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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.

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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
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
up vote 6 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.

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2 by 2 unipotent matrices over a field $k$ is abelian (isomorphic $(k,+)$) – yoyo Dec 14 '11 at 21:25
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 $S_3 \times \mathbb Z$.

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