Equivalence Relation and Group Homomorphism

Question

Let $\phi: G_1 \rightarrow G_2$ be a group homomorphism. Let $K$ = {$g$ | $f(g) = e_2$}, where $e_2$ is the identity element of $G_2$. Let $R$ be the equivalence relation s.t. $aRb$ if $f(a) = f(b)$ for some map $f: X \rightarrow Y$. Show that $g_1 R g_2$ if and only if $g_1 ^{-1} g_2 \in K$.

How do I show that this is true? We start with $\phi(g_1$) = $\phi(g_2$), right? But how does that relate to $K$? What I have so far is:

If $g_1 ^{-1} g_2 \in K$, then $\phi(g_1 ^{-1} g_2)$ = $e_2$. By the homomorphic property, $\phi(g_1 ^{-1} g_2)$ = $\phi(g_1 ^{-1}$)$\phi(g_2)$ = $e_2$. But this is where I get stuck.

Any help would be great. Thank you.

Answer 1

If you want to show it yourself, the trick is seeing that $\phi(g^n) = \phi(g)^n$ (for $n\in\mathbb Z$, and importantly for $n = -1$).

I've included both proofs below:

$f(a) = f(b)\implies b^{-1}a\in K$

If you assume that $f(a) = f(b)$ for some map $f$, then we have that $f(b)\in Y$, a group, so $f(b)^{-1}$ is sensible. We can left multiply by this to get that $f(b)^{-1}f(a) = e_2$, then use the homomorphism property to get that $f(b^{-1}a) = e_2$, so $b^{-1}a\in K$.

$b^{-1}a\in K\implies f(a) = f(b)$

Now, assume that $g_1^{-1}g_2\in K$. Then, we have that $f(g_1^{-1}g_2) = e_2$. We can apply the homomorphism property to get that $f(g_1)^{-1}f(g_2) = e_2$, and now left multiply by $f(g_1)$ to get that $f(g_2) = f(g_1)$, so $g_2Rg_1$, as desired.

For a proof of $\phi(g^n) = \phi(g)^n$, see below:

For $n\in\mathbb N$, you can do a simple induction on the homomorphism property. The same follows for $n\leq 0$ once you've established that $\phi(g^{-1}) = \phi(g)^{-1}$. To see this, note that $e = \phi(e) = \phi(gg^{-1}) = \phi(g)\phi(g^{-1})$. As $\phi(g)$ and $\phi(g^{-1})$ multiply to the identity, they must be inverses. As inverses are unique, we can denote $\phi(g)^{-1} = \phi(g^{-1})$.

Answer 2

Note that $\phi(g^{-1})=\phi(g)^{-1}$ and $\phi(e_1)=e_2$, which should be easy to verify on your own. Applying either of these facts to what you already figured out should immediately give you the conclusion.

And more generally what Mark says is true.