Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 2 / 13F (Groups, Rings and Modules)

Suppose $K=\ker \phi$ for a group homomorphism $\phi :G\to H$. Its identity belongs to $K$, and for $k,l\in K$ we have $\phi (k{l}^{-1})=1$, so it is a subgroup. Moreover, for $g\in G$ and $k\in K$,Hence $gK{g}^{-1}\subseteq K$; applying this also to $g^{-1}$ gives equality, so $K$ is a normal subgroup.

Conversely, if $K$ is a normal subgroup, multiplication of cosets in the quotient group $G/K$ is well-defined. The map $g\mapsto gK$ is a group homomorphism and has kernel of a group homomorphism exactly $K$. This proves both directions without requiring $\phi$ to be injective or surjective.

For the matrix application, reduce all entries modulo $p$:Reduction respects matrix multiplication and the determinant, so this is a group homomorphism. Its kernel of a group homomorphism consists exactly of matrices congruent to the identity modulo $p$, which are the given matrices. Therefore $K=\ker \rho$ is a normal subgroup, the level-$p$ principal congruence subgroup. Surjectivity of reduction is not needed for this argument.