Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Can you think of any examples using specific groups? That might help... – Andrew Aug 18 '12 at 2:43
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

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 $4$.

share|cite|improve this answer

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?

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.