# Thread: Group of order 30

1. ## Group of order 30

I am working on showing that there are exactly 4 non-isomorphic groups of order 30. I'm stuck on showing that this group of order 30, $\displaystyle G$ must have a normal subgroup, $\displaystyle N$ of order 15.
I have $\displaystyle 30=2 \times 3 \times 5$
Then by using Sylow-theorems I have the possibilities for Sylow p-subgroups of G as followed:
$\displaystyle n_2=1,3,5,15; n_3=1,10; n_5=1,6$
Then I use counting argument to show that it can not be the case that $\displaystyle n_5=6$ and $\displaystyle n_3=10$. So, either $\displaystyle n_5=1$ or $\displaystyle n_3=1$. Then it follows that $\displaystyle G$ has either a normal Sylow 3-subgroup or a normal Sylow 5-subgroup. But I can't argue that it must have both of those.
I notice that by Cauchy, G must also have a subgroup of order 2, but didn't see how this helps.

I really appreciate any help.

2. Originally Posted by jackie
I am working on showing that there are exactly 4 non-isomorphic groups of order 30. I'm stuck on showing that this group of order 30, $\displaystyle G$ must have a normal subgroup, $\displaystyle N$ of order 15.
I have $\displaystyle 30=2 \times 3 \times 5$
Then by using Sylow-theorems I have the possibilities for Sylow p-subgroups of G as followed:
$\displaystyle n_2=1,3,5,15; n_3=1,10; n_5=1,6$
Then I use counting argument to show that it can not be the case that $\displaystyle n_5=6$ and $\displaystyle n_3=10$. So, either $\displaystyle n_5=1$ or $\displaystyle n_3=1$. Then it follows that $\displaystyle G$ has either a normal Sylow 3-subgroup or a normal Sylow 5-subgroup. But I can't argue that it must have both of those.
I notice that by Cauchy, G must also have a subgroup of order 2, but didn't see how this helps.

I really appreciate any help.
let $\displaystyle N_1,N_2$ be two subgroups of $\displaystyle G$ with $\displaystyle |N_1|=3, \ |N_2|=5.$ you've proved that either $\displaystyle N_1$ or $\displaystyle N_2$ is normal. thus $\displaystyle N_1N_2$ is a subgroup of $\displaystyle G.$ clearly $\displaystyle |N_1N_2|=15$ and so $\displaystyle [G:N_1N_2]=2.$

now recall that every subgroup of index 2 in any group is normal.