for my master's thesis I am going through M.Hertweck and W.Kimmerle's article on Coleman automorphisms. I encountered the following reasoning, which I can't seem to follow.

We know the following on the group $G$.

Let $G$ be a group such that no chief factor of $G/F^*(G)$ is isomorphic to $C_p$ and $Z(F^*(G))$ is a $p'$-group.

They then start the proof with the following remark, which I can't seem to proof:

"Now if we have some $\sigma \in Aut(G)$ of $p$-power order such that $\sigma|_{F^{*}(G))} = conj(g)|_{F^{*}(G)}$ for some $g \in G$ then $\sigma = conj(xg)$ for some $x \in Z(F^*(G))$ since $C_G(F^*(G)) = Z(F^*(G))$, which is a $p'$-group."

I know that we can view $G/Z(F^*(G))$ as a subgroup of $Aut(F^*(G))$ and that $C_G(F^*(G)) = Z(F^*(G))$, but I don't see how $\sigma$ is uniquely defined by its behaviour on $F^*(G)$ in this particular case. Can someone provide a clue or proof?

PS: We also know that $\sigma$ is a Coleman automorphism (i.e. the restriction of $\sigma$ to a Sylow subgroup of $G$ is inner), but I think this is not necessary to proof the above...


It is a consequence of the fact that $F^*(G)$ is normal. As we have the following lemma:

Let $G$ be a finite group, $H$ a normal subgroup of $G$ and $\sigma \in Aut(G)$ of $p$-power order such that $C_G(H)$ is a $p'$-group and $C_G(H) \subseteq H$. If $\sigma|_H = id|_H$ then $\sigma = id$.

Proof: Let $g \in G$ and $h \in H$, then because $H$ is normal we find $$ \sigma(g^{-1}hg) = g^{-1}hg $$ But when we use the fact that $\sigma$ is a grouphomomorphism we see that $$ h = \sigma(g)g^{-1}hg\sigma(g)^{-1}$$ This tells us that $x_g^{-1}=g \sigma(g)^{-1} \in C_G(H)$ such that $\sigma(g)= x_g^{-1}$. Now one may observe that $$ g = \sigma^{o(\sigma)}(g) = x_g^{o(\sigma)}g $$ But then $x_g$ has $p$-power order, which is impossible unless $x_g = e$. Which means that $\sigma = id$.QED

This theorem can be immediately applied to the problem (after noticing that we may replace $g$ by some $g' = gx$ of $p$-power order, since $g^{o(\sigma)} \in C_G(F^*(G))$ and thus also the $p'$-part of $g$ is in $C_G(F^*(G))$, which if we denote $x$ for the inverse of the $p'$-part of $g$ gives us exactly what is needed.


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