Prove the group of all real translations and all glide reflections with axis$R$ is isomorphic to $\{1,-1\}\times \mathbb R$ 
Let $G$ be the group of all isometries of $C$ consisting of all real translations, and all glide reflections with axis $\mathbb R$. Show that $G$ is isomorphic to $\{1,-1\} \times \mathbb R$.

I feel it difficult, first of all, to write down the group $G$. I think all real translations can be represented as $z \to z+b, \forall b \in \mathbb R$, but what is all glide reflections with axis $\mathbb R$?
Moreover, how can I show the two groups are isomorphic? 
 A: Hint: glide reflections along the real axis are just compositions of translations along the real axis ($z \mapsto a + z$, where $a \in \Bbb{R}$) and complex conjugation ($z \mapsto z^*$).
A: Further to my comment, I'll add what progress I made with this question. Though I'm not entirely sure how correct this is, so if anyone can see any mistakes please point them out.

What we want to do is show that there exists an isomorphism between $G$ and $\{-1,1\}\times \mathbb{R}$ which we can do by creating a function that connects the two groups: $\theta:G\mapsto \{-1,1\}\times \mathbb{R}$ with the property that:
$$ \forall a,b\in G \;\;\; \theta(a\circ b)=\theta(a)*\theta(b) \;\;\;\;\;\;\;\;\; [1] $$
where $*$ is defined: $(a_1,b_1)*(a_2,b_2)=(a_1a_2,b_1+b_2)$ since $\{-1,1\}$ is a multiplicative group and $\mathbb{R}$ is an additive one.
The thing that confused me was how to represent the isometries in $G$. From the other answer it's clear that we can represent any element in $G$ as either $z\mapsto \bar{z}+b$ or $z\mapsto z+b$ (for some $b\in \mathbb{R}$ since both translations are alway in the direction of the real line) since it's either going to be a glide reflection or a translation.
To prove $[1]$ I tried to define $\theta$ using the above reasoning as follows:
$$ \forall a\in G \;\;\; \theta(a) = \begin{cases} (1,b) &\mbox{if translation }\\ 
(-1,b) & \mbox{if glide reflection}\end{cases} $$
now to interpret $\theta(a\circ b)$ you have to just consider the cases of what $a\circ b$ can be since it's just the composition of two isometries.  Since $a,b$ will be either $z\mapsto \bar{z}+b$ or $z\mapsto z+b$ it turns out that only two possibilies exist for their composition:  $z\mapsto \bar{z}+b+b'$ or $z\mapsto z+b+b'$. To illustrate let $a$ be $z\mapsto \bar{z}+b$ and $b$ be $z\mapsto z+b'$ so that $a\circ b$ would be $z\mapsto \overline{z+b'}+b$ which is the same as $z\mapsto \bar{z}+b+b'$ (since $b,b'\in \mathbb{R}$).
So if $a\circ b$ is $z\mapsto \bar{z}+b+b'$ then $\theta$ maps it to $(-1,b+b')$. The r.h.s of $[1]$ would be:
$$ \theta(a)*\theta(b)=(1,b)*(-1,b')=(-1,b+b') $$
which shows they are the same. It doesn't matter which was round you have $a$ or $b$ since $*$ and $\circ$ are commutative. Similarly for $z\mapsto z+b+b'$ would give $(1,b+b')$.
Then we just need to check that it's a bijection which seems intuitively clear. To show it's injective we can consider $\theta(x)=\theta(y)$ so means that $(\pm 1,b_x)=(\pm 1,b_y)$ and thus $b_x=b_y$ so they must be the same isometry and $x=y$. Surjection requires that $(\pm 1,b)$ is mapped to by an isometry in $G$ which it true since both $z\mapsto \bar{z}+b$ and $z\mapsto z+b$ are in $G$. Thus it's an isomorphism.

Hopefully that's sufficient to show it, even if it not very concise, I may have gone little over the top with it! Hopefully someone can check to make sure this a reasonable answer. When I asked this question Servaes said that a possible candidate for $\theta$ could be:
$$ \forall \varphi \in G \;\; \;\;\theta(\varphi)=(-i(\varphi(i)-\varphi(0)),\varphi(0))$$ which I thought was a much nicer way than the piecewise way that I took. 
