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I read this question about central automorphisms. The OP states that he could easily proof that the set of central automorphisms forms a subgroup of the automorphism group, however, I am not able to do this.

A central automorphism is a automorphism $\omega$ such that for every $g \in G$ the element $g^{-1}\omega\left(g\right) \in Z(G)$, where $Z(G)$ is the center of the group $G$.

First of all, the identity map is a central automorphism, so the set of central automorphisms is not empty.

Let $\omega, \theta$ be central automorphisms, then I am trying to show that $\omega \circ \theta^{-1}$ is such an automorphism itself (the subgroup criterion then assures that this set is a subgroup of the automorphis group). Let $x,g \in G$ be two arbitrary elements, then I need to show that $$g^{-1} \omega(\theta^{-1}(g))x = xg^{-1}\omega(\theta^{-1}(g))$$ However, I do not see how to use the information that $\omega, \theta$ are central homomorphisms...

Any hints on how to finish/remarks on my approach are highly appreciated.

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    $\begingroup$ the inverses make it hard to read. Let's just show that $\omega \theta$ is central. Pick $g\in G$. We know that $\theta(g)=gz_1$ for some $z_1\in Z(G)$. Then $\omega(\theta (g))=\omega (gz_1)=\omega (g)\omega (z_1)$. Now $\omega (g)=gz_2$ for $z_2\in Z(G)$. Thus $\omega(\theta (g))=gz_2\omega(z_1)$ and the claim is clear. Can you finish? $\endgroup$
    – lulu
    Sep 11, 2017 at 20:03
  • $\begingroup$ @Lulu: I agree about the inverses, looked better on the paper I was trying on. I think I could show that $\theta^{-1}$ is also a central automorphism: consider $g^{-1}\theta^{-1}(g)$, then $g = \theta(y)$ for some $y \in G$. Moreover, $y^{-1}\theta(y)$ is in the center, so $y^{-1}g$ is in the center. This implies that $g^{-1}y = g^{-1}\theta^{-1}(g)$ is in the center. Correct? $\endgroup$
    – Student
    Sep 11, 2017 at 20:14
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    $\begingroup$ That looks good! $\endgroup$
    – lulu
    Sep 11, 2017 at 20:19
  • $\begingroup$ @Student A little late to the game, but FYI I think it is actually possible to use one-step subgroup test to advantage here - left it as a new answer along with another "conceptual" proof. $\endgroup$
    – Ben
    Apr 12, 2022 at 8:20

2 Answers 2

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It's easier to show, separately, that $\omega\circ\theta$ and $\omega^{-1}$ are central automorphisms.

Since $\omega$ is a central automorphism, we have $$ (\omega\circ\theta)(g) =\omega(\theta(g))\in\theta(g)Z(G).$$ Since $\theta$ is a central automorphism, $\theta(g)\in gZ(G)$, so in fact $(\omega\circ\theta)(g)\in gZ(G)$. Thus, $\omega\circ\theta$ is a central automorphism.

To show that $\omega^{-1}$ is a central automorphism, let $g\in G$, so that $\omega(g)\in gZ(G)$. Then $g\in \omega^{-1}(gZ(G)) = \omega^{-1}(g)Z(G)$, as $Z(G)$ is characteristic. So, $g = \omega^{-1}(g)z$, for $z\in Z(G)$, and hence $\omega^{-1}(g) = gz^{-1}\in gZ(G)$. This shows that $\omega^{-1}$ is a central automorphism.

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    $\begingroup$ Thanks for the ellaborate answer. In my comment on the question, I think I was able to show that $\theta^{-1}$ is also a central automorphism, but I like your way better, since the notation is cleaner. (I also totally forgot about the center being a characteristic subgroup of $G$) Thank you very much! $\endgroup$
    – Student
    Sep 11, 2017 at 20:19
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    $\begingroup$ @Student Glad to help. I see you made some progress while I was typing, so that's great! $\endgroup$
    – James
    Sep 11, 2017 at 20:21
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First an abstract proof:

Surjective homomorphisms of groups stabilize the center, therefore we have a natural homomorphism

$$\text{Aut}(G) \to \text{Aut}(G/Z(G))$$

whose kernel is precisely the central automorphisms, as for $\phi \in \text{Aut}(G)$ then $[\phi(g)] = [g]$ in $G/Z(G)$ is equivalent to the difference $\phi(g)g^{-1}$ being in $Z(G)$. $\square$

Second, it is quite possible to use the one-step subgroup test as in the OP's attempt to your advantage here - just make a substitution $g=\theta(h)$:

$$ \begin{aligned} g^{-1}.(\omega \circ \theta^{-1})(g) &= g^{-1}.\omega(\theta^{-1}(g))\\ &= \theta(h)^{-1}.\omega(h)\\ &= \theta(h^{-1}).\omega(h)\\ &= h.\theta(h^{-1}).h^{-1}.\omega(h) \end{aligned} $$ $\square$

This proof is also nice because each line is an assumption - substitution is bijectivity, next step is homomorphism, last step is centrality.

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