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Let $G$ be a finite group and $p$ a prime divisor for the order of $G$. Show that if $p^{2}$ divides the order of G, then $p$ divides $|Aut(G)|$ .

I know that $G$ has a subgroup of order $p$ and I think that it would be a good idea to consider the following homomorphism to prove that $Aut(G)$ has a subgroup of order $p$ : $$ \phi :G \to Aut(G)$$ $$\phi(g) = C_{g}$$ where $C_{g}$ is the conjugation given by $g$. Also I know that $Ker(\phi)=Z(G)$ then $G/Z(G) \cong Im(\phi)=Inn(G)$, but I can't figure out how to use the fact that $p^{2}$ divides the order of $G$.

Arnaud Mortier
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Lucía Osorio
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  • $|Aut(G)|$ means the order of $Aut(G)$. The group itself is just denoted by $Aut(G)$. – Arnaud Mortier Apr 09 '18 at 13:12
  • Yes, means the order of the automorphism group. – Lucía Osorio Apr 09 '18 at 13:13
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    If $G$ has a central Sylow $p$-subgroup $P$, then $P$ will be contained in the kernel of $\phi$. It looks like you will need something other than inner automorphisms in such a case. – Jyrki Lahtonen Apr 09 '18 at 13:22
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    In the case mentioned by Jyrki Lahtonen, check what automorphisms $P$ itself may have and how to obtain something for $G$. – Hagen von Eitzen Apr 09 '18 at 13:24
  • If $P\unlhd G$ then by Schur-Zassenhaus $G=P\rtimes H$ for another subgroup $H$. Because $P$ is central, this is actually a direct product, and you can describe automorphisms using the idea in Hagen's comment. But, Schur-Zassenhaus feels overkill. I'm not fully in my comfort zone here... – Jyrki Lahtonen Apr 09 '18 at 13:29
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    @JyrkiLahtonen I think the statement can be proven fairly easily for central subgroups. Any element can be uniquely written as a product of commuting elements where one is of $p$-power order and the other is of $p'$-order. That the Sylow $p$-subgroup is central should imply that sending an element to it's $p$-power part is a homomorphism to the Sylow $p$-subgroup, which then splits off. – Tobias Kildetoft Apr 09 '18 at 17:35

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