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?
Best Answer
Or, if you like, you could use the first isomorphism theorem: since the quotient $G/N\cong \phi(G)$ is a group, $N$ is normal.