Let $f: G_1 \rightarrow G_2$ be a group homomorphism, and let $K$ = {$g | f(g) = e_2$}, where $e_2$ is the identity of $G_2$. 1) Show that $K$ is a group. 2) Show that if $K$ = {$e_1$}, then $f$ is injective.
This is what I have:
1) Kinda lost on this one. I know the axioms of a group, but I'm not sure how to apply it to $K$. I would have to show that the identity exists in $K$, the existence of inverses in $K$, and associativity in $K$, right?
2) Since $K$ = {$e_1$}, we only have 1 element, $e_1$, that maps to $e_2$. $0 \rightarrow 0$ is an injective mapping, and therefore $f$ is injective if $K$ = {$e_1$}.
Any help on this would be tremendously helpful. Thank you.
Awesome Problem! This one is fun.
Since $f$ is a homomorphism, $f(e_1) = e_2$, so $e_1 \in K$.
For any $a,b \in K$, we have $f(ab) = f(a)f(b) = e_2e_2 = e_2$, so $ab \in K$, giving us closure.
We have associativity because multiplication in $K$ works the same way as in $G_1$.
For any $a \in K$, $f(aa^{-1}) = e_2 = f(a)f(a^{-1}) = e_2f(a^{-1})$, so $f(a^{-1}) = e_2$ and thus $a^{-1} \in K$.
Thus K is a Group!
Suppose, for some $g,h \in G$, that $f(g) = f(h)$. Then $f(g) = f(g)f(h^{-1})f(h) = f(gh^{-1})f(h)$. But f(g) = f(h), so $f(gh^{-1}) = e_2$. Then $gh^{-1} \in K$, so $gh^{-1} = e_1$. Then $g = h$. Thus, $f$ is injective.
Any questions? I'm happy to clarify in comments.