Fundamental group of Klein Bottle It is well know that the fundamental group of the Klein Bottle $G$ is defined by
$$G=BS(1,-1)=\langle a,b: bab^{-1}=a^{-1}\rangle.$$
I know, for example  that $BS(1,2)$ can be defined as the group
$$BS(1,2)=\langle A,B\rangle $$
where
$$A=\left(
\begin{array}{cc}
 1 & 1 \\
 0 & 1 \\ 
\end{array}
\right), B=\left(
\begin{array}{cc}
 2 & 0 \\
 0 & 1 \\
\end{array}
\right).$$
These matrices satisfy the equation $BAB^{-1}=A^{2}$ and are free : there is not an integer $k$ such that $A^{k}=I$ or $B^{k}=I$. This implies that we obtain an "explicit description" of $BS(1,2)$ as the group generated by $A$ and $B$.
I know that the matrices
$$a=\left(
\begin{array}{cc}
 1 & 1 \\
 0 & 1 \\ 
\end{array}
\right), b=\left(
\begin{array}{cc}
 -1 & 0 \\
 0 & 1 \\
\end{array}
\right)$$
satisfy the relation $bab^{-1}=a^{-1}$ but $b^{2}=I$. This implies that $BS(1,-1)$ is not generated by $a$ and $b$.

My question is: is there an "explicit description" for $G=BS(1,-1)$ with matrices or maybe another couple of objects?

 A: Starting from the representation described by @PaulPlummer as a group of isometries of $\mathbb R^2$, you can obtain a representation as a group of linear transformations of $\mathbb R^3$.
To do this, one uses a standard embedding of the isometry group of $\mathbb R^2$ as a subgroup of $GL(3,\mathbb R)$. Each isometry of $\mathbb R^2$ can be written uniquely as $P \mapsto MP + Q$ for some $M \in O(2,\mathbb R)$ and some $Q \in \mathbb R^2$ (vectors are written in column format). The representing element of $GL(3,\mathbb R)$ is the matrix written in block form as $\pmatrix{M & Q \\ 0 & 1}$. If you then represent a column 2-vector $P$ as a column 3-vector $\pmatrix{P \\ 1}$ then matrix multiplication gives you
$$\pmatrix{M & Q \\ 0 & 1}\pmatrix{P \\ 1} =  \pmatrix{MP+Q \\ 1}
$$
For the Klein bottle group, the two generators are:

*

*A translation $1$ unit to the right:
$$P \mapsto M_1 P + Q_1, \qquad M_1 = \pmatrix{1 & 0 \\ 0 & 1}, \qquad Q_1 = \pmatrix{1 \\ 0}
$$

*A glide reflector, gliding up the $y$-axis by $1$ unit:
$$P \mapsto M_2 P + Q_2, \qquad M_2 = \pmatrix{-1 & 0 \\ 0 & 1}, \qquad Q_2 = \pmatrix{0 \\ 1}
$$
A: Finally , $G$ can be described as the group of $2\times 2$ matrices generated by two matrix $A,B$ such that $BAB^{-1}=A^{-1}$ and $A^{k}\neq I$, $B^{k}\neq I$ for all $k$. Let $A=\left(
\begin{array}{cc}
 a & b \\
 c &  d\\
\end{array}
\right)$ and $B=\left(
\begin{array}{cc}
 \lambda & 0 \\
 0 & \mu \\
\end{array}
\right)$ (to simplify computations). Working whith the equation $BAB^{-1}=A^{-1}$ and assuming that $ad-bc=1$ and $b,c\neq 0$ we have that
$A=\left(
\begin{array}{cc}
 a & b \\
 c &  a\\
\end{array}
\right)$ and $B=\left(
\begin{array}{cc}
 \lambda & 0 \\
 0 & -\lambda \\
\end{array}
\right)$
This family of matrices satisfy the relation $BAB^{-1}=A^{-1}$.
If $A=\left(
\begin{array}{cc}
 2 & 3 \\
 1 &  2\\
\end{array}
\right)$ and $B=\left(
\begin{array}{cc}
 2 & 0 \\
 0 & -2 \\
\end{array}
\right)$, then $BAB^{-1}=A^{-1}$  and $A^{k}\neq I$, $B^{k}\neq I$, for all $k$
Hence $G\equiv \langle A,B\rangle$.
