Fundamental group of Klein Bottle

Solution 1:

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:

1. A translation $1$ unit to the right: $$P \mapsto M_1 P + Q_1, \qquad M_1 = \pmatrix{1 & 0 \\ 0 & 1}, \qquad Q_1 = \pmatrix{1 \\ 0} $$
2. A glide reflector, gliding up the $y$-axis by $1$ unit: $$P \mapsto M_2 P + Q_2, \qquad M_2 = \pmatrix{-1 & 0 \\ 0 & 1}, \qquad Q_2 = \pmatrix{0 \\ 1} $$

Solution 2:

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 $G\equiv \langle A,B\rangle$.