
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 $\displaystyle \mathbb{Z}_{30}$ and $\displaystyle \mathbb{Z}_{10}$ are cyclic groups. Denote by 1 the element cl(1) in $\displaystyle \mathbb{Z}_{30}$. Then 1 is a generator of Z_{30}; any element is of the form $\displaystyle m\cdot1$ for some integer m. So for any homomorphism f from $\displaystyle \mathbb{Z}_{30}$ to any group, f is completely determined by f(1)  $\displaystyle f(m\cdot1)=m\cdot f(1)$. So for a homomorphism f into $\displaystyle \mathbb{Z}_{10}$, there are 10 choices for f(1). I leave it to you to show for any $\displaystyle x\in\mathbb{Z}_{10}, f(m\cdot1)=mx$ is well defined. However, for f to be onto, f(1) must be a generator of $\displaystyle \mathbb{Z}_{10}$. So just ask yourself: how many generators of $\displaystyle \mathbb{Z}_{10}$ 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 $\displaystyle \mathbb{Z}_{30}=\{0,1,2\cdots,29\}$ and $\displaystyle \mathbb{Z}_{10}=\{0,1,2\cdots,9\}.$
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 $\displaystyle G=<a>$ be a multiplicative cyclic group of order 30 with generator a. Then $\displaystyle H=<a^3>$ is the (unique) subgroup of G of order 10. The 4 homomorphism of G onto H are:
$\displaystyle f_1(a^m)=(a^3)^m=a^{3m}$ for all $\displaystyle m\in\mathbb{Z}, a\mapsto a^3$
$\displaystyle f_3(a^m)=((a^3)^3)^m=a^{9m}$ for all $\displaystyle m\in\mathbb{Z}, a\mapsto (a^3 )^3=a^9$
$\displaystyle f_5(a^m)=((a^3)^5)^m=a^{15m}$ for all $\displaystyle m\in\mathbb{Z}, a\mapsto (a^3 )^5=a^{15}$
$\displaystyle f_7(a^m)=((a^3)^7)^m=a^{21m}$ for all $\displaystyle m\in\mathbb{Z}, a\mapsto (a^3 )^7=a^{21}$
If k is any integer,
$\displaystyle f_k(a^m)=((a^3)^k)^m=a^{3km}$ for all $\displaystyle m\in\mathbb{Z}, a\mapsto (a^3 )^k=a^{3k}$
is a homomorphism of G into H. This is onto H only for $\displaystyle k\equiv1, 3, 5, 7 (\mod10)$

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