Let $G$ and $H$ be groups and let $\phi: G \to H$ be a group homomorphism.
Suppose that $f:G\to H$ is bijective.
Then there exists a map $\psi:H\to G$ such that
\[\psi \circ \phi=\id_G \text{ and } \phi \circ \psi=\id_H.\] Then prove that $\psi:H \to G$ is also a group homomorphism.

Proof.

Let $a, b$ be arbitrary elements of the group $H$.
To prove $\psi: H \to G$ is a group homomorphism, we need
\[\psi(ab)=\psi(a)\psi(b).\]

We compute
\begin{align*}
&\phi\left(\, \psi(a)\psi(b) \,\right)\\
&=\phi\left(\, \psi(a) \,\right) \phi\left(\, \psi(b) \,\right) && \text{since $\phi$ is a group homomorphism}\\
&=ab && \text{since $\phi\circ \psi=\id_H$}\\
&=\phi \left(\, \psi (ab) \,\right) && \text{since $\phi\circ \psi=\id_H$}.\\
\end{align*}

Since $\phi$ is injective, it yields that
\[\psi(ab)=\psi(a)\psi(b),\] and thus $\psi:H\to G$ is a group homomorphism.

What’s an Isomorphism?

A bijective group homomorphism $\phi:G \to H$ is called isomorphism.

The above problem guarantees that the inverse map of an isomorphism is again a homomorphism, and hence isomorphism.

The post Inverse Map of a Bijective Homomorphism is a Group Homomorphism first appeared on Problems in Mathematics.