Question

Prove that there are no simple groups of the given order:42

Short answer

It is proved that there is no simple group of order 42.

Step-by-step solution

Step-by-Step Solution Step 1: 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 42 as multiple of its factors as:

$42=2\cdot 3\cdot 7$

ByThird Sylow Theorem, its 7-subgroups should divide 42 and it should be in the form of 1+7k where,$k\ge 0$.

Elements of 1+7k form are 1, 8, 15, 22, 29, 36, and 43.

And divisors of 42 are 1, 2, 3, 7, 6, 14, and 42.

From the above, it can be seen that 1 is the only common number on both lists.

Therefore, 42 has exactly one 7-subgroup and this subgroup is normal byCorollary 9.16.

Consequently, no group of order 42 is simple.

Hence, it is proved that there is no simple group of order 42.