# Prove that a group of order 375 has a subgroup of order 15

#### nhk

Let |G|=375. Since 5divides 375, G has a subgroup H of order 5 by Sylow's 1st theroem. BY Sylow's 3rd theroem, the number of sylow 3-subgroups of G is 1 mod 3 and divideds 125, so their can be either 1 or 25 Sylow 3-subgroups. If there is 1 sylow 3-subgroup K, then K is a normal subgroup of G, so HK is a subgroup of G of order 15.
If their are 25-sylow 3-subgroups, then let K be one of them.
I know need to find out what the conjugates of K are and then use this fact with the normalizer of K to prove that he normalizer is a subgroup of order 15. If i can prove that he conjugates of K are exactly the set of Sylow 3 subgroups I would be done, can someone explain this to me?

#### TheArtofSymmetry

Let |G|=375. Since 5divides 375, G has a subgroup H of order 5 by Sylow's 1st theroem. BY Sylow's 3rd theroem, the number of sylow 3-subgroups of G is 1 mod 3 and divideds 125, so their can be either 1 or 25 Sylow 3-subgroups. If there is 1 sylow 3-subgroup K, then K is a normal subgroup of G, so HK is a subgroup of G of order 15.
If their are 25-sylow 3-subgroups, then let K be one of them.
I know need to find out what the conjugates of K are and then use this fact with the normalizer of K to prove that he normalizer is a subgroup of order 15. If i can prove that he conjugates of K are exactly the set of Sylow 3 subgroups I would be done, can someone explain this to me?
$$\displaystyle |G| = 375 = 5 \times 5 \times 5 \times 3$$. By Cauchy's theorem, G has an element x having an order $$\displaystyle 3$$. Then $$\displaystyle <x>=C_3$$, which is a cyclic subgroup of G having an order 3. By Sylow's first theorem, G has a subgroup of order 5, which is a cylic subgroup of G. Let this group be <y>. I'll leave it to you verify that $$\displaystyle <x> \times <y>$$ is a cyclic subgroup of G having an order 15.

• nhk
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