Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 3 / 6D (Groups)

For $x\in X$,The stabilizer contains the identity; if $g,h$ fix $x$, thenso it is a subgroup.

The orbit-stabilizer theorem statesDefineThe map is well defined and bijective: two elements give the same image exactly when they differ by an element of the stabilizer. Lagrange's theorem then proves the formula.

The orbit of $(1,\dots ,k)$ consists precisely of the ordered $k$-tuples of distinct elements of $\{1,\dots ,n\}$:It has sizeThe stabilizer consists of permutations fixing $1,\dots ,k$ pointwise, while freely permuting the remaining letters:Thusverifying orbit-stabilizer.

Choose $h\in H\setminus \mathrm{Stab}(x)$ and put $y=hx\ne x$. For any $z\ne x$, double transitivity supplies an element $s\in G$ taking the ordered pair $(x,y)$ to $(x,z)$. Since $sx=x$, one hasThen $sh\in H$ sends $x$ to $z$. Together with the identity for $z=x$, this proves that $H$ acts transitively on $X$.

Now let $g\in G$. Transitivity of $H$ gives $h_0\in H$ with $h_{0}x=gx$. Henceso $g\in H$. ThereforeThis is the point-stabilizer maximality property of a doubly transitive group action.