# Equivalence between algebraic statements and geometric relations.

I'm currently trying to read a geometry and symmetry book and came across a little problem that I am having difficulty understanding.
I need to show that if xm=mx then the point X is on the line M, and also that if xa=ax, then the point X is on the plane A.

Note that x is the inversion through the point X and m is the reflection across the mirror M.

Super stuck. Any insight or suggestions would be great. Thanks.

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It's difficult to guess what you mean. Can you give details about where you found this and what sort of algebraic structure you are working in when you say "xm"? –  rschwieb Nov 24 '12 at 1:06
What is $x$? What is $m$? Where does the point $X$ lie (is it in $\mathbb{R}^3$?) –  Brad Nov 24 '12 at 1:14
What does $xm$ mean? This seems like a badly underspecified question. –  Qiaochu Yuan Nov 24 '12 at 2:20
It seems like points $X$, lines $M$ and planes $A$ are associated to some elements $x,m,a$ in some algebraic structure. But I think we have no clue how and which alg.structure.. –  Berci Nov 24 '12 at 14:05
I edited my question, now that I figured out what little x and m were. Sorry for any confusion. –  jan Nov 25 '12 at 20:30

A good way to compare functions is by applying them to something. Given that there is only one point in the problem statement, it is reasonable to apply the function to $x$. At this point, the weakness of your notation becomes a big nuisance, so let us call $r_x$ the reflection with respect to the point $x$ and $r_m$ the reflection with respect to the line/plane $m$.
Now $x$ is the unique fixed point of $r_x$.
So, if $r_x\circ r_m = r_m \circ r_x$, then $r_x(r_m(x))=r_m(r_x(x))=r_m(x)$.
This implies that $r_m(x)$ is a fixed point of $r_x$, so $r_m(x)=x$. This implies that $x$ is a fixed point of $r_m$, so it has to be on the fixed line/plane $m$.