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$\DeclareMathOperator{\Char}{char}$$\DeclareMathOperator{\Aut}{Aut}$ Let $G$ be a group, $H$ a subgroup, such that $H \Char G$. If $\phi \in \Aut(H)$ is there an automorphism $\widehat{\phi} \in \Aut(G)$ such that $\widehat{\phi}|_H$ = $\phi$?

In words, if $\phi$ is an automorphism of $H$, is there necessarily an automorphism $\widehat{\phi}$ of $G$ where $\widehat{\phi}$ maps the elements in $H$ to $H$ in exactly the same way?

I am thinking this is false (consider the non-finite example of $\bar{Q}$ and $\mathbb{C}$), but would like a (several) counter example(s), preferably for the finite case, but I wouldn't mind another infinite case.

The reason why I am not satisfied with $\bar{Q}$ and $\mathbb{C}$ is that, according to some other answers on MO/Math.stack, the automorphisms of $\mathbb{C}$ depends on whether we are willing to use AC or not (as far as I understood those posts), so to me they are not really "concrete" counter-examples, (unless someone can convince me otherwise).

Soyuz
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1 Answers1

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Let $G=\mathbb Z_2\times \mathbb Z_4$ and let $H$ be the subgroup of elements of exponent $2$. $H$ is a Klein group, so the automorphism group of $H$ acts transitively on non-identity elements of $H$. But the automorphisms of $G$ do not act transitively on the non-identity elements of $H$, since the image of the doubling map $G\to G\colon x\mapsto 2x$ is a proper subgroup of $H$ that is characteristic in $G$. It follows that not every automorphism of $H$ is the restriction of an automorphism of $G$.

Keith Kearnes
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  • This is actually also an example that $f \in \operatorname{End}(H)$ can't be extended to $g \in \operatorname{End}(G)$. – Zhang Yuhan Nov 06 '24 at 08:07