Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 2 / 11E (Groups, Rings and Modules)

Write $|G|=p^{r}m$ with $p$ prime and $p\nmid m$. The Sylow theorems assert that a subgroup of order $p^r$ exists; every p-group subgroup is contained in a conjugate of a Sylow subgroup; all Sylow subgroups are conjugate; and their number $n_p$ satisfies $n_p\equiv 1(\mathrm{mod}p)$ and $n_p\mid m$.

For the containment and conjugacy assertions, fix a Sylow subgroup $P$ and let any p-group subgroup $Q$ act by left multiplication on the $m$ left cosets $G/P$. By the orbit-stabilizer theorem, every orbit has size a power of $p$. Since $p\nmid m$, at least one orbit has size one. If $gP$ is such a fixed point, then $qgP=gP$ for all $q\in Q$, equivalently $g^{-1}Qg\subseteq P$. Thus $Q\subseteq gP{g}^{-1}$. If $Q$ too has order $p^r$, this inclusion is equality, proving conjugacy. This is Sylow containment from a coset fixed point.

Since $351=3^3\cdot 13$, the number of Sylow subgroups for $p=13$ divides $27$ and is $1$ modulo $13$, hence is $1$ or $27$. In the first case the unique subgroup is a nontrivial proper normal subgroup.

In the second case the $27$ subgroups of order $13$ have pairwise trivial intersection: two such subgroups sharing a nonidentity element would both be generated by it. They account for $27\cdot 12=324$ nonidentity elements of order $13$. Exactly $27$ elements, including the identity, remain. Every Sylow subgroup for $p=3$ has order $27$, and all its elements lie among these remaining elements, by Lagrange's theorem. It must therefore be precisely that set. The Sylow subgroup for $p=3$ is unique and hence normal. In either case $G$ is not a simple group.

For the general linear group over ${\mathbb{F}}_{31}$, choose a nonzero first column in $31^2-1$ ways and a second column outside its span in $31^2-31$ ways. ThusSince $5^3=125\equiv 1(\mathrm{mod}31)$ and $5\ne 1$, the residue $5$ has order three. The nine diagonal matrices $\mathrm{diag}(5^i,5^j)$, with $0\le i,j<3$, form a subgroup isomorphic to $C_3\times C_3$. It is an abelian group and a Sylow subgroup for $p=3$. Every other Sylow subgroup for $p=3$ is conjugate to it by the Sylow theorems, hence is also Abelian.