The Sylow Theorems

This is standard from p-Groups and Their Properties; we include the short argument since everything else depends on it. Partition $X$ into $P$-orbits. By the Orbit-Stabilizer Theorem, each orbit has size $[P : \mathrm{Stab}_P(x)]$, which divides $|P| = p^a$ for some $a$, hence each orbit size is a power of $p$ (possibly $p^0=1$). An orbit has size $1$ exactly when its unique element is fixed by all of $P$, i.e. exactly the orbits contained in $X^P$. Summing orbit sizes,

and every term in the sum is divisible by $p$ (being $p^{a_i}$ with $a_i \ge 1$). Reducing mod $p$ gives $|X| \equiv |X^P| \pmod p$. $\blacksquare$

We now prove the three Sylow theorems in turn. Throughout, $|G| = p^k m$, $p \nmid m$.

Proof of Existence (Sylow I). We use induction on $|G|$ together with the class equation, a standard combination for $p$-group existence arguments. Base case: if $|G| = 1$ there is nothing to prove (take $k=0$, $P = \{e\}$); more substantively, if $k = 0$ (i.e. $p \nmid |G|$) the trivial subgroup already has order $p^0=1=p^k$. So assume $k \ge 1$. Case 1: $p \mid |Z(G)|$, where $Z(G)$ is the center of $G$. By Cauchy's Theorem applied to the abelian group $Z(G)$, there is an element $z \in Z(G)$ of order $p$. Let $N = \langle z \rangle$; since $z$ is central, $N \trianglelefteq G$. Consider $\bar G = G/N$, which has order $p^{k-1}m$. By induction (this is a smaller group, since $|N|=p>1$), $\bar G$ has a subgroup $\bar P$ of order $p^{k-1}$. Let $P$ be the preimage of $\bar P$ under the quotient map $G \to \bar G$. Then $P$ is a subgroup of $G$ with $|P| = |\bar P|\cdot|N| = p^{k-1}\cdot p = p^k$, as required. Case 2: $p \nmid |Z(G)|$. Apply the class equation:

where the sum runs over representatives $x_i$ of the non-central conjugacy classes, and $C_G(x_i)$ is the centralizer of $x_i$. Since $p \mid |G|$ but $p \nmid |Z(G)|$, there must exist at least one term $[G:C_G(x_i)]$ in the sum with $p \nmid [G:C_G(x_i)]$ — otherwise every term on the right side except possibly $|Z(G)|$ would be divisible by $p$, forcing $p \mid |Z(G)|$ (since $p$ divides the total $|G|$ and all other class-sizes), a contradiction. Fix such an $x_i$ and set $H = C_G(x_i)$. Since $[G:H]$ is not divisible by $p$ and $|G| = |H|\cdot[G:H] = p^k m$, and $p \nmid [G:H]$, we must have $p^k \mid |H|$, so $H$ is a proper subgroup (as $x_i \notin Z(G)$ means $H \ne G$) whose order is still divisible by $p^k$. By induction (applied to the smaller group $H$), $H$ has a subgroup of order $p^k$, and this subgroup is also a subgroup of $G$ of order $p^k$. In both cases we produce a subgroup of order $p^k$, completing the induction. $\blacksquare$

Proof of Conjugacy and Covering (Sylow II). Fix a Sylow $p$-subgroup $P$ (which exists by Sylow I) and let $H \le G$ be any $p$-subgroup. Let $X = G/P$ denote the set of left cosets of $P$ in $G$, so $|X| = [G:P] = m$, which is not divisible by $p$ by hypothesis. Let $H$ act on $X$ by left multiplication: $h \cdot (gP) = (hg)P$. By the orbit-counting lemma applied to the $p$-group $H$ acting on $X$,

Since $p \nmid |X| = m$, we conclude $|X^H| \not\equiv 0 \pmod p$; in particular $X^H \ne \varnothing$, so there exists a fixed coset $gP \in X^H$. This means

i.e. $g^{-1}hg \in P$ for all $h \in H$, i.e. $g^{-1}Hg \le P$, equivalently $H \le gPg^{-1}$. This proves the covering statement. For conjugacy of two Sylow $p$-subgroups $P, Q$: apply the above with $H = Q$ (a $p$-subgroup since $Q \in \mathrm{Syl}_p(G)$) to get $Q \le gPg^{-1}$ for some $g \in G$. But $|gPg^{-1}| = |P| = p^k = |Q|$, so $Q \le gPg^{-1}$ together with equal finite cardinalities forces $Q = gPg^{-1}$. Hence $P$ and $Q$ are conjugate. $\blacksquare$

Proof of Counting (Sylow III). Fix a Sylow $p$-subgroup $P$. By Sylow II, $G$ acts transitively by conjugation on $\mathrm{Syl}_p(G)$ (transitivity is exactly the conjugacy statement just proved), so by the Orbit-Stabilizer Theorem,

where $\mathrm{Stab}_G(P) = \{g \in G : gPg^{-1} = P\} = N_G(P)$, the normalizer of $P$. This proves $n_p = [G:N_G(P)]$. Since $P \le N_G(P) \le G$, we have

so $n_p \mid m$. This proves the divisibility claim. For the congruence $n_p \equiv 1 \pmod p$: let $P$ act on $\mathrm{Syl}_p(G)$ by conjugation (restricting the $G$-action to the subgroup $P$). By the orbit-counting lemma,

where $\mathrm{Syl}_p(G)^P$ is the set of Sylow $p$-subgroups fixed by conjugation by every element of $P$. We claim $\mathrm{Syl}_p(G)^P = \{P\}$. Certainly $P$ itself is fixed (conjugation by any $x \in P$ sends $P$ to $xPx^{-1}=P$). Suppose $Q \in \mathrm{Syl}_p(G)^P$, i.e. $xQx^{-1}=Q$ for all $x \in P$; this says $P \le N_G(Q)$. Now both $P$ and $Q$ are Sylow $p$-subgroups of the group $N_G(Q)$ (they have the correct order $p^k$ and lie inside $N_G(Q)$; note $Q \trianglelefteq N_G(Q)$ by definition of normalizer). Since $Q$ is normal in $N_G(Q)$, it is the unique Sylow $p$-subgroup of $N_G(Q)$: indeed by Sylow II applied inside $N_G(Q)$, $P$ and $Q$ are conjugate in $N_G(Q)$, but $Q$ is normal there, so conjugation fixes $Q$, forcing $P = Q$. Hence $\mathrm{Syl}_p(G)^P = \{P\}$, so $|\mathrm{Syl}_p(G)^P| = 1$, giving

This completes the proof of all three Sylow theorems. $\blacksquare$

If $n_p = 1$ then the unique Sylow $p$-subgroup $P$ satisfies $gPg^{-1} \in \mathrm{Syl}_p(G) = \{P\}$ for every $g\in G$ (since conjugates of Sylow subgroups are Sylow subgroups of the same order), so $gPg^{-1}=P$ for all $g$, i.e. $P \trianglelefteq G$. Conversely if $P \trianglelefteq G$, then for every $g \in G$, $gPg^{-1} = P$, so the conjugacy class of $P$ under $G$-action is $\{P\}$; but by Sylow II this conjugacy class is all of $\mathrm{Syl}_p(G)$, forcing $n_p=1$. $\blacksquare$