Sylow subgroup কী?

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).