I am trying to understand the proof of the following lemma:
Assume $P$ is a nonabelian group of order $p^3$ where $p$ is an odd prime. Assume also that $P$ has exponent $p^2$. Then $O_p(\text{Out}(P)) \in \text{Syl}_p(\text{Out}(P))$.
I can follow most of the steps in the proof:
- first you realize that $P$ contains a unique subgroup $Q<P$ where $Q \cong C_p\times C_p$. So $Q$ is characteristic in $P$. Also you can see that $\Phi(P) = [P, P] = Z(P) \le Q$ and $[P, P] \cong C_p$.
- then one considers the homomorphism $\varphi: \text{Aut}(P)\rightarrow \text{Aut}(P/Q)\times \text{Aut}(Q/[P, P]) \cong C_{p-1}\times C_{p-1}$ and then I have another lemma saying that $\ker{\varphi}$ is a normal $p$-subgroup of $\text{Aut}(P)$, so $\ker{\varphi} \subseteq O_p(\text{Aut}(P))$. You can also argue the converse, that $O_p(\text{Aut}(P)) \subseteq \ker{\varphi}$. And then the proof of the lemma stops.
What I don't understand is why $\ker{\varphi} = O_p(\text{Aut}(P))$ leads to the fact that $O_p(\text{Out}(P)) \in \text{Syl}_p(\text{Out}(P))$. Also, I am not sure how $\varphi$ is actually defined.
Since $Q$ and $[P,P]$ are characteristic in $P$, every $a \in \operatorname{Aut}(P)$ restricts to an automorphism of $P/Q$ and of $Q/[P,P]$ (and of $[P,P]/1$). Since these are all cyclic groups of order $p$, we get that these restrictions of $a$ all have order dividing $p-1$. In particular, none have order a power of $p$ unless they act as the identity on these groups.
Let $\phi(a)$ be the automorphism induces by $a$ on $P/Q$ and $Q/[P,P]$, so that $\phi(a) = (b,c)$ means $b:P/Q \to P/Q : xQ \mapsto a(x)Q$ and $c:Q/[P,P] \to Q/[P,P]:x[P,P] \mapsto a(x)[P,P]$.
Since $\phi(a)$ has order coprime to $p$, if $a$ was in a Sylow $p$-subgroup of $\operatorname{Aut}(P)$, then $\phi(a)=(1,1)$ so $a \in \ker(\phi)$. In other words, $O^{p'}(\operatorname{Aut}(P)) \leq \ker(\phi)$, where $O^{p'}(X)$ is the subgroup of $X$ generated by the Sylow $p$-subgroups of $X$.