
Homomorphisms
How many homomorphisms are there from Z30 onto Z10? How many are there to Z10? Give a formula for any homomorphism you find in the form f(x)= "expression" mod "some number." Be sure to verify that your homomorphism is well defined.
I'm not really sure how to do this at all.

Re: Homomorphisms
Hi,
You need to know that both and are cyclic groups. Denote by 1 the element cl(1) in . Then 1 is a generator of Z_{30}; any element is of the form for some integer m. So for any homomorphism f from to any group, f is completely determined by f(1)  . So for a homomorphism f into , there are 10 choices for f(1). I leave it to you to show for any is well defined. However, for f to be onto, f(1) must be a generator of . So just ask yourself: how many generators of are there?

Re: Homomorphisms
Thank you! Ok as for generators of Z10.....please don't laugh or hate me if I get this wrong but I THINK there are four generators: 1,3,7,9....?

Re: Homomorphisms
Hi again,
Exactly right. For a cyclic group G of order n with G = <x> (multiplicative notation), x^{k} is a generator if and only if k is relatively prime to n.

Re: Homomorphisms
So if those are my four onto homomorphisms, how would I write them in the f(x)...form?

Re: Homomorphisms
Hi again,
Write and
For the homomorphism corresponding to 5, f(x)=5x where the x on the left is an element of \mathbb{Z}_30 and the 5x on the right is the element of \mathbb{Z}_10 computed by "5x mod 10". Example x=17, f(17)=5*16 mod 10=0.
Maybe the following will help (too many integers in the above):
Let be a multiplicative cyclic group of order 30 with generator a. Then is the (unique) subgroup of G of order 10. The 4 homomorphism of G onto H are:
for all
for all
for all
for all
If k is any integer,
for all
is a homomorphism of G into H. This is onto H only for

Re: Homomorphisms
thanks so much!! You have been a big help!