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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.

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  • $\begingroup$ 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
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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 $f^{-1}(ab)=f^{-1}(a)f^{-1}(b).$

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  • $\begingroup$ 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

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