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For a group $G$. The auto commutator subgroup of $G$ is $\langle \,g^{-1}\alpha(g)\mid \alpha\in \operatorname{Aut}(G)\,\rangle$. How to show it is characteristic in $G$

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    Since no element of the group is explicitly mentioned in the definition, working out the image of this under an arbitrary automorphism should result in exactly the same description. Why don't you try? – 2017-01-04
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    The definition of the auto commutator subgroup is incomplete. What is $g$? – 2017-01-04

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Let's call the auto commutator group $A$. It suffices to prove that for any $\varphi \in \operatorname{Aut}(G)$, $\varphi(A) \subseteq A$; or thus that $\varphi(x) \in A$ for an arbitrary $x \in A$. Such $x$ is always of the form $g^{-1}\alpha(g)$ for some $g \in G$, $\alpha \in \operatorname{Aut}(G)$, hence: $$\varphi(x) = \varphi(g^{-1}\alpha(g)) = \varphi(g)^{-1}\varphi(\alpha(g))$$ Now let $h = \varphi(g)$. Then $$\varphi(x) = \varphi(g)^{-1}\varphi(\alpha(g)) = h^{-1}\varphi(\alpha(\varphi^{-1}(h))) = h^{-1}(\varphi\alpha\varphi^{-1})(h)$$ Since $\operatorname{Aut}(G)$ is a group, $\varphi\alpha\varphi^{-1}$ is again an automorphism of $G$ and hence $\varphi(x) \in A$.

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    I think the definition of characteristic subgroup requires showing that $\varphi(A)=A$, but maybe you are using some auxiliary result? – 2017-01-04
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    @MarcvanLeeuwen The sentence beginning "It suffices to prove" is correct, because the inverse of an automorphism is an automorphism. – 2017-01-04