Question

Question: If p and q are distinct primes, prove that there are no simple subgroups of order pq.

Short answer

It is proved that there are no simple subgroups of order pq.

Step-by-step solution

Referring to Corollary 9.16 and 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.

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.

Given that p and q are distinct primes.

Proving that there is no simple group of order pq

Writing pq as multiple of its factors:

$pq=p.q$

Consider Sylow p-subgroups.

ByThird Sylow Theorem, its p-subgroups should divide pq and it should be in the form of 1+pk where $k\ge 0$.

Elements of 1+pk form are 1, 1+p, 1+2p…..

And divisors of pq are 1, p, q and pq.

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

Therefore, pq has exactly one Sylow p-subgroup and this subgroup is normal byCorollary 9.16.

Consequently, no group of order pq is simple.

Hence, it is proved that there is no simple subgroup of order pq.