Let $G$ be generated by $\begin{pmatrix}1&p\\ 0&1\end{pmatrix},\begin{pmatrix}1&0\\ p&1\end{pmatrix}$ for prime $p$. Is $G$ is solvable? A proof.

Question

Here is a problem determining the solvability of the given group.

Let $G$ be a subgroup of $\text{GL}_2(\mathbb{R})$ generated by two elements $$ \begin{pmatrix}1 & p \\ 0 & 1 \end{pmatrix}, \begin{pmatrix}1 & 0 \\ p & 1 \end{pmatrix} $$ where $p$ is a prime integer. Determine whether $G$ is solvable or not.

By letting $A$ the former matrix, the latter is $A^t$. I guess that $G$ is not solvable as it is isomorphic to the free group $F_2$ with two generators. This is because there is no nontrivial relation that $A$ and $A^t$ satisfies: that is, $A^m \neq I$, $(A^t)^n \neq I$ for every nonzero integer $m$ and $n$, and their arbitrary product cannot also form an identity matrix (I think.) However, as $A_5$ is generated by two elements, namely a $5$-cycle and a product of two disjoint transpositions, $F_2$ cannot be solvable as $A_5$ is a factor group of $F_2$, which is not solvable. Hence $G$ is also not solvable.

My question is, is my guess correct? Any comments or suggestions are welcome.

Answer 1

For any prime $r$, the group $SL_2(r)$ is generated by $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$.

Thus $G$ has $SL_2(r)$ as a quotient for any prime $r \neq p$, and so $G$ is not solvable.

Answer 2

Denote $$ x= \begin{pmatrix} 1 & p\\ 0 & 1 \end{pmatrix},\ y= \begin{pmatrix} 1 & 0\\ p & 1 \end{pmatrix}. $$ Since we only need to prove that the group $G=\langle x,y\rangle$ is not solvable, we can reason this way.

If $G$ is solvable, then it has a nontrivial normal abelian subgroup $A$ (for example, the last nontrivial member of the derived series of $G$). Let $$ v= \begin{pmatrix} a & b\\ c & d \end{pmatrix}\in A,\ v\neq I. $$ We have $$ z=x^{-1}vx= \begin{pmatrix} a-pc & *\\ c & d+pc \end{pmatrix}. $$ We don't need an entry at position $(1,2)$ and I didn't calculate it. Since $v,z\in A$ and $A$ is abelian, it follows that matrices $v$ and $z$ commute. The entry of the matrix $zv$ at position $(2,1)$ is $ac+cd+pc^2$. At the same position in the matrix $vz$ there is $ac+cd-pc^2$. Hence $c=0$.

I think the rest is clear, because we still have matrix $y$ in reserve.