4
$\begingroup$

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.

  • 1
    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 sovable2012-08-07

2 Answers 2

1

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

  • 0
    Unique in the big group of order 30? I'm not sure, but why is that important?2012-08-11
1

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.

  • 0
    It is not that difficult, I think Sylow's theorem and Cauchy's theorem should suffice.2012-08-07