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.
 A: Here is how you might go about figuring out the answer to this question without knowing it in advance. As Alan Wang says, the question is equivalent to finding a $2 \times 2$ matrix $X$ of order $n$; that is, $X^n = 1$ and $X^k \neq 1$ for $1 \le k < n$. 
The way we figure out how to take powers of a matrix is by looking at its eigenvalues and eigenvectors; in particular, if $X$ has eigenvalues $\lambda_1, \lambda_2$ then $X^n$ has eigenvalues $\lambda_1^n, \lambda_2^n$, so if $X^n = 1$ this means that $\lambda_1$ and $\lambda_2$ must both be $n^{th}$ roots of unity. In addition, the fact that the characteristic polynomial of $X$ is real means that $\lambda_1$ must be the complex conjugate of $\lambda_2$. So let's try to find $X$ such that its eigenvalues are
$$\lambda_1 = e^{\frac{2\pi i}{n}}, \lambda_2 = e^{-\frac{2\pi i}{n}}.$$
This ensures that $X^k \neq 1$ because its eigenvalues are not equal to $1$.
This would be easy if we needed a complex matrix, but we need a real matrix, so we need to be more careful. It's not hard to show that in order for $X$ to be real, it's necessary and sufficient for the eigenvectors $v_1, v_2$ associated to the above two eigenvalues to also be complex conjugates of each other. 
So let's take $v_1 = \left[ \begin{array}{c} 1 \\ i \end{array} \right], v_2 = \left[ \begin{array}{c} 1 \\ -i \end{array} \right]$. If you compute what $X$ must be from here you'll end up with either Alan Wang's matrix or its inverse. But this way you don't have to know what a rotation matrix is in advance, and in fact you can discover rotation matrices this way. 

The point of doing things this way is that the strategy I am following 1) can be followed without knowing in advance what form the answer takes, using only general facts about linear algebra, and 2) has the property that if it turned out that the answer to the question was no, I would end up with a proof of that fact (which pulling the answer out of thin air does not accomplish). For example, if you ask the same question of $GL_2(\mathbb{Z})$, the answer is already no for $n = 5$, and the reason is that it is not possible for the characteristic polynomial of $X \in GL_2(\mathbb{Z})$ to have a primitive fifth root of unity as a root (exercise). 
A: What you're overlooking is that
$$ \mathbb{C} \subseteq M_{2 \times 2}(\mathbb{R}) $$
We can define an embedding by sending $\mathbb{R}$ to scalar matrices, and $i$ to any square root of $-1$; for example,
$$ a + b i \mapsto \begin{pmatrix} a & b \\ -b & a \end{pmatrix} $$
A: 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 &-\sin2\pi/n \\ \sin2\pi/n & \cos 2\pi/n\end{pmatrix}
A: You can put the $n^{\rm{th}}$ root of unity and/or its powers on the diagonal. For example 
$$\begin{pmatrix}\omega&0\\0&\omega^\alpha\end{pmatrix},$$ 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.
