Prove that the inverse map of a group isomorphism is a group homomorphism

A group isomorphism $f : G \rightarrow K$, where $G, K$ are groups, means that $ f(g_1)f(g_2) = f(g_1g_2) \forall g_1,g_2 \in G $.

Since $f$ is bijective, the inverse map $f^{-1}$ is well defined.

There exists $k\in K : f(g)=k$ and thus $f^{-1}(k) = g$.

How do I go from here? I got stuck in a loop stating that $f^-1(f(g_1))f^-1(f(g_2)) = f^-1(f(g_1g_2)) = k_1k_2$

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    $\begingroup$ Let $g_1=f^{-1}(k_1)$ and $g_2=f^{-1}(k_2)$. We want to show that $f^{-1}(k_1k_2)=g_1g_2$. This follows immediately from $f(g_1g_2)=k_1k_2$. $\endgroup$ – André Nicolas Mar 20 '15 at 15:36

We want to show that $f^{-1}(k_1k_2) = f^{-1}(k_1)f^{-1}(k_2)$.

By bijectivity, we have there are unique $g_1,g_2 \in G$ with:

$g_1 = f^{-1}(k_1), g_2 = f^{-1}(k_2)$.

It is immediate that $f(g_1) = k_1,f(g_2) = k_2$.

Since $f$ is a homomorphism, $f(g_1g_2) = f(g_1)f(g_2) = k_1k_2$.

Therefore, by bijectivity, $f^{-1}(k_1k_2) = g_1g_2 = f^{-1}(k_1)f^{-1}(k_2)$


The secret is to work in $K$.

$f(f^{-1}(ab))=ab$, and $f(f^{-1}(a)f^{-1}(b))=f(f^{-1}(a))f(f^{-1}(b))=ab$

Hence $f(x)=f(yz)$, so since $f$ is one-to-one, $x=yz$.


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