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**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.