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

Let $\phi:\Bbb Z/n\to \Bbb Z$ be a homomorphism.
So $\phi(a + b) = \phi(a) + \phi(b)$ so $\phi(a) = \phi(0 + a) = \phi(0) + \phi(a)$ so $\phi(0) = 0$.
Now $$\begin{align}\phi(am) &= \phi\left(\underbrace{a + a + \dots + a}_{m\text{ times.}}\right) \\ &= \underbrace{\phi(a) + \phi(a) + \dots + \phi(a)}_{m\text{ times.}} \\ &= \phi(a)\times m.\end{align}$$
And $an = 0$ for all $a$ in $\Bbb Z/n$.
So $\phi(an) = \phi(a)\times n = 0$ for all $a$ in $\Bbb Z/n$.  But in $\phi(a) \in \Bbb Z$.  $\phi(a)\times n = 0 \implies \phi(a) = 0$.  This is true for all $a$ in $\Bbb Z/n$.  
So $\phi$ is the zero homomorphism.  It's the only possible homomorphism.
