# What kind of isometry is a composition of a glide reflection with itself? Justify

Is there a simple algebraic proof?

Thanks!

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Well, if you compose a glide reflection with itself, what you get is a translation - which is an isometry. It's not hard to see this on euclidean spaces; even easier in the plane. But to do this with an algebraic proof? Try do proof this using only geometric arguments; if you have a glide reflection then it has a line in which it reflects; this line must divide the plane in two components, in such a way that when you apply the reflection again, the original geometric object which is being reflected is in the original "side" of the plane, but translated. – Marra Dec 12 '12 at 22:36
yeah i checked geometrically, i am trying to solve it algebraically as well by using the fact that Mk composed with Tab = Tab composed with Mk where AB is parallel to the line of reflection k. Tab is a translation and Mk is a reflection across line k AND Mk composed with Tab is glide reflection. – UH1 Dec 12 '12 at 23:07

Let the glide reflection be $T$. Without loss of generality the reflection part of $T$ is reflection in the $y$-axis. Assume that the translation part is by the vector $(a,b)$.
The reflection takes $(x,y)$ to $(-x,y)$. The translation part now takes us to $(-x+a,y+b)$. So $T$ takes $(x,y)$ to $(-x+a,y+b)$.
Do it again. The reflection part takes $(-x+a,y+b)$ to $(x-a,y+b)$, and the translation part takes this to $(x,y+2b)$. So $T^2$ is translation parallel to the $y$-axis by $2b$.
I was being sloppy, had vectors in mind but did not saay so. Have rewritten. A glide reflection is a reflection followed by a translation. The algebra above is that if $T$ is a glide reflection, then $T^2$ is a translation parallel to the axis of reflection. – André Nicolas Dec 12 '12 at 23:34