Sylow Theorems Question

Question

Let $p$ be the smallest prime that divides the order of a finite group $G$. If $H$ is a Sylow $p$-subgroup of $G$ and is cyclic. Prove that $N(H)=C(H)$.

Answer 1

You should have these two elements, among others, at your disposal:

1. The order of a cyclic group of order a power of a prime $p$.
2. The fact that if $a \in N_{G}(H)$ (here $H$ is an arbitrary subgroup of the group $G$), then $x \mapsto a^{-1} x a$ is an automorphism of $H$.

Answer 2

In general, if $H$ is a subgroup of $G$, then $N_G(H)/C_G(H)$ can be embedded in $Aut(H)$ (let $N_G(H)$ act on $H$ by conjugation, the kernel of the action is clearly $C_G(H)$). Now if $H$ is cyclic (or more general, abelian), then $H \subseteq C_G(H) \subseteq N_G(H) \subseteq G$. If in addition $H$ is a Sylow $p$-subgroup, say of order $p^a$, then $|Aut(H)|=p^{a-1}(p-1)$ and moreover $p$ cannot divide $[N_G(H):C_G(H)]$, since from the inclusions of subgroups above it follows that this last index divides $[G:H]$. But $[N_G(H):C_G(H)]$ must also divide $|Aut(H)|$, so we conclude it divides $(p-1)$. If $p$ is the smallest prime dividing $|G|$, this can only be the case if $[N_G(H):C_G(H)]=1$ and you are done.

Note that is this case the Sylow $p$-subgroup $H$ satisfies the condition of the famous Burnside Normal $p$-Complement Theorem: $H\subseteq Z(N_G(H))$ (observe that this is equivalent to $C_G(H)=N_G(H)$). One can conclude that $H$ has a normal complement, i.e. $G$ has a normal subgroup $N$ such that $G=HN$ and $H \cap N = 1$. The proof relies on transfer theory or put more general, on cohomology theory.