Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This isn't exactly a homework problem-- it's on a sample exam.

My first instinct is to look to matrix groups, since they are very often non-abelian and infinite, but I haven't had any luck.

share|improve this question
2  
I think your instinct was right in considering matrix groups. Maybe you could check that the subgroup of invertible upper triangular matrices is solvable (start with $2$ by $2$ matrices to get the idea). –  Joel Cohen Dec 14 '11 at 21:07
5  
If you know dihedral groups, try the infinite dihedral group. –  Mikko Korhonen Dec 14 '11 at 21:16
    
Thank you all. I'll look into these suggestions. –  user18297 Dec 14 '11 at 21:31
    
@yoyo: Does that work? Or do you mean something other than unit norm quaternions? Those contain $SU(2)$, and that is simple, so hardly solvable? –  Jyrki Lahtonen Dec 14 '11 at 21:38
    
@yoyo: True, but my main point was that in which way will the unit quaternions form a solvable group? –  Jyrki Lahtonen Dec 15 '11 at 6:23

2 Answers 2

up vote 5 down vote accepted

I confess that the only example that comes to mind is the one Joel mentions in the comments: the subgroup $B$ of $GL_n(K)$, where $K$ is an infinite field and $n \geq 2$, consisting of invertible upper triangular matrices. Let's work this out when $n = 2$. Then \[ B = \left\{\begin{pmatrix} a & b \\ 0 & d \end{pmatrix} \Bigg|\ a, d \in K^*, b \in K\right\}. \] This is infinite because $K$ is, and if $d$ is an element of $K^*$ not equal to $1$ then \[ \begin{pmatrix} 1 & 0 \\ 0 & d \end{pmatrix} \qquad \text{and} \qquad \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \] do not commute. There's a homomorphism $B \to K^* \times K^*$ sending a general element as above to the diagonal $(a, d)$. It's surjective and the kernel is the normal subgroup \[ U = \left\{\begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix} \Bigg|\ b \in K\right\} \] of $B$. And as yoyo says in the comments, $U$ is isomorphic to the additive group of $K$. We get an abelian tower \[ B \supset U\supset \{I\} \] If $n > 2$ then there are more steps in the tower; I think this example is written out in the general case in first chapter of Lang's Algebra. For $n = 3$ our $U$ is the Heisenberg group, which is interesting enough.

share|improve this answer
4  
2 by 2 unipotent matrices over a field $k$ is abelian (isomorphic $(k,+)$) –  yoyo Dec 14 '11 at 21:25
2  
Change the top-left 1 to be an arbitrary invertible ring element and you get AGL(1,K), a non-abelian solvable group. –  Jack Schmidt Dec 14 '11 at 22:04
    
@Jyrki This was meant to be the first step beneath the triangular group. I'll edit to make this clearer. –  Dylan Moreland Dec 14 '11 at 22:17
    
[I just didn't want to write out $\begin{pmatrix}* & * \\ 0 & *\end{pmatrix}$, and I think that caused trouble in the end.] –  Dylan Moreland Dec 14 '11 at 22:29
    
Things are now written out in gory detail. I thought the original answer was alright under a charitable interpretation of my words, but it's hopefully better now. –  Dylan Moreland Dec 14 '11 at 22:50

How about $S_3 \times \mathbb Z$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.