My question reads:

Let G be a group and let Aut G be the set of all automorphisms of G. Prove that Aut G is a group under the operation of composition for functions.

I am not too sure if I am proving this Aut G itself is a group. Would this mean then that I would have to make sure it fulfills all the group requirements? If so I am not too sure how to go about it.

Proving closure:

I would think I need to pick say f,g in Aut G then use the fact that these are automorphisms. From here, I am not too sure how to proceed.

  • $\begingroup$ You need to check that Aut $G$ satisfies the properties of a group. Do you know what the definition of a group is? $\endgroup$ – Joshua Ruiter Feb 10 '17 at 20:25
  • $\begingroup$ @JoshuaRuiter yes, I need closure, associativity, identity and inverse. I guess my main question is if you have an automorphism is it already a homomorphism? I am just learning this so I am confused $\endgroup$ – Sam Feb 10 '17 at 20:30
  • $\begingroup$ The definition of automorphism is an isomorphism from the group to itself. So by definition, it is a homomorphism, and it is invertible. That gives you a big hint as to what should be the inverse of automorphism. Hopefully you also know that the composition of homomorphisms is a homomorphism. $\endgroup$ – Joshua Ruiter Feb 10 '17 at 20:32
  • $\begingroup$ @JoshuaRuiter Hmm okay makes more sense. And no I did not know that last fact but that would be helpful for showing closure. Then is the idea I have for closure okay? I would pick say f and g in AutG which are automorphims and also homomorphism and since we can compose I can show closure? $\endgroup$ – Sam Feb 10 '17 at 20:41
  • $\begingroup$ Then the inverse of an automorphism is itself? $\endgroup$ – Sam Feb 10 '17 at 20:42

Let $G$ be a group. By definition $\text{Aut}(G)=\left\{f:G\rightarrow G\mid f \mbox{ is an isomorphism of }G\right\}$. Given two automorphisms $f,g\in \text{Aut}(G)$, we can consider the composition $g\circ f$. We claim that $\text{Aut}(G),\circ$ is a group. We have to check all axioms.

First of all we need to show that $g\circ f$ is again an automorphism, i.e. a homomorphism that is bijective. Now since $g$ and $f$ are bijective, $g\circ f$ is bijective. Moreover, $$(g\circ f)(ab)=g(f(ab))=g(f(a)f(b))=g(f(a))g(f(b))=(g\circ f)(a)(g\circ f)(b)$$ for all $a,b\in G$. Hence $g\circ f$ is a group homomorphism.

Secondly we need to show that $\circ$ is associative, i.e. $(h\circ g)\circ f=h\circ (g\circ f)$. Just evaluate both morphisms at $a\in G$ and see that both expressions coincide due to the associativity of $G$.

Thirdly we need to check that there is a neutral element for $\circ$. Clearly $Id_G:G\rightarrow G:a\mapsto a$ is an automorphism. Since $f\circ Id_G=Id_G\circ f$ for all $f\in \text{Aut}(G)$, $Id_G$ is the neutral element.

Last but not least, we have to check that each $f\in \text{Aut}(G)$ has an inverse for $\circ$. Consider the inverse function $f^{-1}$. Clearly $f^{-1}\circ f=Id_G=f\circ f^{-1}$. So it remains to show that $f^{-1}$ is a group morphism. Now it's a very good exercise to prove this last statement.

EDIT: Let's prove the last statement. Suppose that $f:G\rightarrow G$ is a group isomorphism. We need to show that $f^{-1}$ is a group morphism. Let $a,b\in G$. By definition there exist a unique $x,y\in G$ such that $f(x)=a$ and $f(y)=b$. Hence $$f^{-1}(ab)=f^{-1}(f(x)f(y))=f^{-1}(f(xy))=xy.$$ Similarly $$f^{-1}(a)f^{-1}(b)=f^{-1}(f(x))f^{-1}(f(y))=xy.$$ Hence $f^{-1}(ab)=f^{-1}(a)f^{-1}(b).$

  • $\begingroup$ how do you know that g and f are a bijection? In the first part you are showing closure? $\endgroup$ – Sam Feb 11 '17 at 0:02
  • $\begingroup$ what do you mean by group morphism in the last section? $\endgroup$ – Sam Feb 11 '17 at 0:29
  • $\begingroup$ $g$ and $f$ are elements of $\text{Aut}(G)$. Thus by definition they are group isomorphisms, i.e. they are group morphisms and invertible functions by definition. $\endgroup$ – Mathematician 42 Feb 11 '17 at 11:06
  • $\begingroup$ Okay I think I get it more now. Basically we have to show that under each condition we still get a homomorphism? $\endgroup$ – Sam Feb 11 '17 at 14:27
  • $\begingroup$ What do you mean with 'under each condition'? $\endgroup$ – Mathematician 42 Feb 12 '17 at 13:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.