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


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.

  • $\begingroup$ @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

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