Solution
= Solution
If $K\triangleleft G$, the quotient map $G\to G/K$ is surjective with kernel $K$. Conversely, every kernel is normal because $\phi(gkg^{-1})=\phi(g)1\phi(g)^{-1}=1$.
= Solution
If $K\triangleleft G$, the quotient map $G\to G/K$ is surjective with kernel $K$. Conversely, every kernel is normal because $\phi(gkg^{-1})=\phi(g)1\phi(g)^{-1}=1$.