I held back for a while before deciding to pose this question. I've done this before but couldn't quite 're-reasoned'.

Question: If G is a group, prove that Aut(G) is a group.

Let $\alpha, \beta \in Aut\left ( G \right )$. By definition:

the group operation is preserved.

$\alpha \left ( x_{1}x_{2} \right )=\alpha \left ( x_{1} \right )\alpha \left ( x_{2} \right )$

$\beta \left ( x_{1}x_{2} \right )=\beta \left ( x_{1} \right )\beta \left ( x_{2} \right )$

$\alpha, \beta$ is a bijection from G to G.

By the two-step subgroup test, it suffices to show that $\alpha \beta \in Aut\left ( G \right )$ whenever, $\alpha, \beta \in Aut\left ( G \right )$


$\alpha^{-1} \in Aut\left ( G \right )$ whenever $\alpha$ is.

A bit of help to get me going?

  • $\begingroup$ Nitpick: You're being asked to show that $\operatorname{Aut}(G)$ is a group, not that it is a subgroup (of which group would that be?) -- so applying only a subgroup test would seem to be a bit fast. $\endgroup$ Mar 13, 2017 at 13:50
  • $\begingroup$ @HenningMakholm That would be a subgroup of all bijections $G\to G$. $\endgroup$
    – freakish
    Mar 13, 2017 at 13:51
  • $\begingroup$ I agree. I have tried to establish Aut of G is a group via the group axiom but the closure property is giving me some issues. $\endgroup$ Mar 13, 2017 at 13:51
  • 1
    $\begingroup$ @Mathematicing It is necessary to do it that way, though! Note that the composition of two homomorphisms is a homomorphism, and the composition of two bijections is a bijection. $\endgroup$ Mar 13, 2017 at 13:52
  • $\begingroup$ Thank you for the reminder. It suffices. $\endgroup$ Mar 13, 2017 at 13:53

1 Answer 1


First of all note that the multiplication in $Aut(G)$ is given by function composition. And maybe lets also recall what $Aut(G)$ is:

$$Aut(G)=\{f:G\to G\ |\ f\mbox{ is an invertible homomorphism}\}$$

So let $\alpha\in Aut(G)$. We will show that $\alpha^{-1}\in Aut(G)$. So obviously $\alpha^{-1}$ is invertible with $\alpha$ as its inverse so it is enough to show that $\alpha^{-1}$ is a group homomorphism. Pick $a,b\in G$ and put $x=\alpha^{-1}(a)$ and $y=\alpha^{-1}(b)$. Then since $\alpha$ is a homomorphism we get


Now act with $\alpha^{-1}$ on both sides to get

$$xy = \alpha^{-1}(ab)$$

and by definition of $x,y$:


which shows that $\alpha^{-1}\in Aut(G)$.

Now let $\alpha, \beta\in Aut(G)$. We will show that $\alpha\circ\beta\in Aut(G)$. First of all it is a homomorphism. Indeed for any $a, b\in G$ we have

$$(\alpha\circ\beta)(ab)=\alpha(\beta(ab))=\alpha(\beta(a)\beta(b))=\alpha(\beta(a))\alpha(\beta(b))=(\alpha\circ\beta)(a)\ (\alpha\circ\beta)(b)$$

Secondly we need to show that $\alpha\circ\beta$ is invertible. But you can easily verify that $$(\alpha\circ\beta)^{-1}=\beta^{-1}\circ\alpha^{-1}$$

Also if you start by proving that

$$inv(G)=\{f:G\to G\ |\ f\mbox{ is invertible}\}$$

is a group under the function composition then these two properties are enough to show that $Aut(G)$ is a subgroup (hence a group) of $inv(G)$.


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