Sylow's third theorem: numerical constraints

The Sylow count divides the prime-to-$p$ part and is $1$ modulo $p$.Suggest a slogan

Let $G$ be a finite group and $p$ a prime, write $\abs G=p^a m$ with $p\nmid m$, and let $n_p$ be the number of Sylow $p$-subgroups of $G$. Then \[ n_p\mid m, \qquad n_p\equiv1\pmod p, \qquad n_p=[G:N_G(P)] \] for every Sylow $p$-subgroup $P$, where $N_G(P)$ is the normalizer of $P$.

Let $G$ act by conjugation on the set $\Syl_p(G)$ of Sylow $p$-subgroups. By Sylow's second theorem the action is transitive, and the stabilizer of $P$ is $N_G(P)$, so the orbit-stabilizer theorem gives $n_p=[G:N_G(P)]$. Since $P\leq N_G(P)$, $n_p$ divides $[G:P]=m$.

Now restrict the action to $P$. If $Q\in\Syl_p(G)$ is fixed by $P$, then $P\leq N_G(Q)$, so $P$ and $Q$ are Sylow $p$-subgroups of $N_G(Q)$ in which $Q$ is normal; by Sylow’s second theorem in $N_G(Q)$ they are conjugate there, so $P=Q$. Thus $P$ is the only fixed point, and every other $P$-orbit has size a positive power of $p$, so $n_p\equiv1\pmod p$.

Guide Appearances

Sylow theory

Dependencies

T-QYDVHOrbit-stabilizer theoremuses
T-EF2MZSylow's second theorem: Sylow subgroups are conjugateuses

Backlinks

ST-DKYXZCount Sylow subgroups before classifying a finite groupuses