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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?

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Can you think of any examples using specific groups? That might help... –  Andrew Aug 18 '12 at 2:43
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There is always a group homomorphism (no need to add the word function) $z: H \rightarrow G$ sending everything to the identity of $G$, and it is never the inverse of $f$ (except in one case that you should be able to guess). This should make you cautious of the kind of statement you ask about. –  Marc van Leeuwen Aug 18 '12 at 4:15
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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 $4$.

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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?

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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.

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