# Group of order 30

• Nov 4th 2009, 09:55 PM
jackie
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, $G$ must have a normal subgroup, $N$ of order 15.
I have $30=2 \times 3 \times 5$
Then by using Sylow-theorems I have the possibilities for Sylow p-subgroups of G as followed:
$n_2=1,3,5,15;
n_3=1,10;
n_5=1,6$

Then I use counting argument to show that it can not be the case that $n_5=6$ and $n_3=10$. So, either $n_5=1$ or $n_3=1$. Then it follows that $G$ 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.
• Nov 4th 2009, 10:19 PM
NonCommAlg
Quote:

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, $G$ must have a normal subgroup, $N$ of order 15.
I have $30=2 \times 3 \times 5$
Then by using Sylow-theorems I have the possibilities for Sylow p-subgroups of G as followed:
$n_2=1,3,5,15;
n_3=1,10;
n_5=1,6$

Then I use counting argument to show that it can not be the case that $n_5=6$ and $n_3=10$. So, either $n_5=1$ or $n_3=1$. Then it follows that $G$ 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 $N_1,N_2$ be two subgroups of $G$ with $|N_1|=3, \ |N_2|=5.$ you've proved that either $N_1$ or $N_2$ is normal. thus $N_1N_2$ is a subgroup of $G.$ clearly $|N_1N_2|=15$ and so $[G:N_1N_2]=2.$

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