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If the fundamental region for a torus T is a square, show that T has an isometry of order 4 induced by a rotation of the plane.

Secondly, if the fundamental region is a rhombus with angles $\pi/3$ and $2\pi/3$, show that T has an isometry of order 6 induced by a rotation of the plane.

We have defined the torus T as T=$R^2$$ / \Gamma$, where $\Gamma$ is generated by two linearly independent translations, $t_1$ and $t_2$. The fundamental region of a T is a parallelogram, but in this special case we are assuming that this parallelogram is actually a square. I have no clue where to start with this problem, so any help is appreciated.

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Suppose $R^2$ is generated by $u,v$ and $\Gamma$ is generated by the translations $t_u$ and $t_v$. The linear map $f$ defined by the matrix $A=\pmatrix{0 & -1\cr 1& 0}$ in $u,v$ is the matrix of the rotation of angle ${\pi\over 2}$. Since $f\Gamma f^{-1}=\Gamma$, $f$ induces an automorphism of $T^2$. Its order is 4.

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    I don't know why the matrix A is the matrix of the rotation $\pi/2$ – 2017-02-08
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    $\pmatrix{ cos(\theta) & -sin(\theta) \cr sin(\theta) & cos(\theta)}$ is the matrix of the rotation of angle $\theta$. Take $\theta=\pi/2$. – 2017-02-08
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    Got it. That part makes sense now. How does this lead us to conclude that the order of f is 4 though? – 2017-02-08