I am given that for all reflections $g$ there are infinitely many lines $L$ satisfying $g(L) = L$ which makes perfect sense (just take lines perpendicular to the axis of reflection). I am asked to show that if $g$ is a glide-reflection, there is only 1 line satisfying that condition.
Attempt: glide reflections are given by $g(z) = a\overline{z} + b$ such that $a\overline{b} + b \neq 0.$ So the restriction is that for all z on the line the output of g should be again on the line. So, $g(z) = lz, l \in \mathbb{R}.$ Equating the two conditions we get $a\overline{z} -lz + b = 0,$ and so we need a line of solutions to this equation. There is a theorem that states there only exists a line of solutions if $\|a\| = \|-l\|$ and $a\overline{c} = \overline{-l}c.$ But since $l \in \mathbb{R}, l = 1$ or $l = -1.$ So, we get that for all the points $z$ on our line, $g(z) = z$ or $g(z) = -z.$ Okay, i am sort of stuck here. I know that i have yet to used the glide condition $a\overline{b} + b \neq 0.$ Thanks in advance.
Just show that any point on the line of reflection $r$ is mapped to a different point on $r$. Now, any other line $l$ intersecting $r$ can't be mapped to itself, since the image intersection point $A$ is on $r$ bot not equal to $A$, so is not on $l$. Thus, any line that is mapped to itself must be parallel to $r$. Any such line not equal to $r$ is not mapped to itself because its image is on the other side of $r$, thus the only invariant line under $g$ is $r$.