Question

Question: If p, q, r are primes with $p<q<r$, prove that a group of order pqr has a normal Sylow r-subgroup and, hence, is not simple.

Short answer

It is proved that the group of order pqrhas a normal Sylow r-subgroup and hence, is not simple.

Step-by-step solution

Referring to Corollary 9.16 and Theorem 9.17 (Third Sylow Theorem).

Corollary 9.16

Let G be a finite group and K is a Sylow p-subgroup for some prime p. Then, K is normal in G if and only if K is the only Sylow p-subgroup.

Theorem 9.17 (Third Sylow Theorem)

The number of Sylow p-subgroups of finite group G divides and is of the form 1+pk for some non-negative integer k.

Proving that the group of order pqr has a normal Sylow r-subgroup and hence, is not simple

Writing pqr as multiple of its factors as:

$pqr=p·q·r$

According to Theorem 9.17, r-subgroups should divide pqr and should be in the form of 1+rk where$,k\ge 0$ .

Elements of 1+rk are 1, 1+r and 1+2r.

And divisors of pqare 1, p, q, r, pq, qr, rq and pqr.

Since p, q,and r are three distinct primes, from the above it can be seen that 1 is the only common number on both lists.

Therefore, pqr has exactly one Sylow r-subgroup and this subgroup is normal byCorollary 9.16.

Consequently, no group of order pqr is simple.

Hence, it is proved that the group of order pqr has a normal Sylow r-subgroup and hence, is not simple.