Prove that $(T, \cdot)$ is a group and find all of its subgroups 
Let $T = \{ A \in M_2(\mathbb{R}) \mid A^tA = I_2 \}$. Prove that $(T, \cdot)$ is a group, and then find all of its finite subgroups.

For the first point of this problem, we need to check the group axioms:
The group is associative, since the multiplication of matrices on $M_2(\mathbb{R})$ is also associative. The identity element is, obviously, $I_2$. Now, we need to find the inverse element:
Let $X'$ be the inverse element, so $XX' = X'X = I_2$, for any $X \in T$.
Since $X \in T$, we know that $X^tX = I_2$. So $X^tX = X'X$, and by knowing that $X$ is invertible, we conclude that $X' = X^t$.
All we need to do now, is check if $X^t \in T$, that is $XX^t = I_2$.
Now, this is where I have some doubts. From the equality $XX' = I_2$, we can easily prove that $X^t \in T$, but I'm not really sure if this is a good proof, or if it is correct.
For the second point of the problem, I've only managed to found the cyclic subgroups and I don't know how to find other subgroups.
Thank you!
 A: Let's check the group axioms:


*

*closure. Suppose $A,B \in T$. Then $(AB)^t(AB) = B^tA^tAB = B^tIB = B^tB = I$

*identity. As you pointed out, $I\in T$ is the identity.

*inverse. If $A\in T$ then $A^tA = AA^t = I$, so $A^t$ is the inverse of $A$. We know that $A^t \in A$ because $(A^t)^tA^t = AA^t = T$.

*associative. This follows since multiplication in $M_2(\mathbb{R})$ is associative.


In fact, the group $T$ is just $O(2)$, the group of orthogonal $2\times2$ matrices. This is also the group of linear isometries of $\mathbb{R}^2$, which are rotations and reflections. Thus the finite subgroups of $T$ are $C_n$, cyclic groups of order $n$ generated by a real matrix similar to $\left(\begin{array}{cc} e^{\frac{2i\pi}{n}} & 0 \\ 0 & e^{-\frac{2i\pi}{n}} \end{array} \right)$ and $D_{2n}$, dihedral groups of order $2n$, generated by a real matrix similar to $\left(\begin{array}{cc} e^{\frac{2i\pi}{n}} & 0 \\ 0 & e^{-\frac{2i\pi}{n}} \end{array} \right)$ and $\left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$. See if you can prove that these are in fact all the subgroups of $T$ there are.
