Question

Prove that there are no simple groups of the given order:200

Short answer

It is proved that there is no simple group of order 200.

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 200

Writing 200 as multiple of its factors as:

$200=2^3\cdot 5^2$

By Third Sylow Theorem, its 5-subgroups should divide 200, and it should be of the form of 1+5k, where $k\ge 0$.

Elements of 1+5k form are1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51,…201.

And divisors of 200 are 1, 2, 4, 5, 8, 10, 25, 50,100, and 200.

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

Therefore, 200 has exactly one 5-subgroup and this subgroup is normal byCorollary 9.16.

Consequently, no group of order 200 is simple.

Hence, it is proved that there is no simple group of order 200.