Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 8

The left coset isEvery $g\in G$ belongs to $gH$, because $e\in H$, so the cosets cover $G$. If $gH$ and $kH$ intersect, choose $g{h}_1=k{h}_2$ in the intersection. Then $k^{-1}g=h_2{h}_1^{-1}\in H$, so $g=kh$ for some $h\in H$. Thus $gH=khH=kH$.

Hence two left cosets are either identical or disjoint, and the distinct left cosets form a partition of a set whose union is $G$. Finiteness is unnecessary for this partition argument. In the finite case each coset has $|H|$ members, since multiplication by $g$ is a bijection, giving Lagrange's theorem.

Denote a matrix by $A(x,y,z)$, with $y$ the upper-right entry. Direct matrix multiplication givesThis proves closure. Associativity is inherited from matrix multiplication. The identity is $A(0,0,0)$, and the inverse isBoth products with this matrix are the identity. Thus these real matrices form the Heisenberg group, in the coordinate convention with central coordinate $y$.

Let $N=\{A(0,b,0):b\in \mathbb{R}\}$. Its multiplication adds $b$, its identity is $b=0$ and its inverse negates $b$, so it is a subgroup. MoreoverThus every element of $N$ is central, and $gN{g}^{-1}=N$. Therefore $N$ is a normal subgroup. In fact it is the full center of a group, as described by the center quotient of the real Heisenberg group.

Perform the product and inverse formulas from part (b) in the commutative ring $\mathbb{Z}/4\mathbb{Z}$. They involve only addition, multiplication and negatives, so closure and inverses remain valid; no division by a nonunit is needed. Associativity again comes from matrix multiplication, and $A(0,0,0)$ remains the identity. The Heisenberg group over a commutative ring therefore hasThese are residues modulo four, not arithmetic in the finite field of four elements.

For the additional subgroup task, write an element of $H$ as $A(t,y,t)$. Its product and inverse areBoth remain in $H$, and the product is unchanged on exchanging the factors because $ts=st$. The identity is included, so $H$ is an Abelian subgroup. There are four independent choices each for $t,y$, givingThus there are four distinct left cosets. Explicitly the surjective group homomorphism $A(x,y,z)\mapsto x-z(\mathrm{mod}4)$ has kernel $H$; the four cosets are the level sets of that difference, with representatives $A(d,0,0)$, $d=0,1,2,3$. This is the Abelian plane subgroup of the Heisenberg group modulo four.