# Group of order 21 - prove it's not comutative but solvable

• Aug 26th 2010, 09:55 AM
Group of order 21 - prove it's not comutative but solvable
Let G be a group of order 21, with at least 3 elements of order 3. Prove that G is not commutative, but solvable.

I believe it's got something to do with the 3-Sylow sub-group of G, and there can be either 1 3-Sylow-subgroup or 7 3-Sylow-subgroup. How do I continue this (Surprised)?

Thanks!
• Aug 26th 2010, 11:02 AM
tonio
Quote:

Let G be a group of order 21, with at least 3 elements of order 3. Prove that G is not commutative, but solvable.

I believe it's got something to do with the 3-Sylow sub-group of G, and there can be either 1 3-Sylow-subgroup or 7 3-Sylow-subgroup. How do I continue this (Surprised)?

Thanks!

A commutative group of order $\displaystyle 21=3\cdot 7$ has normal Sylow subgroups of order 7 and 3 and thus one of

each, from where it follows that if our group has three different groups of order 3 it can't be commutative.

Nevertheless, such a group does have one single Sylow 7-subgroup $\displaystyle P$ which is then normal, and then

we get that both $\displaystyle P\,\,\,and\,\,\,G/P\cong C_3$ are solvable groups (even abelian), and thus also $\displaystyle G$ is.

Tonio
• Aug 26th 2010, 02:10 PM
hmm.. but if n_3 divides 7=21/3, and n_3=1mod(7), then n_3 is either 1 or 7. Why do you claim it has three different groups of order 3? How comes it doesn't have only one?

And, how do you use the fact that there are at least 3 elements of order 3 in the group, in order to find out there exist three different groups of order 3?

Thanks :)
• Aug 26th 2010, 07:00 PM
tonio
Quote:

hmm.. but if n_3 divides 7=21/3, and n_3=1mod(7), then n_3 is either 1 or 7. Why do you claim it has three different groups of order 3? How comes it doesn't have only one?

My bad: i misread the question. You said there're three different elements of order 3 and I read "three different subgroups of order 3". It never minds: if there are more than 2 elements of order 3 then there are more than 1 subgroup of order 3 and thus the argument stays the same.
And you're right: there are either 1 or 7 subgroups of order 3 in a group of order 21...

Tonio

And, how do you use the fact that there are at least 3 elements of order 3 in the group, in order to find out there exist three different groups of order 3?

Thanks :)

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• Aug 26th 2010, 10:08 PM
I still can't seem to understand how that gets me closer to the solution - my question stays the same :)
• Aug 27th 2010, 04:35 AM
tonio
Quote:

solvable then G is solvable, or even easier: look at the series $\displaystyle 1\leq P \leq G$ ...