Question

Question: If p and q are two distinct primes, prove that there is no simple group of order${\mathit{p}}^{\mathbf{2}}\mathit{q}$ .

Short answer

It is proved that there is no simple group of order$p^{2}q$.

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$\left|G\right|$ and is of the form 1+pk for some non-negative integer k.

Proving that there is no simple group of order 42

Writing$p^{2}q$ as multiple of its factors as:

role="math" localid="1656649407820" $p^{2}q=p^2.q$

Consider Sylow p-subgroups.

ByThird Sylow Theorem, its p-subgroups should divide$p^{2}q$ and it should be in the form of 1+pk, where $k\ge 0$.

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

And divisors of$p^{2}q$ are 1, p, q, $p^2$, pq, and$p^{2}q$.

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

Therefore,$p^{2}q$has exactly one Sylow p-subgroup and this subgroup is normal byCorollary 9.16.

Consequently, no group of order$p^{2}q$ is simple.

Hence, it is proved that there is no simple group of order$p^{2}q$.