Question

Question: If G has order ${\mathit{p}}^{\mathbf{k}}\mathit{m}$with$\mathit{m}\mathbf{<}\mathit{p}$ , prove that G is not simple.

Short answer

It is proved that G is not simple.

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.

Given that, G has order$p^{k}m$ with$m<p$ .

Proving that group G is not simple

Writing$p^{k}m$ as multiple of its factors as:

$p^{k}m=m.(p.p........p)ktimes$

Consider Sylow p-subgroups.

ByThird Sylow Theorem, its p-subgroups should divideand 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$p^{k}m$are 1, m,mp,$m{p}^2$ …$m{p}^k$, and p,$p^2$ ….$p^k$.

Since p and m are two distinct primes, by considering different values of p and k, it can be seen that 1 is the only common element on both lists.

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

Consequently, no group of order$p^{k}m$is simple.

Hence,it is proved that G is not simple.