1
$\begingroup$

How many homomorphism are there from the cyclic group $\mathbb{Z}_6$ to $\mathbb{Z}_2\times \mathbb{Z_4}$

Consider a homomorphism $\phi: \mathbb{Z}_6 \rightarrow \mathbb{Z}_2 \times \mathbb{Z}_4$

Since all homomorphisms maps the identity of the first group to that of the second,

so there are 8 homomorphisms are there i am right ...some one help me please

$\endgroup$
2
$\begingroup$

A homomorphism of a cyclic group is uniquely determined by where it sends the generator, i.e. which element you choose to be $\phi(1)$. Also, remember that whatever $\phi(1)$ you choose, you must have $$0=\phi(0)\\=\phi(1+1+1+1+1+1)\\=\phi(1)+\phi(1)+\phi(1)+\phi(1)+\phi(1)+\phi(1)$$Not every element in $\Bbb Z_2\times\Bbb Z_4$ fulfills this, but each one of those is a valid candidate. Count them and you have your answer.

$\endgroup$
  • $\begingroup$ @Arthur...according your idea in homomorphism a generator of elements send to generator $\endgroup$ – Inverse Problem Oct 30 '17 at 6:33
  • $\begingroup$ can you give some more clarification .....@Arthur $\endgroup$ – user293581 Oct 30 '17 at 6:39
  • 1
    $\begingroup$ @rajendra For instance, $\phi(1)=(0,2)$ is valid, because $6\cdot (0,2)=(0,0)$. So we count that one. On the other hand, $\phi(1)=(1,1)$ is not valid, because $6\cdot (1,1)=(0,2)\neq (0,0)$. There are $8$ elements in $\Bbb Z_2\times \Bbb Z_4$. Some of them are valid, some are not. Check each one, and count the valid ones. It should take less than a minute. $\endgroup$ – Arthur Oct 30 '17 at 6:47
  • $\begingroup$ Elements of $\mathbb Z_2 \times \mathbb Z_4$ have order 1, 2, or 4, while $\phi(1)$ must have order dividing 6. So exactly those elements of $\mathbb Z_2 \times \mathbb Z_4$ with order 1 or 2 determine a homomorphism. Thus there is the one trivial homomorphism, and 3 non-trivial homomorphisms, with image isomorphic to $\mathbb Z_2$. $\endgroup$ – fredgoodman Oct 30 '17 at 19:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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