Prove: $G = HN$, and $H \cap N =$ {$e$}, (Isometries in $\mathbb{R^2}$) I'm trying to solve this problem:
Let $G = E(\mathbb{R^2})$ be the group of all isometries of $\mathbb{R^2}$, so $G$ consists of translations, rotations about the origin and reflections in a line through the origin.
Let $N$ be the group of all translations in $\mathbb{R^2}$. 
Finally, let $H$ be $O_2(\mathbb{R})$, the orthogonal group, that consists of all rotations and reflections in $\mathbb{R^2}$
Prove:
a) $G = HN$, with $HN =$ {$hn | h \in H, n \in N$}
b) $H \cap N =$ {$e$} with $e$ the identity element
I already did this:
I proved that $N$ is a normal subgroup of $G$, and that $H$ is a subgroup of G. For a) I think it is enough to prove that every $g \in G$ can be written as $hn$, with $h \in H, n\in N$. I think one way to do this is to prove it for every combination of elements in $G, H$ and $N$, but maybe there is a shorter way to prove it.
For b), I do really have no idea of how the proof should be.
Could you help me with solving this problem?
Thanks in advance
 A: To each point $(x, y)$ of $\mathbb R^2$ associate the vector 
$$
\begin{bmatrix}
x\\y\\1
\end{bmatrix}.
$$
Note that a difference-of-points, which I'll call a "vector", has the form 
$$
\begin{bmatrix}
a\\b\\0
\end{bmatrix}.
$$
Then a transformation $T$ in $G$ can be written as matrix multiplication of the form
$$
\begin{bmatrix}
x\\y\\1
\end{bmatrix} \mapsto
\begin{bmatrix}
p & q & a \\
r & s & b \\
0 & 0 & 1
\end{bmatrix} \cdot
\begin{bmatrix}
x\\y\\1
\end{bmatrix} = M \cdot \begin{bmatrix}
x\\y\\1
\end{bmatrix} 
$$
Proof: Compute $T(0,0) = (a, b)$ to get two of the matrix entries. Let
$$
S(x, y) = T(x, y) - (a, b)
$$
Then $S$ is evidently linear, so it's represented by a $2 \times 2$ matrix 
$$K = \begin{bmatrix}
p & q \\
r & s 
\end{bmatrix}$$ in the standard basis. That completely determines the matrix above. 
Now to say that $S$ is linear and an isometry tells us that the columns of $K$ (which are the result of transforming the standard basis) must be unit vectors, and perpendicular. So $K$ is an orthogonal $2 \times 2$ matrix. If its determinant is $1$, then we must have $q = -r$ and $p = s$, and $p^2 + r^2 = 1$. Letting $\theta = atan2(r, p)$, we get $p = \cos \theta; r = \sin \theta$, and $K$ is evidently a rotation matrix (through angle $\theta$). 
On the other hand, if the determinant is $-1$, then we must have $q = r$ and $p = -s$, so that 
$$
K \cdot \begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}
$$
has the form we described, hence $K$ is a product of a rotation and a reflection. 
In short: $O(2)$ consists of all products of rotations and reflections, as does the $O(2)$ subgroup of $M_{33}$ consisting of matrices like 
$$
\begin{bmatrix}
p & q & 0 \\
r & s & 0 \\
0 & 0 & 1
\end{bmatrix}
$$
And the matrix 
$$
\begin{bmatrix}
p & q & a \\
r & s & b \\
0 & 0 & 1
\end{bmatrix} 
$$
can be written as 
$$
\begin{bmatrix}
1 & 0 & a \\
0 & 1 & b \\
0 & 0 & 1
\end{bmatrix} \cdot
\begin{bmatrix}
p & q & 0 \\
r & s & 0 \\
0 & 0 & 1
\end{bmatrix} 
$$
The left matrix corresponds to a translation; the right one to a rotation/reflection. There's your factorization into $HN$. 
