What are the subgroups of D12 and how can they be proven to be isomorphic to Dm?

[TheForumLord]

Homework Statement

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?

Homework Equations

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...

Thanks in advance!

 

[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.

 

[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 really 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!

 

[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.

 

[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

 

[boboYO]

yep that's right.

 

[TheForumLord]

Thanks a lot!

You have any idea about the first one?