Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2017 / Paper 2 / 9

Write $A(a,b,c)$ with $a$ in entry $(1,2)$, $b$ in entry $(1,3)$ and $c$ in entry $(2,3)$, and perform all arithmetic in the finite field ${\mathbb{F}}_5$. Direct multiplication givesThus the set is closed. Associativity comes from matrix multiplication. The identity element is $A(0,0,0)$, and solving the product formula for it givesBoth orders of multiplication with this inverse give the identity. Hence these matrices form a finite group, an instance of the Heisenberg group over a prime field. Each coordinate has five independent possibilities, with distinct triples giving distinct matrices, soThe explicit coordinate convention matters because the central coordinate is $b$ here; other descriptions of the Heisenberg group over a prime field may place the central coordinate last.

A member of $H$ is $A(t,b,t)$. The multiplication law from part (a) becomesIt stays in $H$ and is unchanged when the two factors are exchanged, since $ts=st$. The identity lies in $H$, and $A(t,b,t{)}^{-1}=A(-t,t^2-b,-t)$ also lies in $H$. Thus $H$ is an Abelian subgroup, namely the Abelian plane subgroup of a finite Heisenberg group in this coordinate convention.

There are five choices each for $t,b$, soThis verifies Lagrange's theorem: the subgroup order divides the group order, and the number of distinct left cosets is five.

Explicitly, the map $\phi (A(a,b,c))=a-c$ is a surjective group homomorphism to the additive group of ${\mathbb{F}}_5$, by the product formula. Its kernel of a group homomorphism is $H$. Thus $H$ is even a normal subgroup; the five left cosets are the level sets $a-c=d$ for $d=0,1,2,3,4$, with representatives $A(d,0,0)$. Each contains 25 elements, providing a direct count as well.

Squaring with the product law givesEquality to the identity matrix requires $2a=2c=0$ in ${\mathbb{F}}_5$. Since 2 has a multiplicative inverse modulo 5, this forces $a=c=0$. The middle condition then gives $2b=0$, hence $b=0$. ThereforeThis agrees with Lagrange's theorem: a group of odd order has no element of order two.

The subset specified by $b\ne 0$ is not a subgroup. It omits the identity element. It also fails closure: both $A(0,1,0)$ and $A(0,4,0)$ lie in it, but their product is $A(0,0,0)=I$, whose $b$ coordinate is zero.