Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 1 / 9E / c

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.