Let $G$ be the group of all isometries of $\mathbb{C}$ consisting of all real translations, and all glide reflections with axis $\mathbb R$. Show that $G$ is isomorphic to $\{1,-1\} \times \mathbb R$.
This question has been asked before but it hasn't had any answers yet and I wanted to add what I'd tried without taking over the other question. It seems that it's intuitively true since you can let the $−1$ or $1$ indicate whether it's a glide reflection or not (respectively) and then the real number defines the amount it's translated. I just can't turn that it into something rigorous.
It seems that we'd need to define $\theta:G\rightarrow \{-1,1\}\times \mathbb{R}$ and then we'd need to show that $$\forall a,b\in G \;\;\; \theta(a\circ b)=\theta(a)*\theta(b) \;\;\;\;\;\;\;\; [1]$$ where from intuition it seems we could then define $*$ by: $$ (a_1,b_1)*(a_2,b_2)=(a_1a_2,b_1+b_2) $$ since, for example, if we compose two glide reflections (both across the same line) then it's clear that both reflections cancel out which corresponds to $(-1,b_1)*(-1,b_2)=(1,b_1+b_2)$. Then we can just add the translations.
So with that defined it seems that $[1]$ should be true since we have $a\circ b$, which is just another element of $G$ (since it's a group), and so there must exist an element of $\{-1,1\}\times\mathbb{R}$ that it's mapped to by $\theta$. The R.H.S composes the two outside of $G$ but (hopefully) with the same effect.
The problem I have is I'm not sure if this is correct, and even if it is I'm not sure how to put this is a more compact and rigorous way (like, if it's correct, I'm not sure how you'd show that $\theta$ is a bijection between the two groups). Any pointers would be great! Thanks
EDIT - added further to comments: since all $a$ will be either $z\mapsto z+b$ or $z\mapsto \bar{z}+b$ then: $$\forall a\in G \;\;\; \theta(a) = \begin{cases} (1,b) &\mbox{translation }\\ (-1,b) & \mbox{glide reflection}\end{cases} $$