Let $p$ be a fixed prime and $\Gamma _2(p)$ the multiplicative groups of all matrices $ \begin{bmatrix}{a}&{b}\\{c}&{d}\end{bmatrix}$ where $a,b,c,d$ are integers, $a,d$ are equal to 1 module p and $b,c$ are multiples of $p$.
We consider $u = \begin{bmatrix}{1}&{p}\\{0}&{1}\end{bmatrix}$ and $v = \begin{bmatrix}{1}&{0}\\{p}&{1}\end{bmatrix}$
Now, we want to show that the subgroup $\left<{u,v}\right>$ is isomorphic with $F_a(u,v)$, where $F_a(u,v)$ denotes the free group with free basis $\left\{{u,v}\right\}$. In the proof of this fact, author says it suffices to show that $W\neq e_2$, where $W$ is an alternating product of non-zero powers of $u$ and $v$, and $e_2$ is the identity $2\times 2$ matrix.
My question is, how come is that $W\neq e_2$ implies $\left<{u,v}\right>$ is isomorphic with $F_a(u,v)$?
