I'm reading David R. Finston and Patrick J. Morandi's book Abstract Algebra: Structure and Application and in section 10.5.2 "Point Groups of Wallpaper Patterns" page 167 I'm confused during a theorem proof, where it asserts that in Wallpaper Pattern, a rotation in its Point Group in lattice basis is in O2(Z):
Theorem 10.42. Let $W$ be a wallpaper pattern with lattice $L_{v,w}$ and point group $G_0$: Then $G_0$ is one of the ten groups $C_1,C_2,C_3,C_4,C_6,D_1,D_2,D_3,D_4$, or $D_6$.
Proof of the Theorem. From Corollary 10.36, $G_0$ is either cyclic or dihedral. If $G_0$ contains no rotation, then $G_0 \cong C_1$ or $G_0 \cong D_1$, i.e., consists of just the identity or the identity and a reflection. Otherwise $G_0 \cong C_n$ or $G_0 \cong D_n$ with $n>1$ and contains a rotation $r$ through some minimal angle $\frac{2\pi}{n}$. The matrix for $r$ with respect to the standard basis is then $\begin{pmatrix}\cos \frac{2\pi}{n} & -\sin \frac{2\pi}{n}\\ \sin \frac{2\pi}{n} & \cos \frac{2\pi}{n}\end{pmatrix}$ while the matrix for $r$ with respect to the lattice basis $\{v,w\}$ is in $O_2(\mathbb Z)$ and these matrices are similar. Since similar matrices have the same trace, it follows that $2 \cos \frac{2\pi}{n}$ is an integer, i.e., $\cos \frac{2\pi}{n} \in \{0, \pm 1, \pm \frac12\}$ and $n\in \{1,2,3,4,6\}$. Finally, if $G_0$ contains no reflection, then $G_0 \cong C_i$ for one of these $i$. Otherwise $G_0 \cong D_i$.
I agree with the proof except one part made me lost -- why "the matrix for $r$ with respect to the lattice basis $\{v,w\}$ is in $O_2(\mathbb Z)$"?
Take an example, if $L_{v,w}$ is a hexagon lattice, take $v=\begin{pmatrix}1 \\ 0\end{pmatrix}$ and $w=\begin{pmatrix}\frac12 \\ \frac{\sqrt{3}}{2}\end{pmatrix}$, and $r$ as rotating counterclockwise $\frac\pi 3$, i.e. rotating $v$ to $w$ and $w$ to $\begin{pmatrix}-\frac12 \\ \frac{\sqrt{3}}{2}\end{pmatrix}$. Then in the basis of $\{v,w\}$, $v=\begin{pmatrix} 1 \\ 0\end{pmatrix}$, $w=\begin{pmatrix} 0 \\ 1 \end{pmatrix}$; so $r$ rotates $v$ to $w$ and $w$ to $\begin{pmatrix} -1 \\ 1 \end{pmatrix}$; so the matrix is $r_{v,w}=\begin{pmatrix} 0 & -1 \\ 1 & 1 \end{pmatrix}$ , it doesn't sound that $r_{v,w}\in O_2(\mathbb Z)$?
Before hand, in pg 166 it's only proved that
Lemma 10.40. Let $W$ be a wallpaper pattern with lattice $L_{v,w}$ and point group $G_0$. Then $G_0$ acts on $L_{v,w}$ in the sense that if $A\in G_0$ and $u\in L_{v,w}$, then $Au\in L_{v,w}$. In particular, with respect to the basis $\{v,w\}$ of $\mathbb R^2$, the elements of $G_0$ lie in $GL_2(\mathbb Z)$.
This only proves in $r_{v,w}\in GL_2(\mathbb Z)$, not $O_2(\mathbb Z)$.
So, where did I miss?