Few group theory questions I am trying to solve the following;
First, given $G$ is a group and $H$ a subgroup of $G$, what can we say about the relation $a \cong b$ if $b^{-1}a \in H$ 
I can show that it is reflexive as the identity is always in the subgroup.
if $a \cong b$ then $b^{-1}a \in H$ and so $(b^{-1}a)^{-1}=a^{-1}b \in H$  so $b \cong c$.
Now I must determine if $G$ being abelian is required for this to be transitive.
My thought would be that if $ a \cong b $ then $b^{-1}a \in H $ and if $b \cong c$ then $c^{-1}b \in H$ and by properties of subgroups so to is $c^{-1}bb^{-1}a=c^{-1}a$ which would seem to imply $a \cong c$, but I don't know if I am somehow doing something that couldn't be done if $G$ was not abelian. What do you guys think?
And second part asks to discuss the possible homomorphism (group) from $\mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Z}$ and it says for $n \ge 1$, is every homomorphism the zero one. Either yes, no, or it depends on $n$.
My thoughts are that only the zero homomorphism is possible, since if $\phi$ was a homomorphism then $\phi(ab)=\phi(a)\phi(b)$ and this would quickly result in having $ab=n=0$ but $\phi(a)$ and $\phi(b)$ not being $0$.
If $n$ was prime, then I am not so sure, but wouldn't this still end up mapping many things to the identity etc?
Thank you all

*

*Correct, commutativity is not needed and not used in your proof.

*The group operation is rather addition. 
Hint: Where can the equivalence class $[1]_{n\Bbb Z}$ be mapped by a homomorphism to $\Bbb Z$?

