# Thread: homorphic images of Z_6

1. ## homorphic images of Z_6

I'm really not understanding how to figure this out. Thanks in advance for help.

*homomorphic, sorry about the spelling in the title

2. Originally Posted by tas10
I'm really not understanding how to figure this out. Thanks in advance for help.

*homomorphic, sorry about the spelling in the title
Yeah, we have no idea either. Mostly because you haven't actually asked a question.

3. Originally Posted by tas10
I'm really not understanding how to figure this out. Thanks in advance for help.

*homomorphic, sorry about the spelling in the title
As your group is abelian, every subgroup is normal. The first isomorphism theorem tells you that normal subgroups correspond precisely to homomorphic images. So, it is sufficient to find all subgroups of your group.

So, what are the subgroups of $\displaystyle \mathbb{Z}_6$?

Also, you may want to prove that the homomorphic image of a cyclic group is cyclic.

These two facts will allow you to classify all homomorphic images.

So, if $\displaystyle \mathbb{Z}_6$ has a subgroup of order n then it will have $\displaystyle \mathbb{Z}_{6/n}$ as a homomorphic image. Does that make sense?...