Question

Prove that there are no simple groups of the given order:231

Short answer

It is proved that there is no simple group of order 231.

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 231

Writing 231 as a multiple of its factors as:

$231=3\cdot 7\cdot 11$

By Third Sylow Theorem, its 11-subgroups should divide 231 and it should be of the form of 1+11k, where $k\ge 0$.

Elements of 1+11k form are 1, 12, 23, 34, 45, 56, 67, 78, and 232.

And divisors of 231 are 1, 3, 7, 11, 21, 33, 77, and 231.

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

Therefore, 231 has exactly one 11-subgroups and this subgroup is normal byCorollary 9.16.

Consequently, no group of order 231 is simple.

Hence, it is proved that there is no simple group of order 231.