Exercise 4.5.16 (A group of order $pqr$ is not simple)

Proof. Suppose that $|G|=\mathit{𝑝𝑞𝑟}$, where $p,q$ and $r$ are primes with $p<q<r$, and let $n_p,n_q,n_r$ be the numbers of corresponding Sylow subgroups. By Sylow’s Theorem,

Assume for the sake of contradiction that $n_p\ne 1,n_q\ne 1$ and $n_r\ne 1$. Then

Moreover, $n_r=1+\mathit{𝑘𝑟}$, where $k\ge 1$ since $n_r\ne 1$. Therefore $n_r>r>q>p$, thus

and by (1) and (2),

Any two Sylow subgroups have trivial intersection $\{1\}$. Therefore, using (4), there are $n_r\cdot (r-1)=\mathit{𝑝𝑞}(r-1)$ elements of order $r$ in $G$. Similarly there are $n_q\cdot (q-1)\ge p(q-1)$ elements of order $q$ and $n_p\cdot (p-1)\ge q(p-1)$ elements of order $p$ (and of course, one element of order $1$, the identity of $G$). This shows that

because $p>1$ and $q>1$. Since $|G|=\mathit{𝑝𝑞𝑟}$ by hypothesis, this is a contradiction, which proves that

Hence $G$ has a normal Sylow subgroup for either $p,q$ or $r$ (so $G$ is not a simple group). □