Suppose $N \leq G$ is a subgroup of some group and it is the kernel of some group homomorphism $\phi$. Prove $N \unlhd G$
We want to show that $gNg^{-1}=N$ for every $g \in G$, given that $N:=$ ker $\phi$ some $\phi:G \rightarrow H$ a homomorphism. To show $gNg^{-1} \subset N$, its enough to show for every $x \in gNg^{-1}$ is in the kernel of $\phi$. We have that
\begin{align} \phi(x)&= \phi(gng^{-1}) && \text{for some $n \in N$}\\ &=\phi(g)\phi(n)\phi(g)^{-1} && \text{as $\phi$ is a homomorphism}\\ &=\phi(g)e_H\phi(g)^{-1} && \text{as $n \in N:=$ ker $\phi$}\\ &=\phi(g)\phi(g)^{-1} && \text{definition of mult by identity}\\ &=e_H && \text{definition of inverses} \end{align}
thus $x \in$ ker $\phi$ and thus $x \in N$. Is this enough of a direct proof that $N$ is normal if its the kernel of some group homomorphism?