Let $f \in$ Hom(G,H). Show that Kerf($f$) is a subgroup of $G$ and is normal.

Question

Question: Let $f \in$ Hom(G,H). Show that Ker($f$) is a subgroup of $G$ and is normal.

Here's my proof: Let $x,y \in$ Ker($f$). Then $f(xy^{-1}) = f(x)f(y^{-1}) = ef(y)^{-1} = f(y)^{-1} = e^{-1} = e$. So then $xy^{-1} \in$ Ker $f$ and Ker(f) $<$ G.

As for normality, let $g \in G$, $n \in$ Ker($f$). Then $f(gng^{-1}) = f(g)f(n)f(g^{-1}) = f(g) e f(g^{-1}) = f(g) f(g)^{-1} = e \implies gng^{-1} \in$ Ker($f$). So Ker($f$) $\unlhd G$

Is this proof correct?

Answer 1

I thinks the answer is simple: It is plainly a perfect proof.