Question

Question: If $\left|G\right|=375$, prove that G has a subgroup of order 15.

Short answer

It is proved that G has a subgroup of order 15.

Step-by-step solution

Referring to Corollary 9.14 (Cauchy’s Theorem), Corollary 9.16, and Theorem 9.17 (Third Sylow Theorem)

Corollary 9.14 (Cauchy’s Theorem)

If G is a finite group whose order is divisible by a prime p. Then, G contains an element of order p.

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.

Theorem 9.17 (Third Sylow Theorem)

The number of Sylow p-subgroups of finite group G divides and is of the form 1+pk for some non-negative integer k.

Proving that G has a subgroup of order 15

Writing 375 as multiple of its factors as:

$375=3·5^3$

Considering the Sylow 3-subgroup of G.

According toTheorem 9.17, the Sylow 3-subgroup must divide G and should be in the form of 1+3k.

Therefore, G has only 1 unique Sylow subgroup, and fromCorollary 9.16, it is normal in G.

Again, considering the Sylow 5-subgroup of G.

According toTheorem 9.17, G has only 1 unique Sylow subgroup.

Since the product of two subgroups is a subgroup, G has Sylow 3-subgroup and Sylow 5-subgroup. As 3 and 5 both are prime then, according toCorollary 9.14, there must be a subgroup of order 15.

Hence, it is proved that G has a subgroup of order 15.