Proof, wheather a subset of a Group is a Subgroup I have to check, weather the following subset of a group is also a subgroup:
$$U = \left\{ \begin{pmatrix} a & -b \\ \overline{b} & \overline{a} \end{pmatrix} \in GL(2, \mathbb{C}) \bigg\vert |a|^2 + |b|^2 = 1, \ a,b \in \mathbb{C} \right\} \subset (GL(2,\mathbb{C}), \cdot)$$
I did the following:
1) $\forall x,y \in U: x\cdot y \in U$
$$\begin{pmatrix} a & -b \\ \overline{b} & \overline{a} \end{pmatrix} \cdot \begin{pmatrix} a' & -b' \\ \overline{b'} & \overline{a'} \end{pmatrix} = \begin{pmatrix} aa'-b\ \overline{b'}  & -ab'-b\ \overline{a'} \\ \overline{b}a'+\ \overline{ab'} & -\overline{b}b' +\overline{a'a} \end{pmatrix}$$
For $|a|^2 + |b|^2 = 1$ and $|a'|^2 + |b'|^2 = 1$. To be honest - I am unable to check weather that is $\in U$ - how can this be done? 
2) $e\in U$
Let $a = 1$ and $b = 0$ we have $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = e$ and $|1|^2 + |0|^2 = 1 \Rightarrow e \in U$.
3) $\forall x \in U: x^{-1} \in U$
What I did was just calculating the inverse matrix but I am stuck:
$$\left( \begin{array}{cc|cc} a & -b & 1 & 0 \\ \overline{b} & \overline{a} & 0 & 1 \end{array}\right)$$
$$a,\ \overline{a}\ne0,\ I \cdot \frac{1}{a}; II \cdot \frac{1}{\overline{a}} $$
$$\left( \begin{array}{cc|cc} 1 & -\frac{b}{a}  & \frac{1}{a} & 0 \\ \frac{\overline{b}}{\overline{a}} & 1 & 0 & \frac{1}{\overline{a}} \end{array}\right)$$
And well - now what ever I add, multiply or subtract I am not able to get the unit matrix on the left side. Is there anything I've overlooked? Because since 
$$\det\left| \begin{array}{cc} a  & -b \\ \overline{b} & \overline{a} \end{array} \right| = a\overline{a} + b\overline{b} = |a|^2 + |b|^2 = 1 \ne 0$$
every matrix in $U$ should be invertible, right?  
Thank you very much for your help!
FunkyPeanut

For part (1):
$$\begin{pmatrix} a & -b \\ \overline{b} & \overline{a} \end{pmatrix} \cdot \begin{pmatrix} a' & -b' \\ \overline{b'} & \overline{a'} \end{pmatrix} = \begin{pmatrix} aa'-b\ \overline{b'}  & -ab'-b\ \overline{a'} \\ \overline{b}a'+\ \overline{ab'} & -\overline{b}b' +\overline{a'a} \end{pmatrix}$$
You want here that 
$$
(aa'-b\ \overline{b'})(\overline{aa'-b\ \overline{b'}}) + (ab'+b\ \overline{a'})(\overline{ab'+b\ \overline{a'}}) =1
$$
but this is not hard to check. You "just" have to write it all out.
For part (3), note that the determinant of any matrix in your subset is $1$ (as you have shown). Recall that the inverse of a matrix in $SL(2,\mathbb{C})$
$$
\pmatrix{a & b \\ c & d} 
$$
is
$$\pmatrix{d & -b \\ -c & a}
$$
Now that means that the inverse of one of the matrices
$$
\begin{pmatrix} a & -b \\ \overline{b} & \overline{a} \end{pmatrix}
$$
is
$$
\begin{pmatrix} \overline{a} & b \\ -\overline{b} & a \end{pmatrix}
$$
But this is clearly again in the subset.
