Question

Prove that there are no simple groups of the given order:255

Short answer

It is proved that there is no simple group of order 255.

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 255

Writing 255 as multiple of its factors as:

$255=3\cdot 5\cdot 17$

By Third Sylow Theorem, its 17-subgroups should divide 255 and it should be of the form of 1+17k, where $k\ge 0$

Elements of 1+17k form are 1, 18, 35, 52, 69, 86….256.

And divisors of 255 are 1, 3, 5, 15, 17, 51, 85, and 255.

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

Therefore, 255 has exactly one 17-subgroups and this subgroup is normal byCorollary 9.16.

Consequently, no group of order 255 is simple.

Hence, it is proved that there is no simple group of order 255.