Question

Let n be a composite positive integer and p a prime that divides n. Assume that 1 is only divisor of n that is congruent to 1 modulo p. If G is a group of order n, prove that G is not simple.

Short answer

It is proved that group 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 n is a composite positive integer and p a prime that divides n.

Assume that 1 is only divisor of n that is congruent to 1 modulo p and G is a group of order n.

Proving that group G is not simple

Since G is a group of order n, therefore, according toThird Sylow Theorem,G divides n.

According to our assumption, 1 is the only divisor of n.

So, fromCorollary 9.16, group G has a unique Sylow p-subgroup which is normal to G.

This implies G is not simple