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Let $G$ and $H$ be groups. Then we have the set $X = \{f: G \to H : f $ is an isomorphism$\}$. I know that if $H = G$ this is a group under function composition, but I don't think this is true if $H\neq G$ (as function composition doesn't make sense then). I would like to see if there is a way to make it into a group under some other operation.

Here is my idea. Let $\phi: G\to H$ be an isomorphism. Then define the operation on $X$ by $$ (f\star g)(x) = f(\phi^{-1}(g(x))). $$ Question 1: I think this makes the set $X$ into a group, is that true? For example, the identity element if $\phi$.

Question 2: Is there any other obvious way to make $X$ into a group?

(I am just in basic algebra. I don't know about the isomorphism theorems, just the basic definitions of a group and definition of an isomorphism.)

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Yes, this is a group. It is isomorphic to the automorphism group of $G$.

This isomorphism is because $\phi^{-1}g\in\operatorname{Aut}(G)$ so $f\star g$ corresponds to the product $f\cdot \phi^{-1}g$ (a straight change of notation, and an isomorphism is simply a change or notation). For example, $f\star \phi f^{-1}=1$ and $f\star \phi=f$.

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  • $\begingroup$ Thanks. Could you elaborate on $f\star \phi f^{-1} = 1$? $\endgroup$ – John Doe Oct 21 '15 at 13:30
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    $\begingroup$ By $fg(x)$ I mean $f(g(x))$ (while some people - me included usually! - mean $g(f(x))$), so $(f\star \phi f^{-1})(x)=f(\phi^{-1}(\phi f^{-1}(x)))=x$ for all $x$. $\endgroup$ – user1729 Oct 21 '15 at 13:31
  • $\begingroup$ Great. That makes sense! $\endgroup$ – John Doe Oct 21 '15 at 13:33

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