Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 2 / 9E / Solution

A Sylow subgroup of a finite group $G$ is a subgroup whose order is the largest power of a prime $p$ dividing $|G|$. The Sylow theorems say that such subgroups exist, that every $p$-subgroup lies in one, that all Sylow $p$-subgroups are conjugate, and that their number $n_p$ divides $|G|/p^a$ and obeys $n_p\equiv 1(\mathrm{mod}p)$.

For the last congruence, let one Sylow $p$-subgroup $P$ act by conjugation on the set of all Sylow $p$-subgroups. Every orbit other than a fixed point has size divisible by $p$. If $Q$ is fixed, then $P,Q\le N_G(Q)$. They are Sylow subgroups of this normalizer, so they are conjugate within $N_G(Q)$; because every element of $N_G(Q)$ fixes $Q$ under conjugation, this forces $P=Q$. Thus there is exactly one fixed point and $n_p\equiv 1(\mathrm{mod}p)$.

Now let $H<A_n$ have index $m$ with $1<m<n$. The coset action gives a homomorphismIts kernel is a normal subgroup of the simple group $A_n$. It cannot be all of $A_n$, since the action is transitive and nontrivial, so it is trivial. This would embed $A_n$ into $S_m$, contrary to $|A_n|=n!/2>m!=|S_m|$. Hence no such subgroup exists.

Suppose finally that a group $G$ of order $90$ were simple. The Sylow count satisfiesand simplicity excludes $n_5=1$, so $n_5=6$. Conjugation on these six subgroups gives a nontrivial homomorphism $G\to S_6$, which simplicity makes injective. Its image lies in $A_6$, because the composite with the sign homomorphism $S_6\to \{\pm 1\}$ must be trivial. It would therefore be a subgroup of $A_6$ of index $360/90=4$, contradicting the result just proved. Thus

Solved by gpt-5.6-sol high.