Math Is Fun Forum

laipou

Member

Registered: 2009-09-19

Posts: 24

non-abelian groups of order 27

Question:
Let G be a non-abelian group of order 27.
1.Show that every subgroup of order 9 contains the center.
2.How many subgroups are there?

I think I have to use the Sylow thms,but I don't know how.

Avon

Member

Registered: 2007-06-28

Posts: 80

Re: non-abelian groups of order 27

I don't think Sylow's theorems can be used to solve these problems.

Some results that I think are more useful are:
If G/Z(G) is cyclic then G is abelian.
If G is a finite p-group then Z(G) is non-trivial.
If G is a p-group and H is a subgroup of index p in G then H is a normal subgroup of G.

I assume that part 2 is supposed to be "How many subgroups of order 9 are there?".

laipou

Member

Registered: 2009-09-19

Posts: 24

Re: non-abelian groups of order 27

Thanks for the hints.
Now I know 1.The center of G(Z(G)) must be of order 3.
                  2.Every subgroup of order 9 is abelian.
By 1 and 2,I know every subgroup(H) of order 9 contains the center(otherwise HZ(G) is abelian and has order 27 ->|)

How many subgroups of order 9 are there? <-- I still have no idea.
Need more suggestions.

Last edited by laipou (2009-09-22 17:09:19)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: non-abelian groups of order 27

First prove that no element can be of order 9.  Then for any element of order 3, take a look at the set:

<Z(G), x>

How many elements in that subgroup?  How many different ways are there to do this (distinctly!)?  And finally, could there be any subgroup of order 9 that does not take this form?

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Avon

Member

Registered: 2007-06-28

Posts: 80

Re: non-abelian groups of order 27

First prove that no element can be of order 9.

Z9 has an automorphism with order 3, so there is a semi-direct product of Z9 and Z3 that is non-abelian.

To answer the second part you should consider the subgroups of G/Z(G).

laipou

Member

Registered: 2009-09-19

Posts: 24

Re: non-abelian groups of order 27

Thanks for helping.
I think now I know how to solve this problem.