# Why are these two free groups isomorphic?

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 $F_a(u,v)$?

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What author says that? Book, article, link...? –  DonAntonio Apr 21 '13 at 19:06

OTOH, it looks straightforward: a group generated by a set $\,X\,$ is free on that set iff any normal word (i.e., without subwords of the form $\,x\cdot x^{-1}\,$ and putting together powers of the same letter, i.e. $\,x\cdot x^2=x^3\,$ ) cannot be the trivial word (i.e., the unit element in the group), as then there are no non-trivial relations in that group...