Finding $(r^{'})^{1800}$ where $r^{'}$ is rotation We are in $\mathbb{R}^{3}$.
We have $r^{'}=t_{(1,0,-1)}\circ r$ where $r$ is a linear rotation by $\frac{\pi }{6}$ around the axis $\mathbb{R}(1,1,1)$.
I need to find $(r^{'})^{1800}$.
I realize that $(1,0,-1)$ is orthogonal to the axis of rotation and that $1800=150\times 12$ so $r^{1800}$ is the identity matrix. 
But apparently $(r^{'})^{1800}$ is identity too, why? What happened to the translations?
Thanks in advance.
 A: We have that $r'^{1800}=r^{1800}+(r^{1799}+r^{1798}+...+r+1)(1,0,-1)$
But $(r-1)(r^{1799}+r^{1798}+...+r+1)=r^{1800}-1=0$.
Since $(1,0,-1)$ is orthogonal to the axis of rotation of $r$ then the sum $(r^{1799}+r^{1798}+...+r+1)(1,0,-1)$ is also orthogonal to that axis.
That means that $(r^{1799}+r^{1798}+...+r+1)(1,0,-1)$ is not in the kernel of $r-1$. The kernel of $r-1$ are the points fixed by the rotation, i.e. the axis of rotation. 
Since still $(r-1)(r^{1799}+r^{1798}+...+r+1)(1,0,-1)=0$ it must be that $(r^{1799}+r^{1798}+...+r+1)(1,0,-1)=0$.
Therefore $r'^{1800}=r^{1800}$.
A: It turns out that $(r')^{12}$ is the identity function. Since $1\,800=150\times12$, then...
Actually, it is simpler to do the computations supposing that $r'=t_{(1,0,0}\circ r$, where $r$ is a linear rotation of $\frac\pi6$ radians around the $\mathbb{R}(0,1,0)$ axis (see it as a chage of basis). Then$$r'(x,y,z)=\left(\frac{\sqrt{3}}2x-\frac12z+1,y,\frac12x+\frac{\sqrt{3}}2z\right)$$and so$$(r')^3(x,y,z)=\left(-z+\frac{\sqrt{3}+3}2,y,x+\frac{\sqrt{3}+1}2\right).$$Therefore$$(r')^6(x,y,z)=\left(1-x,y,-z+\sqrt{3}+2\right),$$and so, as I wrote, $(r')^{12}$ is the identity.
