Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 3 / 5D (Groups)

The orbit and stabiliser areSuch an action can be faithful: the natural action of $S_3$ on three points is faithful, although every point stabiliser has order two.

If $y=gx$, thenIndeed, $h$ fixes $x$ exactly when $gh{g}^{-1}$ fixes $gx$. Conjugation by $g$ is therefore the required isomorphism.

For $1\le k<n$, any nonidentity permutation moves some $k$-subset: choose a moved point and complete a subset so that it contains that point but not its image. Thus the kernel is trivial. For $k=n$ the action is trivial, so it is faithful only in the degenerate case $n=1$.

If $N◃S_n$, then $N\cap A_n◃A_n$, hence the intersection is either $1$ or $A_n$. In the latter case $N$ is $A_n$ or $S_n$. In the former, $N$ injects into $S_n/A_n\cong C_2$, so $|N|\le 2$. A normal subgroup of order two would be central, but $Z(S_n)=1$ for $n\ge 3$. Therefore the normal subgroups are