Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 1 / 10G (Groups, Rings and Modules)

Write $|G|=p^{n}m$ with $p\nmid m$. The Sylow theorems state:

Proceed by induction on $|G|$. If $p\mid |Z(G)|$, Cauchy's theorem for finite groups gives a central subgroup $C$ of order $p$. By induction $G/C$ has a subgroup of order $p^{n-1}$; its inverse image in $G$ has order $p^n$.

If $p\nmid |Z(G)|$, the class equation shows that some noncentral conjugacy class has size not divisible by $p$. For a representative $x$, its class size is $[G:C_G(x)]$, so $p^n\mid |C_G(x)|$. This centralizer is proper because $x$ is noncentral, and induction gives it, hence $G$, a subgroup of order $p^n$. This proves Sylow's first theorem.

For $|G|=12$, the third Sylow theorem gives $n_3\equiv 1(\mathrm{mod}3)$ and $n_3\mid 4$, so $n_3=1$ or $4$. In the first case the Sylow $3$-subgroup is unique.

If $n_3=4$, conjugation acts transitively on the four Sylow $3$-subgroups. Each normalizer has order $|G|/4=3$ and is the subgroup it normalizes. The action kernel lies in every such subgroup, whose intersection is trivial, so $G$ embeds in $S_4$. Its image has order $12$ and hence index two. The unique index-two subgroup of $S_4$ is $A_4$, so