Let $\varphi$ be a group homomorphism from $G$ to $H$. Let $\varphi(G)$ be its image and $K$ its kernel. Show that $K$ is a normal divisor of $G$.

Question

I have been stuck on this for a few days maybe someone here can figure it out. I haven't done a proof with kernels before. Any guidance would be greatly appreciated.

Let $\varphi$ be a group homomorphism from $G$ to $H$. Let $\varphi(G)$ be its image and $K = \{x \in G: \varphi(x) = e\}$ its kernel.

1. Show that $K$ is a normal divisor of $G$.
2. Define $G/K$.
3. Define a function $f\colon G/K \to H$ by $f(xK) = \varphi(x)$. Prove that $f$ is an isomorphism between $G/K$ and $φ(G)$.

Answer 1

I'll get you started with part 1. First, recall that a subgroup $N$ of $G$ is normal if for every $g \in G$ we have $gNg^{-1} \subseteq N$.

Then observe that for every $g \in G$ and for every $k \in K$ $$ \varphi(gkg^{-1}) = \varphi(g) \varphi(k) \varphi(g)^{-1} = \varphi(g) \varphi(g)^{-1} = e $$ which by definition of $K$ means that $gkg^{-1} \in K$, hence $K$ is indeed normal.

Can you prove part 3 now?