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. 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.
 A: 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.
A: 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.
