Sylow Theorems

By the class equation (Theorem 8.5): $\lvert P\rvert = \lvert Z(P)\rvert + \sum_i [P:C_P(x_i)]$, summing over noncentral conjugacy class representatives $x_i$. Each $x_i$ is noncentral, so $C_P(x_i) \neq P$, meaning $[P:C_P(x_i)] > 1$. Since $[P:C_P(x_i)]$ divides $\lvert P\rvert=p^n$ (Lagrange) and is a divisor of a prime power greater than 1, it must itself be divisible by $p$. So every term in the sum $\sum_i[P:C_P(x_i)]$ is divisible by $p$, hence the whole sum is divisible by $p$. Since $p$ also divides $\lvert P\rvert = p^n$ (as $n\geq1$), rearranging the class equation gives $\lvert Z(P)\rvert = \lvert P\rvert - \sum_i[P:C_P(x_i)]$, a difference of two multiples of $p$ — so $p$ divides $\lvert Z(P)\rvert$. Since $e\in Z(P)$ always, $\lvert Z(P)\rvert\geq1$; combined with $p\mid\lvert Z(P)\rvert$, we get $\lvert Z(P)\rvert \geq p > 1$.

Suppose $G/Z(G) = \langle gZ(G)\rangle$ for some $g\in G$. Take any $a,b\in G$; each coset $aZ(G)$ and $bZ(G)$ is some power of $gZ(G)$, so $a=g^iz_1$ and $b=g^jz_2$ for integers $i,j$ and $z_1,z_2\in Z(G)$. Then:

using that $z_1\in Z(G)$ commutes past $g^j$, and similarly:

using that $z_1,z_2\in Z(G)$ also commute with each other. So $ab=ba$ for arbitrary $a,b\in G$: $G$ is abelian. (And if $G$ is abelian, $Z(G)=G$ by Chapter 05's Exercise 5.4, forcing $G/Z(G)=\{e\}$.)

By Theorem 9.1, $Z(G) \neq \{e\}$, so by Lagrange's Theorem, $\lvert Z(G)\rvert \in \{p, p^2\}$ (a nontrivial divisor of $p^2$). If $\lvert Z(G)\rvert = p^2$, then $Z(G)=G$, and $G$ is abelian, done. If $\lvert Z(G)\rvert=p$, then $\lvert G/Z(G)\rvert = p^2/p=p$, which is cyclic by Corollary 4.6; by the Lemma above, $G$ is abelian, forcing $Z(G)=G$ — contradicting $\lvert Z(G)\rvert=p<p^2$. So this second case cannot occur, and $\lvert Z(G)\rvert=p^2$ is the only possibility: $G$ is abelian.

By strong induction on $\lvert G\rvert$. Write $\lvert G\rvert=p^nm$, $\gcd(p,m)=1$. If $n=0$, the trivial subgroup $\{e\}$ works.

Case 1: $p$ divides $\lvert Z(G)\rvert$. By Cauchy's Theorem (Theorem 8.6), $Z(G)$ has an element $z$ of order $p$. Let $N=\langle z\rangle$; since $N\subseteq Z(G)$, every $g\in G$ satisfies $gNg^{-1}=N$ (elements of $N$ commute with everything, so conjugation fixes them individually), so $N\trianglelefteq G$. Then $\lvert G/N\rvert = \lvert G\rvert/p = p^{n-1}m$, strictly smaller than $\lvert G\rvert$. By the induction hypothesis, $G/N$ has a subgroup $\bar Q$ of order $p^{n-1}$. Let $Q$ be the preimage of $\bar Q$ under the quotient map $\pi:G\to G/N$ (a subgroup of $G$, since preimages of subgroups under homomorphisms are subgroups — the same style of check as Theorem 6.2's proof). Since $N=\ker\pi \subseteq Q$, Lagrange applied within $Q$ gives $\lvert Q\rvert = \lvert N\rvert\cdot\lvert Q/N\rvert = p\cdot p^{n-1} = p^n$ (using $Q/N \cong \bar Q$ via $\pi$ restricted to $Q$). So $Q$ is a Sylow $p$-subgroup of $G$.

Case 2: $p$ does not divide $\lvert Z(G)\rvert$. By the class equation, $\lvert G\rvert = \lvert Z(G)\rvert + \sum_i[G:C_G(x_i)]$. If $p$ divided every term $[G:C_G(x_i)]$, it would divide the whole sum, and since $p\mid\lvert G\rvert$, rearranging would force $p\mid\lvert Z(G)\rvert$ — contradicting this case's assumption. So some noncentral $x_i$ has $p \nmid [G:C_G(x_i)]$. Since $\lvert G\rvert = [G:C_G(x_i)]\cdot\lvert C_G(x_i)\rvert$ with $p^n \mid \lvert G\rvert$, and since $p$ is prime and shares no factor with $[G:C_G(x_i)]$, the entire $p^n$ must sit in the other factor: $p^n$ divides $\lvert C_G(x_i)\rvert$. Since $x_i$ is noncentral, $C_G(x_i)\neq G$, so $\lvert C_G(x_i)\rvert < \lvert G\rvert$. By the induction hypothesis, $C_G(x_i)$ has a subgroup of order $p^n$ — a Sylow $p$-subgroup of $C_G(x_i)$, and therefore also a subgroup of $G$ of order $p^n$: a Sylow $p$-subgroup of $G$.

Since $P \le N_G(P) \le G$, indices multiply along the chain (Chapter 04, Section 05): $[G:P] = [G:N_G(P)]\cdot[N_G(P):P]$. The left side is $[G:P] = \lvert G\rvert/\lvert P\rvert = p^nm/p^n = m$. So $[G:N_G(P)]$ divides $m$. By the Example above, $[G:N_G(P)]$ is exactly the number of conjugates of $P$.

Since $Q\le N_G(R)$ and $R\trianglelefteq N_G(R)$ (normalizers always make their subgroup normal), the Second Isomorphism Theorem (Theorem 6.6, applied inside the ambient group $N_G(R)$) gives:

So $\lvert QR\rvert = \lvert R\rvert\cdot\dfrac{\lvert Q\rvert}{\lvert Q\cap R\rvert}$. Both $\lvert Q\rvert$ and $\lvert R\rvert$ are powers of $p$ (by hypothesis), so $\lvert QR\rvert$ is a power of $p$ as well. But $QR\le G$ (Theorem 6.6), so $\lvert QR\rvert$ divides $\lvert G\rvert$ by Lagrange, and $\lvert R\rvert=p^n$ is already the largest power of $p$ dividing $\lvert G\rvert$. Since $\lvert QR\rvert \geq \lvert R\rvert$ (as $R\subseteq QR$) and $\lvert QR\rvert$ is a $p$-power dividing $\lvert G\rvert$, we must have $\lvert QR\rvert = \lvert R\rvert = p^n$. Then $\lvert Q\rvert/\lvert Q\cap R\rvert = \lvert QR\rvert/\lvert R\rvert = 1$, so $Q\cap R = Q$, meaning $Q\subseteq R$.

Fix a Sylow $p$-subgroup $P$ (which exists by Theorem 9.3), and let $S=\{gPg^{-1}:g\in G\}$, the set of conjugates of $P$. By the Fact in Section 03, $\lvert S\rvert = [G:N_G(P)]$ divides $m$, so $p\nmid\lvert S\rvert$.

Let $Q$ be any Sylow $p$-subgroup of $G$ (not assumed related to $P$ yet). Let $Q$ act on $S$ by conjugation: $q\cdot(gPg^{-1}) = q(gPg^{-1})q^{-1}$ — this lands back in $S$, since $q(gPg^{-1})q^{-1} = (qg)P(qg)^{-1}$, still a conjugate of $P$ by the element $qg\in G$. By the Orbit-Stabilizer Theorem, every orbit of this $Q$-action has size dividing $\lvert Q\rvert=p^n$, hence every orbit size is a power of $p$ (possibly $p^0=1$).

Since the orbits partition $S$ and $\lvert S\rvert$ is not divisible by $p$, not every orbit can have size divisible by $p$ (else their sum, $\lvert S\rvert$, would be too) — so at least one orbit has size exactly $1$. That means some $R=gPg^{-1}\in S$ satisfies $q\cdot R = R$ for every $q\in Q$, i.e. $qRq^{-1}=R$ for all $q\in Q$, i.e. $Q\le N_G(R)$.

Since $R$ is a conjugate of the Sylow $p$-subgroup $P$, $\lvert R\rvert=\lvert P\rvert=p^n$: $R$ is itself a Sylow $p$-subgroup. Now Lemma 9.4 applies directly ($Q$ is a $p$-subgroup with $Q\le N_G(R)$, $R$ Sylow): $Q\le R$. Since $\lvert Q\rvert=\lvert R\rvert=p^n$ (both Sylow), $Q=R=gPg^{-1}\in S$.

Since $Q$ was an arbitrary Sylow $p$-subgroup, every Sylow $p$-subgroup equals some conjugate of $P$: $\mathrm{Syl}_p(G) = S$, and since conjugacy is transitive (composing conjugating elements), all Sylow $p$-subgroups are conjugate to one another.

(b) By Theorem 9.5, $\mathrm{Syl}_p(G)$ is exactly the set of conjugates of a fixed Sylow $p$-subgroup $P$, so $n_p = \lvert\mathrm{Syl}_p(G)\rvert$ equals the count from the Fact in Section 03, which divides $m$.

(a) Fix a Sylow $p$-subgroup $P$, and let $P$ act on $\mathrm{Syl}_p(G)$ by conjugation. By Orbit-Stabilizer, every orbit size divides $\lvert P\rvert=p^n$, so every orbit size is a power of $p$.

$P$ itself is a fixed point of this action: for $h\in P$, $hPh^{-1}=P$ (conjugating $P$ by its own elements just permutes $P$, landing back on $P$ as a set — this is exactly closure inside $P$). So $\{P\}$ is an orbit of size 1.

Claim: it is the only size-1 orbit. Suppose $R\in\mathrm{Syl}_p(G)$ is also fixed by every $h\in P$, i.e. $hRh^{-1}=R$ for all $h\in P$, i.e. $P\le N_G(R)$. By Lemma 9.4 (with $Q=P$), $P\le R$; since $\lvert P\rvert=\lvert R\rvert=p^n$, $P=R$.

So $\mathrm{Syl}_p(G)$ splits into the single fixed orbit $\{P\}$ and other orbits, each of size a positive power of $p$ (hence divisible by $p$, since a size-1 orbit besides $\{P\}$ is now ruled out). Counting elements: