Does GL(2,R) contain cyclic subgroup of order n ? Does GL(2,R) contain cyclic subgroup of order n ?  GL(2,R) is a General Linear group of order 2. 

I just can not figure out this. Can you tell me the answer with explanation?

I know that the group contains a Infinite cyclic subgroup generated by a matrix whose (2,1) element is 0 and others are 1, and a cyclic subgroup of order 2.
Your question is equivalent to finding an element of order $n$ in $GL(2,\Bbb{R})$.  

Try the matrix \begin{pmatrix}\cos 2\pi/n &amp;-\sin2\pi/n \\ \sin2\pi/n &amp; \cos 2\pi/n\end{pmatrix}
You can put the $n^{\rm{th}}$ root of unity and/or its powers on the diagonal. For example 
$$\begin{pmatrix}\omega&amp;0\\0&amp;\omega^\alpha\end{pmatrix},$$ where $\omega^n=1$ and $\gcd(\alpha,n)=1$.

Moreover, as order of the generator is finite, using basic linear algebra arguments you can also prove that any such cyclic subgroup will have a generator which is a diagonal matrix. Further any such diagonal generator will be of the form as given above.