_4 \trianglerighteq A_4 \trianglerighteq V \trianglerighteq \{e\}$ , where $V = \{e, (1\,2)(3\,4), (1\,3)(2\,4), (1\,4)(2\,3)\}$ is the Klein group of double transpositions (normal in $S_4$ : a union of conjugacy classes), with abelian quotients $\Z/2\Z$ , $\Z/3\Z$ , $V$ . In Chapter 4, “the general equation of degree $n$ is solvable by radicals” will literally mean “ $S_n$ is a solvable group”. Whence the importance of the next definition.

Definition 1.31

A group $G \neq \{e\}$ is simple if its only normal subgroups are $\{e\}$ and $G$ . A nonabelian simple group is not solvable: $D(G) \trianglelefteq G$ is not $\{e\}$ (else $G$ abelian), so $D(G) = G$ and the derived series 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 for $n \geq 5$ .

Proof. Let $N \trianglelefteq A_n$ , $N \neq \{e\}$ . By Lemma 1.32 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, and the only normal subgroups of $S_n$ are $\{e\}$ , $A_n$ and $S_n$ .

Proof. $A_n$ is nonabelian simple, hence not solvable (Definition 1.31); a group containing a non-solvable subgroup is not solvable (Proposition 1.29). 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 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 iff all its composition factors are cyclic of prime order.

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

Uniqueness: induction on $\abs G$ , the case $G$ simple being clear. Take two composition series, with penultimate terms $M \trianglelefteq G$ and $N \trianglelefteq G$ (so $G/M$ , $G/N$ are simple). If $M = N$ , conclude by induction applied to $M$ . Otherwise $MN$ , normal in $G$ and strictly containing $M$ , equals $G$ ( $M$ is maximal normal: any normal $M \subsetneq L \subsetneq G$ would map to a proper nontrivial normal subgroup of the 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 of $K$ . Then $M$ carries two composition series: its original one, and the series of $K$ followed by $K \trianglelefteq M$ (the quotient $M/K \cong G/N$ is 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: if all factors are $\Z/p_i\Z$ , the series is a chain with abelian quotients, so $G$ is solvable (Proposition 1.29). Conversely, a composition factor of a solvable group is solvable (a quotient of a subgroup) and simple; a solvable simple group 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, with a well-defined parts list — an arithmetic of groups in which the simple groups are the primes, and where how the parts are glued (extension data, as in the semidirect product) replaces mere multiplication. The classification of the finite simple groups — 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. — The ten subgroups of the dihedral group $D_4 = \langle r, s \mid r^4 = s^2 = e,\ srs^{-1} = r^{-1}\rangle$ . The three subgroups of index $2$ (middle row) are normal, as is the center $\langle r^2\rangle$ (highlighted); the four reflection subgroups fall into two conjugacy classes of two. Chains from bottom to top give composition series, 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. (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

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

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), 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 $I + E_{13} \neq I$ .

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 $\operatorname{Inn}(G) \trianglelefteq \operatorname{Aut}(G)$ , with $\operatorname{Inn}(G) \cong G/Z(G)$ .

Solution

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). (c) If $H, K \trianglelefteq G$ and $H \cap K = \{e\}$ , show that $hk = kh$ for all $h \in H$ , $k \in K$ .

Solution

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 $[h,k] = hkh^{-1} k^{-1}$ equals $(hkh^{-1})k^{-1} \in K$ ( $K$ normal) and $h(kh^{-1}k^{-1}) \in H$ ( $H$ 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 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

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.

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

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, intersect trivially (coprime orders), and $\abs{P_3P_5} = 15$ (Exercise 1.4(a)): by Proposition 1.24, $G \cong \Z/3\Z \times \Z/5\Z \cong \Z/15\Z$ (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 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. (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

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, 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 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.

Exercise 1.8 ★★

(a) Let $H \leq G$ of index $n$ . Show that the 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 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. (c) Show that every transitive action of $G$ on a set $X$ is isomorphic to the action on a coset space: there is a bijection $X \to G/G_x$ commuting with the actions.

Solution

Solution of Exercise 1.8.

(a) The 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 (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.

(c) Let the 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 is all of $X$ ), and it intertwines the actions: $\Phi(h \cdot gG_x) = \Phi(hgG_x) = (hg) \cdot x = h \cdot \Phi(gG_x)$ .

Exercise 1.9 ★★

(a) Show that $D(G)$ is the smallest normal subgroup 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

Solution of Exercise 1.9.

(a) $D(G)$ is normal with abelian quotient (Proposition 1.29); 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 factors $f = \bar f \circ \pi$ through $G^{\mathrm{ab}}$ , uniquely since $\pi$ is surjective.

(b) Commutators are even permutations, so $D(S_n) \subseteq A_n$ . Conversely every $3$ -cycle is a commutator:

\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): $D(S_n) = A_n$ 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 grow”), and deduce that every maximal subgroup of a $p$ -group is 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

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).

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 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 of $\sigma$ in $S_5$ and in $A_5$ . (b) Deduce that $A_5$ is simple: a normal subgroup is a union of conjugacy classes, contains $e$ , and has cardinality dividing $60$ . (c) Show that a simple group of order $60$ necessarily has $n_5 = 6$ .

Solution

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 $N$ is a union of conjugacy classes 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 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, contradicting simplicity ( $1 < 5 < 60$ ). Hence $n_5 = 6$ .

Exercise 1.12 ★★

(Normalizers of Sylow subgroups 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 is contained in a proper normal subgroup of $G$ , and that a maximal subgroup containing $N_G(P)$ is self-normalizing.

Solution

Solution of Exercise 1.12.

(a) $P$ is normal in $H = N_G(P)$ by definition of the normalizer, 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 Sylow subgroup 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 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 plus the analysis below of $p^2$ : there remain $6, 8, 10, 12, 14, 15$ .

Part I — Tools.

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 $\mathbb F_2$ .)
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 (discuss by the maximal order of an element).
Let $\varphi, \varphi' \colon K \to \operatorname{Aut}(H)$ be two actions. Show that if $\varphi' = \varphi \circ \alpha$ with $\alpha \in \operatorname{Aut}(K)$ , then $H \rtimes_{\varphi} K \cong H \rtimes_{\varphi'} K$ .
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 $pq$ .

Let $\abs G = 2p$ with $p$ an odd prime. Show that $G$ has a normal subgroup $N = \langle r \rangle$ of order $p$ and an element $s$ of order $2$ outside $N$ .
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 $6$ , $10$ , $14$ .
More generally, let $\abs G = pq$ with $p < q$ primes. Show that if $p \nmid q-1$ then $G$ is cyclic (Example 1.22), 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 (cyclicity of $\mathbb F_q^\times$ ). Conclude for order $15$ .

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

Show that $G$ has an element $r$ of order $4$ (use question 1) and that $N = \langle r\rangle$ is normal.
Let $s \notin N$ . Show that $s^2 \in N$ (Exercise 1.1(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 $s$ ?).
In the case $s^2 = e$ , show $G \cong D_4$ .
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

$\mathrm i \mapsto \begin{pmatrix} \iu & 0\\ 0 & -\iu \end{pmatrix}, \qquad \mathrm j \mapsto \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}.$
Show that every nontrivial subgroup of $Q_8$ contains $-1$ ; deduce that every subgroup of $Q_8$ is normal, that $D_4 \not\cong Q_8$ (count elements of order $2$ ), and that $Q_8$ is not a semidirect product of two proper subgroups.

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

Show $n_3 \in \{1, 4\}$ , $n_2 \in \{1, 3\}$ , and that $n_3 = 4$ forces $n_2 = 1$ (count elements of order $3$ ).
Suppose $n_3 = 4$ . The conjugation 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 and contain all squares — Exercise 1.1; count the squares in $S_4$ ); and conclude $G \cong A_4$ .
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.
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.

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 $p^3$ .

Show that $\abs{Z(G)} = p$ , that $G/Z(G) \cong (\Z/p\Z)^2$ (a cyclic quotient by the center forces abelianity: Exercise 1.2), 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 $[x, y] = xyx^{-1}y^{-1}$ is central and of order dividing $p$ .
(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 $z$ .)
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.
(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 $H_p$ .
(Exponent $p^2$ ) Suppose some $r \in G$ has order $p^2$ , and set $N = \langle r\rangle$ , normal (index $p$ : Exercise 1.10). 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 $p$ ).
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.

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.
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

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). 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 (abelian), $\abs{\langle x\rangle \langle y\rangle} = 8$ : Proposition 1.24 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): 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 (Exercise 1.1). 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(a)): by Proposition 1.26, $G \cong \Z/p\Z \rtimes_\varphi \Z/2\Z$ 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). 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. 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). 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 $\varphi, \varphi'$ 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 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.

9. $s^2 \in N$ by Exercise 1.1(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), $\langle \mathrm i\rangle, \langle \mathrm j\rangle, \langle \mathrm k\rangle$ (index $2$ ) and $Q_8$ : all normal ( $\{e\}$ and the center trivially, index $2$ by Exercise 1.1, $Q_8$ 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 $H \rtimes K$ 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 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(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 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), $\varphi \colon P_2 \to \operatorname{Aut}(\Z/3\Z) = \{\pm\mathrm{id}\} \cong \Z/2\Z$ .

$\varphi$ trivial: direct products $\Z/3\Z \times \Z/4\Z \cong \Z/12\Z$ 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 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) 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 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: $G/Z \cong (\Z/p\Z)^2$ . Since $G/Z$ is abelian, every commutator 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 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 $z = [y, x]$ 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 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. 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) 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 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: 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 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 $p \nmid q - 1$ .)

1.1 Quotient groups and the isomorphism theorems

1.2 Group actions

1.3 The Sylow theorems

1.4 Products, direct and semidirect

1.5 Solvable groups; simplicity of AnA_nAn​

1.6 Exercises

1.7 Problem: the groups of order at most 15

1.5 Solvable groups; simplicity of $A_n$