Prove that isomorphisms are symmetric

Question

Prove that isomorphisms are symmetric

Now if $(G, *)$ and $(H, \circ)$ are two groups. A map $\gamma : G \to H$ such that $\gamma (x * y) = \gamma(x) \circ \gamma(y)$ for all $x, y \in G$ is a homomorphism, and a bijective homomorphism is an isomorphism, and we notationally put $G \cong H$ if a isomorphism exists between $G$ and $H$.

I need to prove that $G \cong H \implies H \cong G$.

Now we know that $\gamma : G \to H$ is an isomorphism, and it seems like $\gamma^{-1} : H \to G$ should be an isomorphism as well. Proving that $\gamma^{-1}$ is bijective is trivial, however proving that it is an homomorphism doesn't seem to be as easy, since $\gamma^{-1}(\gamma(x * y)) = \gamma^{-1}(\gamma(x) \circ \gamma(y)) \implies (x * y) = \gamma^{-1}(\gamma(x) \circ \gamma(y))$ and since we have no idea how $\circ$ affects $\gamma^{-1}$ we can't deduce anything further. (please correct me if I'm wrong)

However I found a proof online, that I'm not so sure about

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In the proof above how do we know that $$s_1 \circ s_2 = \phi^{-1}(t_1) \circ \phi^{-1}(t_2) \ \ \ \ \ \ \ \ (1)$$ and how do we know that the $t_1$ and $t_2$ in $(1)$ above is the same as $t_1$ and $t_2$ in equality $(2)$ below? $$\phi^{-1}(t_1 * t_2) = s_1 \circ s_2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$

Answer 1

What we would like to show, based on what we know already, is that for any $h_1,h_2 \in H$, that:

$\gamma^{-1}(h_1 \circ h_2) = \gamma^{-1}(h_1) \ast \gamma^{-1}(h_2)$.

Now, since $\gamma$ is bijective, we know there exists unique $g_1,g_2 \in G$ with:

$\gamma(g_1) = h_1, \gamma(g_2) = h_2$.

So:

$\gamma^{-1}(h_1 \circ h_2) = \gamma^{-1}(\gamma(g_1) \circ \gamma(g_2))$

and since $\gamma$ is a homomorphism, $\gamma(g_1) \circ \gamma(g_2) = \gamma(g_1\ast g_2)$, thus:

$\gamma^{-1}(h_1 \circ h_2) = \gamma^{-1}(\gamma(g_1) \circ \gamma(g_2)) = \gamma^{-1}(\gamma(g_1\ast g_2)) = g_1 \ast g_2.$

Now, from $\gamma(g_1) = h_1, \gamma(g_2) = h_2$, applying $\gamma^{-1}$ to both these equations gives:

$g_1 = \gamma^{-1}(\gamma(g_1)) = \gamma^{-1}(h_1)$, and:

$g_2 = \gamma^{-1}(\gamma(g_2)) = \gamma^{-1}(h_2).$

So, finally, we have: $\gamma^{-1}(h_1 \circ h_2) = g_1 \ast g_2 = \gamma^{-1}(h_1) \ast \gamma^{-1}(h_2)$, which is what we sought to prove.

Answer 2

Those are good questions. I think you have to ask them in part because the proof you're quoting is too condensed for a beginning student of abstract algebra and should use more words. Here's what's going on in the last three lines.

For any $s \in S$, $$ \phi^{-1}(\phi(s)) = s . $$ That's just the meaning of "inverse function". We can call $\phi(s)$ by the name $t$ if we like. Then that equation becomes $$ \phi^{-1}(t) = s . $$

That explains (1) and (2) in your question: you're just using different names for $s_1$ and $s_2$.

I'd write the argument this way.

Given $t_1$ and $t_2$ in $T$ I need to prove $$ \phi^{-1}(t_1 \circ t_2) = \phi^{-1}(t_1) \circ \phi^{-1}(t_2) . $$

Since $\phi$ is a bijection, the two sides of that equation will be equal if they have equal values when I apply $\phi$. Then for the left member:

$$ \phi(\phi^{-1}(t_1 \circ t_2)) = t_1 \circ t_2 \quad (\text{meaning of } \phi^{-1}) $$ and on the right: $$ \begin{align} \phi(\phi^{-1}(t_1) \circ \phi^{-1}(t_2)) & = \phi(\phi^{-1}(t_1)) \circ \phi(\phi^{-1}(t_2))) \quad (\phi \text{ is a morphism})\\ & = t_1 \circ t_2 \quad (\text{meaning of } \phi^{-1}) \end{align}. $$ Done.

Answer 3

I can't see the image but I'll try to guess the proof enouncement. Let guess $\phi:H\rightarrow G$ is an isomorphism and let prove $\phi^{-1}$ is also. Its easy to see that $\phi^{-1}$ is bijective. As we know $\phi$ is bijective, given $t_1,t_2\in G$ take $s_i=\phi^{-1}(t_i)$, so $t_i=\phi(s_i)$, which results in $(1)$. We know that $\phi(s_1\circ s_2 )=t_1*t_2$ as $\phi$ is homomorphism, so taking $\phi^{-1}$ we get $(2)$. Joining $(1)$ and $(2)$ proves that $\phi^{-1}$ is homomorphism