Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 3 / 6D / a / Solution

For either coefficient system $\mathbb{Z}$ or ${\mathbb{F}}_p$, multiplying two matrices keeps all entries in that system, and multiplicativity of the determinant preserves determinant one. The identity matrix has determinant one, and for a determinant-one matrix the matrix inverse isIts entries again belong to the same coefficient system and its determinant is one. Together with associativity of matrix multiplication, these facts establish all the group axioms. Thus both sets are special linear groups.

Reduce each entry modulo five to defineReduction respects sums and products, so $\overline{MN}=\overline{M}\overline{N}$. It also respects the determinant, ensuring that the image has determinant one. Hence $\pi$ is a group homomorphism. Its kernel of a group homomorphism consists exactly of determinant-one integer matrices congruent to the identity modulo five:The determinant-one condition remains part of this set; the four integers cannot be chosen independently without it. A kernel of a group homomorphism is a normal subgroup, since $\pi (gk{g}^{-1})=\pi (g)e\pi (g{)}^{-1}=e$ for $k\in \ker \pi$. This is the principal congruence subgroup of level five.