Proving that a set of matrices is an abelian group 
Prove that the set of matrices in the form of
   $\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha
 &\cos \alpha \end{array}\right]$ (while $\alpha \in R$) with the
  operation of matrix multiplication is an abelian group.

Could someone please point me some direction? I know how to multiply matrices or what's the definition of an abelian group but... How can I apply it here? And show such thing, for such a large group of possible alphas?
EDIT: OK, so thanks to your advices, I tried to move a little but I think I don't get the idea behind it... On Wikipedia, one can find that "For all a, b in A, the result of the operation a • b is also in A." - does that mean I have to try multiply every term from the matrix by every other? I mean: $cos \alpha \cdot sin \alpha$ and so on? Or did I get it all wrong?
 A: The addition formulas for $\sin$ and $\cos$ yield
$$
\begin{align}
&\begin{bmatrix}
\cos(\alpha)&-\sin(\alpha)\\
\sin(\alpha)&\cos(\alpha)
\end{bmatrix}
\begin{bmatrix}
\cos(\beta)&-\sin(\beta)\\
\sin(\beta)&\cos(\beta)
\end{bmatrix}\\
&=
\begin{bmatrix}
\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)
&-\sin(\alpha)\cos(\beta)-\cos(\alpha)\sin(\beta)\\
\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)
&\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)
\end{bmatrix}\\
&=\begin{bmatrix}
\cos(\alpha+\beta)&-\sin(\alpha+\beta)\\
\sin(\alpha+\beta)&\cos(\alpha+\beta)
\end{bmatrix}
\end{align}
$$
The rest of the group requirements are pretty immediate after this.
A: Go to your list of group axioms, and get cranking.  Trig identities will be your friend.
A: That it is an group is obvious, the set is $SO(2)$, the only thing you need to prove is that it is abelian. 
Just compute the product to see it.
The special orthogonal group $SO(n)$ is the group of all matrices such that 
$$O \cdot O^T = O^T \cdot O= I$$ and $\det(O)=1$. 
So we have 
$$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot \begin{pmatrix} 
a & c\\ b & d \\ \end{pmatrix}=\begin{pmatrix} 1 & 0\\ 0 & 1 \\ \end{pmatrix}$$
So we have
\begin{align*}
1&=a^2 +b^2 \\
0&= ac + bd\\
0&= ac + bd\\
1&= c^2+d^2 \\ 
\end{align*}
and $$\det 
\begin{pmatrix} a & b\\c & d \\ \end{pmatrix}=ad -bc=1$$
As 
$$1=\sin^2(\alpha)+\cos^2(\alpha)=a^2 +b^2 $$
As $a\in(-1,1)$ you can chose $\alpha$ such that $\cos(\alpha)=a$ and we know that $b=\pm \sin(\alpha)$. It follows that $d=\cos(\alpha)$ and $c=-b$.
So your set ist just that group. Now we only need to prove it is a group.
At first we show that the general linear group is a group. The general linear group are all invertible $n\times n$ matrices and called $GL_n$ together with the matrix multiplication. 
That the matrix multiplication is associative is obvious as the composition of functions is associative and the matric multiplication is the composition of two linear functions. The neutral element ist the identiy matrix, and every element has an inverse as we definied the group to be all invertible matrices, and as the inverse of a matrix is invertible the inverses are in $GL_n$ too. We need to prove that the set is closed under multiplication, as we know that 
$$\det(A\cdot B)=\det(A)\cdot \det(B)$$
and $\det(A)\neq 0 \neq \det(B)$ we see it is closed under multiplication. 
Hence $GL_n$ is a group, now we show that $SO(n)$ is a subgroup of $GL_n$.
It is closed under multiplication as 
$$(AB)^T = B^T A^T$$
and $$B^T A^T A B = I$$ 
when $A^T A=I$ and $B^T B=I$
and $1\cdot 1 =1$.
Hence $SO(n)$ is a group with matrix multiplication.
And as
$$\begin{pmatrix} \cos(\alpha) & -\sin(\alpha)\\ \sin(\alpha) & \cos(\alpha)\\ \end{pmatrix} \cdot \begin{pmatrix} \cos(\beta) & -\sin(\beta)\\ \sin(\beta)& \cos(\beta)\\ \end{pmatrix} = \begin{pmatrix} \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta) 
& -(\cos(\alpha)\sin(\beta)+\cos(\beta)\sin(\alpha)\\  (\cos(\alpha)\sin(\beta)+\cos(\beta)\sin(\alpha)& \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta) \end{pmatrix}$$
A: An idea for another approach: define the abelian group
$$T:=\{z\in\Bbb C\;;\;|z|=1\iff z=e^{i\alpha}\;,\;\alpha\in\Bbb R\}\;,\;\text{and the map}\;\;\phi: T\to SO_2(\Bbb R)\;\text{defined by}$$
$$\phi(e^{i\alpha}):=\begin{pmatrix}\cos\alpha&\!\!-\sin\alpha\\\sin\alpha&\;\cos\alpha\end{pmatrix}$$
Check the above is an onto set map and  it induces an operation on $\,SO_2(\Bbb R)\,$ taking product of matrices (as expected) and making this last set into a group which is then trivially abelian.
