Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 3 / 6D (Groups)

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.

The general linear group $G{L}_2({\mathbb{F}}_5)$ consists of invertible two-by-two matrices over the finite field ${\mathbb{F}}_5$, under matrix multiplication. Such a matrix has two linearly independent columns. There are $5^2-1=24$ choices for a nonzero first column, and $5^2-5=20$ choices for a second column outside its one-dimensional span. ThusThe determinant is a group homomorphismwhere ${\mathbb{F}}_5^{\times }=\{1,2,3,4\}$ is the multiplicative group of nonzero field elements. It is surjective because $\mathrm{diag}(t,1)$ has determinant $t$. Its kernel of a group homomorphism is $S{L}_2({\mathbb{F}}_5)$. The first isomorphism theorem gives a quotient of order four, whenceThe nonzero squares form the subgroup $\{1,4\}$ of ${\mathbb{F}}_5^{\times }$. Its inverse imageis a subgroup with index two. Indeed, composing the determinant with the quotient map to ${\mathbb{F}}_5^{\times }/\{1,4\}$ makes $H$ the kernel of a surjective group homomorphism to a group of order two. Thus $|H|=240$.