Property of the group homomorphism and inverse homomorphism Suppose that there are two groups $G$ and $H$, and there is a group homomorphism function $f: G \rightarrow H$. If there is a group homomorphism function $z: H \rightarrow G$ that is not an inverse of $f$, can $G$ and $H$ be said to be in bijection?  
 A: Being in bijection has nothing to do with the group structure. Using trivial homomorphisms and some finite and infinite groups, it easy to see what are the possibilities. I list some explicit examples.
Consider $G = \mathbb{Z}$ and $H = \mathbb{Z} / 2\mathbb{Z}$. There is a group homomorphism $\Phi : G \rightarrow H$ and $\Psi : H \rightarrow G$ defined as the trivial homomorphism sending everything to the identity. Clearly $G$ and $H$ are not in bijection. 
Similarly $G = \mathbb{Z} / 4 \mathbb{Z}$ and $H = \mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 2\mathbb{Z}$. Again letting $\Phi$ and $\Psi$ be trivial group homomorphism. In this case, they two are in bijection since they are finite group of order $4$. 
A: Let $G$ be $Z_6$, and let $H$ be $Z_2$.
Consider the homomorphism $f:G\to H$ which takes 0,2, and 4 in $Z_6$ to 0 in $Z_2$, and 1,3, and 5 in $Z_6$ to 1 in $Z_2$.
Let $z:H\to G$ be the homomorphism which takes every element in $Z_2$ to 0 in $Z_6$.  
$z$ is not an inverse of $f$.
Can $G$ and $H$ be said to be in bijection?
A: Let $G$ and $K$ be finite groups.
Let $f:G \rightarrow G\times K$ be the canonical injection.
Let $g:G\times K \rightarrow G$ be the projection.
If $K \neq 1$, $G$ and $G\times K$ are not isomorphic.
