Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 3 / 5D (Groups)

A left group action is a map $(g,x)\mapsto g\cdot x$ with $e\cdot x=x$ and $(gh)\cdot x=g\cdot (h\cdot x)$. The group orbit and stabiliser subgroup areThe identity belongs to $G_x$. If $g,h\in G_x$, then $(gh)\cdot x=g\cdot x=x$; also $g^{-1}\cdot x=g^{-1}\cdot (g\cdot x)=x$. Thus $G_x$ is a subgroup.

For the orbit-stabilizer theorem, the map $g{G}_x\mapsto g\cdot x$ from left cosets to the group orbit is a bijection: $g\cdot x=h\cdot x$ if and only if $h^{-1}g\in G_x$, which is precisely $g{G}_x=h{G}_x$. Each left coset has $|G_x|$ elements, since multiplication by $g$ is a bijection from $G_x$ to $g{G}_x$, and the left cosets partition $G$. For finite $G$ this proves

The identity permutation fixes every multivariate polynomial. With permutation composition $(\sigma \tau )(i)=\sigma (\tau (i))$, substitution givesThus this is a left variable-permutation action on polynomials. The substitution acts on variable names; using the displayed definition avoids reversing the composition order.

For the given multivariate polynomial, the group orbit corresponds to the three partitions of four indices into two unordered pairs:These three polynomials are distinct, and every pairing occurs under a permutation. The orbit-stabilizer theorem givesThe eight elements of the stabiliser subgroup independently interchange the members of each pair, and may also interchange the two pairs.

Applying a permutation to the elements of an $r$-element subset preserves its cardinality. The identity fixes each subset andso this is the symmetric group action on subsets. For two $r$-element subsets $U,W$, choose a bijection $U\to W$ and a bijection between their complements; together they define a permutation of the full $n$-element set. Hence the action is transitive.

A member of the stabiliser subgroup must permute $U$ internally and its complement internally, with independent choices. ThusThis also agrees with the orbit-stabilizer theorem; $0!=1$ handles $r=n$.