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Let Nil be the subgroup of $GL_3(\mathbb{R})$ given by matrices of the form $$ \left( \begin{array} 11 & x & z\\ 0 & 1 & y\\ 0 & 0 & 1 \end{array} \right) $$ with $x,y,z \in \mathbb{R}$. The collection of matrices with $x = 0$ comprises a normal subgroup isomorphic to $\mathbb{R}^2$, and so there is a short exact sequence $$ 1 \longrightarrow \mathbb{R}^2 \longrightarrow \text{Nil} \longrightarrow \mathbb{R} \longrightarrow 1; $$ in other words, Nil is an extension of $\mathbb{R}$ by $\mathbb{R}^2$.

I have been told that this fact explains why Nil does not contain a rank-two free group as a subgroup, but I have not been able to piece together the details of the argument. Can somebody help me out here? Generally, if $G$ is an extension of $H$ by $K$, under what hypotheses on $H$ and $K$ can one conclude that $G$ does not contain a rank-two free group?

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up vote 3 down vote accepted

This short exact sequence shows $\mathrm{Nil}$ is solvable, but the free group is not. For example, there are quotient groups of the free group $\langle x,y\rangle$ that are not solvable: $x\mapsto (12), y\mapsto (12345)$ maps the free group onto th unsolvable group $S_5$, so the free group itself cannot be solvable. Hence it cannot be a subgroup of the solvable group $\mathrm{Nil}$.

To answer your last question: $H,K$ solvable is a sufficient condition for $G$ not to contain a rank-two free group.

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Indeed, virtually soluable is sufficient - look up the Tits alternative. – user1729 Jul 27 '11 at 10:49

In fact, for any $k>0$, an extension of $H$ by $K$ has a rank-$k$ free group as a subgroup if and only at least one of $H$ and $K$ does.

The proof follows from the Schreier formula for the free rank of a subgroup of a free group of finite index, and from the result that any nontrivial normal subgroup of infinite index in a free group is free of infinite rank.

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