Let $\varphi:G\to H$ and $\psi:H\to K$ be homomorphisms. Show that $\ker(\varphi)\unlhd\ker(\psi\circ\varphi).$

Question

Let $φ: G → H$ and $ψ: H → K$ be two homomorphisms.

(a) Show that $ψ ◦ φ: G → K$ is a homomorphism.

(b) Show that $ker(φ)$ is a normal subgroup of $ker(ψ ◦ φ)$.

SOLUTION

For any $h ∈ ker(φ)$ and $g ∈ ker(ψ ◦ φ)$, the conjugate $ghg^{−1}$ is in ker(φ):

$φ(ghg^{−1}) = φ(g)φ(h)φ(g^{−1}) = φ(g)φ(g^{−1}) = e$, $e$ being the identity element.

I've managed to do question (a) but struggling with question (b). The solution in my textbook is given below but I'm struggling to understand it. I thought for g and $ghg^{-1}$ to be conjugate $g$ and $h$ both had to be in the same group.

Answer 1

$\operatorname{ker} \varphi$ is a normal subgroup of $G$ and $\operatorname{ker} \varphi \subseteq \operatorname{ker}(\psi \circ \varphi)$, so $\operatorname{ker} \varphi$ is also normal in $\operatorname{ker}(\psi \circ \varphi)$ by the following lemma:

Let $N \unlhd G$ and $N \subseteq K$ where $K \leq G$. Then $N \unlhd K$.