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We have a function $\varphi:G\rightarrow H$ is an isomorphism, show its inverse $\varphi^{-1}:H\rightarrow G$ is also an isomorphism

I am fine with showing it to be a homomorphism and surjective, this is my attempt at showing injectivity, I'm unsure if they would be sufficient

Assume: $\forall h\in H, \exists g\in G$ s.t. $\varphi(g)=h$ and $\forall g_1, g_2\in G, \varphi(g_1)=\varphi(g_2)\Rightarrow g_1=g_2$

Injectivity - Need to show $\varphi^{-1}(g_1)=\varphi^{-1}(g_2)\Rightarrow g_1=g_2$

$\varphi^{-1}(g_1)=\varphi^{-1}(g_2)\Rightarrow g_1=\varphi(\varphi^{-1}(g_2))$ from our surjectivity assumption on $\varphi$ we know there exists such a $\varphi^{-1}(g_2)\in H$ and from our injectivity assumption on $\varphi$ we know that this solution is unique.

[Sufficient? What more could be added? How could I do this clearer?]

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Your question has nothing to do with group theory; the fact that inverse functions are necessarily bijective is a matter of set theory. And if you know, as a suppose, that in order to have an inverse function, a function $f$ must be bijective, it is pretty obvious that the inverse $f^{-1}$ will always be bijective. After all the requirement for an inverse is that both $f^{-1}\circ f$ and $f\circ f^{-1}$ are identity functions of their respective domains; then from the symmetry of this requirement it is immediate that $f$ is also the inverse of $f^{-1}$. But then $f^{-1}$ always has an inverse function ($f)$, so it must be bijective.

One can of course deduce from this a more detailed argument for why $f^{-1}$ must be injective. Recall that the have an inverse $f$ must be injective, since if $f(x)=f(y)$ then applying the supposedly existing inverse $f^{-1}$ to both sides one finds $f^{-1}(f(x))=f^{-1}(f(y))$ which becomes $x=y$. Now switching roles of $f$ and $f^{-1}$, if one has $f^{-1}(x)=f^{-1}(y)$ then applying $f$ to both sides gives $x=y$.

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A simple proof can be given easily. Just note that $\varphi\circ \varphi^{-1}=\text{id}$ implies $\varphi^{-1}$ is injective since $\text{id}$ is injective and again $\varphi^{-1}\circ \varphi=\text{id}$ implies that $\varphi^{-1}$ is surjective since $\text{id}$ is surjective . Hence $\varphi^{-1}$ is bijective.

I have used the following two theorems,

  1. $f\circ g$ injective implies $g$ injective.

  2. $f\circ g$ surjective implies $f$ surjective

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  • $\begingroup$ Good use of both theorems :) $\endgroup$ – GniruT Nov 27 '15 at 13:41
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I think there might be some issues in your notation (you mixed up some variables). We're talking about $\varphi^{-1}:H \rightarrow G$. For injectivity of $\varphi^{-1}$, we want to show that for all $h_1, h_2 \in H$, we have $\varphi^{-1}(h_1)=\varphi^{-1}(h_2) \Rightarrow h_1 = h_2$.

So $$\color{blue}{\varphi^{-1}(h_1)=\varphi^{-1}(h_2)} \Rightarrow \varphi(\varphi^{-1}(h_1))=\varphi(\varphi^{-1}(h_2)) \Rightarrow \color{blue}{h_1 = h_2}$$

thus

$$\varphi^{-1}(h_1)=\varphi^{-1}(h_2) \Rightarrow h_1 = h_2.$$

And that's it.

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I think just applying the map $\phi$ to both sides of the equation $\phi^{-1}(g_1) = \phi^{-1}(g_2)$ to get $g_1 = g_2$ is fine. You have probably defined $\phi^{-1}(g_1)$ as "the element that $\phi$ maps to $g_1$" so $\phi(\phi^{-1}(g_1)) = g_1$ is just true by definition.

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