Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 8

The quotient group $G/N$ consists of the cosets $gN$, with multiplicationFor well-definedness, replacing representatives by $g{n}_1$ and $h{n}_2$ changes their product to $gh(h^{-1}{n}_{1}h)n_2$. Since $N$ is a normal subgroup, the final two factors lie in $N$, so the product coset does not change.

Associativity follows from that in $G$; the identity element is the coset $N$, and $(gN{)}^{-1}=g^{-1}N$. Hence these cosets form a group. By Lagrange's theorem, every subgroup $H\le G$ has order dividing $|G|$, and its subgroup index is $[G:H]=|G|/|H|$. In particularNormality is needed for the quotient multiplication, whereas Lagrange’s counting formula holds for every subgroup.

The four displayed matrices fail closure: for example ${\sigma }_x{\sigma }_y=i{\sigma }_z$ is not one of them. Thus they do not form a group by themselves.

Let ${\sigma }_0=I$ and ${\mu }_4=\{1,-1,i,-i\}$. The enlarged set is the single-qubit Pauli groupThere are 16 distinct matrices: $I,{\sigma }_x,{\sigma }_y,{\sigma }_z$ are linearly independent, so two different ones cannot be scalar multiples, and the four phases are distinct for each nonzero matrix.

The Pauli matrices satisfyHere $\delta$ is the Kronecker delta and $ϵ$ the Levi-Civita symbol. Thus any product of two Pauli matrices is another Pauli matrix or the identity times a phase in ${\mu }_4$. Multiplying by the two original phases preserves membership in ${\mu }_4$, proving closure of all 16 matrices.

Associativity is inherited from matrix multiplication, the identity is $I$, andThis proves every group axiom. The resulting group is non-Abelian, since reversing ${\sigma }_x{\sigma }_y$ changes its sign.

For every positive integer $m$, the group laws give $(g^{-1}{)}^m=(g^m{)}^{-1}$. ConsequentlyThe sets of positive exponents returning the two elements to the identity element are identical. If this set is nonempty, its least member is the order of a group element in both cases. If it is empty, neither element has finite order. Therefore, including the infinite-order case,For a direct divisibility argument in the finite case, if $g$ has order $n$, then $(g^{-1}{)}^n=e$, so its inverse’s order divides $n$. Reversing the roles shows $n$ divides that order, forcing equality. This is the general fact that an element and its inverse have the same order; it also reflects $⟨g⟩=⟨g^{-1}⟩$ as cyclic subgroups.