# Homework Help: ABSTRACT- Dihedral Group

1. Feb 4, 2010

### TheForumLord

1. The problem statement, all variables and given/known data
1. Let Dn be the dihedral group of order 2n, n>2 .
A. Prove that each non-commutative sub-group of Dn isomorphic to Dm for some m.
B. Who are all the non-commutative subgroups of D12?

2. Let G be the group of all the matrices from the form:
1 a c
0 1 b
0 0 1

where a,b,c are in Z and the binary action is matrix multiplication.
What is the center of this group?
2. Relevant equations
3. The attempt at a solution
I'll be glad to receive some guidance in these two questions... All the ways I can think about are very not elegant ... There are probably some elegant ways out there...

2. Feb 4, 2010

### CompuChip

Have you learned about generators yet?
Because that is how I would handle this question (every subgroup of Dn is generated by some set of elements from Dn).

For the second question, let A be an arbitrary matrix and Z an element from the center. Then you can write
$$A = I + \begin{pmatrix} 0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \end{pmatrix}$$
and
$$Z = I + \begin{pmatrix} 0 & x & y \\ 0 & 0 & z \\ 0 & 0 & 0 \end{pmatrix}$$
where I is the 3 x 3 identity matrix. Now it should be true that AZ = ZA for all values of a, b and c in Z.

3. Feb 4, 2010

### TheForumLord

Well, let's see:
If we'll define :
$$D_{n} = ( 1, s , s^{2} ,..., s^{n-1} , a, sa, s^{2}a ,..., s^{n-1}a )$$
Then the generators of Dn are a and s (where s is a rotation, a = symmetry ) ... How can we continue from here?

About the second one:
According to your guidace, we only need to figure out for which x,y,z:
$$\begin{pmatrix} 0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \end{pmatrix}*\begin{pmatrix} 0 & x & y \\ 0 & 0 & z \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & x & y \\ 0 & 0 & z \\ 0 & 0 & 0 \end{pmatrix} * \begin{pmatrix} 0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \end{pmatrix}$$

Hence:
$$\begin{pmatrix} 0 & 0 & xc \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & az \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

Which implies: xc=az ... We can't find any further conditions on x,y,z and I realy don't think that the condition "xc=az" is all we need...

Hope you'll be able to give me some furhter help

Thanks a lot!

Last edited: Feb 4, 2010
4. Feb 4, 2010

### boboYO

For the matrix question:

What you did is you found out that if xc=az then the matrices commute. But that's not what the question is asking.

We're trying to find the elements that commute with -every- other element. For example the identity matrix commutes with every element. Are there any other ones? It should be quite easy, you've already done most of the work.

5. Feb 4, 2010

### TheForumLord

Well... We've found a condition for x and z ... But we can show that for different choices for a,c - A and Z won't commute...
Hence - > x=z=0 and y can be defined to be any integer...Hence- all of the elements in the center are from the form:
$$Z = I + \begin{pmatrix} 0 & 0 & y \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$ where y is an integer.... Am I right in this one?

Thanks

6. Feb 4, 2010

### boboYO

yep that's right.

7. Feb 5, 2010

### TheForumLord

Thanks a lot!

You have any idea about the first one?