Question

give an example of a nontrivial homomorphism Ï• for the given groups,if exists.
Ï•:Z12->Z5

Answer 1

\[\phi:Z_{12}\rightarrow Z_{5}\]

Answer 2

I don't think there is a nontrivial homomorphism from \(\phi:\mathbb{Z}_{12}\rightarrow \mathbb{Z}_{5}\) because 5 and 12 are coprime, but I'm rusty so I'm re-reading about it, will be back in a few minutes to confirm.

Answer 3

That took way more searching than expected: "Suppose \(G\) and \(H\) are finite groups whose orders are relatively prime. Then, if \(\phi\) is a homomorphism, the order of \(\phi(G)\) divides the order of \(G\). Since \(\phi(G)\) is a subgroup of \(H\), the order of \(\phi(G)\) divides the order of \(H\). By our assumption of relatively prime order, we obtain that \(\phi(G)\) must be trivial, and thus, the homomorphism is the trivial homomorphism."

Answer 4

So you can see since \(\mathbb{Z}_{12}\) is order 12 and \(\mathbb{Z}_{5}\) is order 5, they have coprime orders, and therefore the only homomorphism between them is the trivial homomorphism.

Answer 5

in the answer given:
......the number of cosets of Ker(\[\phi\]) must
then be 5. But the number of cosets of a subgroup of a finite group is a divisor of the order of the
group, and 5 does not divide 12.

Answer 6

So Z12 -> Z4 would be OK, right? (Division by 4)

Answer 7

it's easy if just look at the order..but don't know how to prove it...

Answer 8

Well it's really just a corollary of Lagrange's Theorem. Do you need a rigorous formal proof?

Answer 9

can you explain this...."By our assumption of relatively prime order, we obtain that Ï•(G) must be trivial, and thus, the homomorphism is the trivial homomorphism."

Answer 10

Do you follow why the order of the image has to divide both the order of G and the order of H? That's the key part. Once you have that, it becomes clear that the homomorphism is trivial because, since G and H have coprime orders, the only order that divides them both is 1, and the group with order 1 is the identity group, making it a trivial homomorphism.

Answer 11

ok2...i get it...thanks...

Answer 12

And yes, @estudier, there are nontrivial homomorphisms from Z12 to Z4. Specifically, x → x mod 4 should work, and I'm pretty sure x → x mod 3 would also work.

Answer 13

Glad I could help @wiah, I'm studying group theory myself so it's nice to have some questions about it on here to play with.