Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 1 / 9E (Groups, Rings and Modules)

Let $k=|G/H|$. The coset action gives a nontrivial homomorphism $G\to S_k$ whose kernel is normal. Simplicity makes it injective. The sign map is trivial on the nonabelian simple image, so $G$ embeds in $A_k$. For $k\le 4$, $A_k$ is solvable, as are its subgroups, whereas a nonabelian simple group is not. Hence $k\ge 5$.

Let $S$ act by conjugation on the Sylow $p$-subgroups. The only fixed point is $S$: if $S$ normalizes another $T$, then $ST$ is a $p$-subgroup, forcing $S=T$. For $T=gS{g}^{-1}\ne S$, a stabilizer element in $S$ normalizes $T$, hence lies in $T$ by the same argument; the hypothesis then makes it trivial. Every nontrivial orbit therefore has size $|S|$, so

Sylow gives $n_7\mid 24$ and $n_7\equiv 1(\mathrm{mod}7)$. Simplicity excludes $n_7=1$, so $n_7=8$. Distinct order-seven subgroups intersect trivially, hence there areelements of order seven.

If all distinct Sylow 2-subgroups met trivially, part (b) would give $n_2\equiv 1(\mathrm{mod}8)$. But Sylow gives $n_2\mid 21$ and $n_2$ odd, so $n_2\in \{3,7,21\}$ after excluding normality, none congruent to one modulo eight. Thus some distinct pair has nontrivial intersection. That intersection is a nontrivial finite 2-group, so Cauchy's theorem supplies an element of order two.