I just got my midterm back, and the professor isn't going to post solutions, so I was wondering if the MSE community can help me understand a multiple-part question. The problem was:
Let $\phi$ : $G_1 \rightarrow G_2$ be a group homomorphism. Let $K$ = {$g | \phi(g) = e_2$}, where $e_2$ is the identity of $G_2$.
a) Show that $K$ is a group. (Proven and shown)
b) Define a relation $R$ on $X$ by $aRb$ if $f(a) = f(b)$. Show that $g_1Rg_2$ if and only if $g_1 ^{-1}g_2 \in K$.
c) Show that if $K$ = {$e_1$}, then $\phi$ is injective.
I understood part a) and was able to prove it successfully, but could not do b) or c) at all. I was completely lost on b), and for c), I was unable to show that $\phi : G_1 \rightarrow G_2$ is injective. Any help with me understanding b) and c) would be tremendously helpful. Thank you!
p.s. I don't know how to make those yellow block things for questions/problems, so if anyone could help me edit that in that'd be great. Thanks!
Always remember these two relationships with homomorphisms: $$\phi(a) \cdot_{G_2} f(b)=f(a \cdot_{G_1} b)$$ $$\phi(a^{-1})=\phi(a)^{-1}$$
Now, let's do part b: $$g_1 R g_2 \iff \phi(g_1)=\phi(g_2)$$ We have to get $e$ somewhere in here and since the expression we're trying to prove has an inverse on the left side, the most natural way to get $e$ is by multiplying both sides by the inverse of the left side of the equation: $$g_1 R g_2 \iff e_{G_2}=\phi(g_1)^{-1}\cdot_{G_2}\phi(g_2)$$ Use the second equation at the top: $$g_1 R g_2 \iff e_{G_2}=\phi(g_1^{-1})\cdot_{G_2}\phi(g_2)$$ Use the first equation at the top: $$g_1 R g_2 \iff e_{G_2}=\phi(g_1^{-1}\cdot_{G_1}g_2)$$
For part c, look at the second part of this proof. Notice how it uses $f(g_1)=f(g_2) \iff g_1^{-1}\cdot g_2 \in K$, which is very similar to the statement proven in part b.