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What are the steps in showing a group of order 30 is solvable/non-solvable?

I don't know how to proceed. All I know is that the group either has a group of order $5$ or $3$. I don't need all the steps for this problem, just an outline what to do.

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It has both subgroups of order $5,3$ as they are prime dividing the order of the group (not to say this is relavent to the question), the subgroup of order $5$ is probably the way to go since it is solvable and the quotient is also sovable – Belgi Aug 7 '12 at 8:54
up vote 1 down vote accepted

Some ideas:

1) Show that such a group always has one unique group of order 3 or one unique group of order 5

2) Using the above show that such a group always has a subgroup of order 15

3) Now use the following: if a group $\,G\,$ has a normal sbgp. $\,N\,$ s.t. both $\,N\,$ and $\,G/N\,$ are solvable, then $\,G\,$ is solvable

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This is a good strategy. – Bombyx mori Aug 7 '12 at 9:07
How do you show (2)? Can I use the argument that I can find a subgroup $N$ with order $15$ to make a composition series $G \geq N \geq \{e\}$ which is solvable. Is this right? – apple mcdonald Aug 11 '12 at 14:43
The product of the normal subgroup of order 3 or 5 by the other one is a subgroup of order 15, and yes: you can do directly that abelian series since the only group of order 15 is the (abelian, of course) cyclic one. – DonAntonio Aug 11 '12 at 14:49
Oh yes, that's a corollary I read somewhere. And that subgroup of order 15 is unique right? – apple mcdonald Aug 11 '12 at 15:26
Unique in the big group of order 30? I'm not sure, but why is that important? – DonAntonio Aug 11 '12 at 16:04

There are only four groups of order 30. They are $$S_{3}\times \mathbb{Z}_{5}, D_{5}\times \mathbb{Z}_{3}, C_{30}, D_{15}$$

$C_{30}$ is abelian so must be solvable. $S_{3}$ has a unique normal subgroup of order 3, and its quotient must be abelian. $D_{5}$ and $D_{15}$'s quotient with their normal subgroups must be abelian if you think of them as semidirect product. In fact, all groups of order 30 come as semidirect product of $\mathbb{Z}_{3}\times \mathbb{Z}_{5}$ with $\mathbb{Z}_{2}$. So they must be solvable. See this lecture note.

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Showing these are all the groups of order $30$ must be a difficult task... – Belgi Aug 7 '12 at 9:14
It is not that difficult, I think Sylow's theorem and Cauchy's theorem should suffice. – Bombyx mori Aug 7 '12 at 16:44

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