group homomorphism with complex composition Let G be the group $(1,-1,i,-i)$ with multiplication of complex numbers as composition. Let H be the quotient group $Z/4Z$. Then the number of nontrivial group homomorphisms from H to G is=?
H will be ${0,1,2,3}$. but I am having trouble understanding the concept of homomorphism.
 A: HINT 1 $G$ and $H$ are isomorphic.
HINT 2 What are the identities of $G$ and $H$? What must happen to the identity? What are the (normal) subgroups of $\mathbb{Z}/4$? They correspond to kernels of homomorphisms $G \to H$. 
A: Hints for the problem:
I'll follow your notation above of $H=\{0,1,2,3\}$, keeping in mind those are all modulo 4. I'll use $a,b\in H$ for arbitrary elements.
You mentioned that you're having trouble with what a homomorphism is. We can describe what it will do in this case. A homomorphism $\phi:\Bbb Z/4\Bbb Z\to G$ is going to satisfy the following condition by definition:
$$
\phi(a + b)=\phi(a)\cdot\phi(b)
$$  
(Addition is the operation in $H$, and multiplication in $G$.
You probably already know $\Bbb Z/4\Bbb Z$ is cyclic and generated by $1$.
A homomorphism from $H$ is completely determined by what it does with the generator $1$.
Since $1$ has order 4, it can be mapped to anything in $G$ whose order divides 4. If $1\mapsto 1\in G$, you have the trivial homomorphism. 
What are the other possibilities?
