Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 2 / 11E (Groups, Rings and Modules)

Let $\mathcal{S}$ be the set of all subsets of $G$ with $p^n$ elements. First show that $|\mathcal{S}|=\left(\genfrac{}{}{0ex}{}{p^{n}r}{p^n}\right)$ is not divisible by $p$. In the polynomial ring over the field of $p$ elements, $(1+X{)}^p=1+X^p$, since the intermediate binomial coefficients are divisible by $p$. Iterating givesComparison of coefficients of $X^{p^n}$ yields $\left(\genfrac{}{}{0ex}{}{p^{n}r}{p^n}\right)\equiv r\not\equiv 0(\mathrm{mod}p)$.

Let $G$ act on $\mathcal{S}$ by left translation. Because its orbit sizes sum to a number not divisible by $p$, some orbit has size prime to $p$. Choose a subset $S$ in that orbit, with stabilizer $H$. The orbit-stabilizer theorem gives the orbit size $[G:H]$, so $p\nmid [G:H]$. On the other hand $H$ acts freely on the elements of $S$: if $hs=s$, cancellation in the group gives $h=1$. Every $H$-orbit in $S$ therefore has $|H|$ elements, and $|H|$ divides $|S|=p^n$. Sinceand the second factor has no factor of $p$, the first must contain all $p^n$. Hence $\boxed{|H|=p^n}$, proving the first of the Sylow theorems. This is Sylow existence by subset action and includes the trivial case $n=0$.

The remaining Sylow theorems state that every $p$-subgroup of a finite group is contained in a Sylow subgroup, and that all Sylow $p$-subgroups are conjugate. If $|G|=p^{n}r$ with $p\nmid r$ and $n_p$ is the number of Sylow $p$-subgroups, thenIn particular $n_p=1$ if and only if that Sylow subgroup is a normal subgroup.

Relabel the two primes so that $p<q$. By the Sylow theorems, the number $n_q$ of Sylow $q$-subgroups divides $p$, so it is either $1$ or $p$. But $n_q\equiv 1(\mathrm{mod}q)$ and $1<p<q$, ruling out $n_q=p$. Thus there is a unique Sylow subgroup of order $q$, which is a normal subgroup. It is nontrivial and proper because $1<q<pq$. Consequently no group of order $pq$ is simple. No additional assumption on divisibility of $q-1$ by $p$ is required.