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 $f^{-1}(f(g_1))f^{-1}(f(g_2)) = f^{-1}(f(g_1g_2)) = k_1k_2$
Best Answer
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, $f^{-1}(k_1k_2) = g_1g_2 = f^{-1}(k_1)f^{-1}(k_2)$