# Algebraic proof that isometries are surjective

I'm really struggling with an algebraic proof that all isometries $f$ in $\mathbb{R}^2$ are surjective. I start off by letting A be an arbitrary point in $\mathbb{R}^2$. Let $B=f(A)$. Now consider that $f(B)=B'$ must be located in a circle of radius $d(A,B)$, centered at $B$. I call this circle $C_B$ and also form a circle of the same radius around $A$, called $C_A$. Now that there are two intersecting circles, I define a point $C=C_A \cap C_B$. $\vartriangle ABC$ is equilateral, so $f(\vartriangle ABC)$ must also be equilateral, and I believe that this will uniquely identify $f(C)=C'$. Now I think my next step is to consider where $f(C')$ maps? I'm guessing ultimately if I keep doing this I will arrive at a point $D$ in which $f(D)$ will map to $A$ but I'm not not seeing the entire picture. Thanks for any assistance

Here's another line which may or may not suit you, depending on your context.

Proposition: the isometries are generated by translations and linear orthogonal transformations.

Sketch: pick a scalene triangle with one vertex on the origin. After an isometries has been performed, it is easy to find first the unique translation that aligns the corresponding pair of vertices to the origin, then the unique linear orthogonal transformation aligning the other two corners.

To tackle your question: Translations are clearly onto: for a translation by v, and arbitrary x, $x=(x-v)+v$.

Linear orthogonal transformations are 1-1 (since they preserve distance) hence onto.

So, the group of transformations these generate together are onto, since composition of two onto maps results in onto maps.

Almost. Pick $A,B\in \Bbb R^2$ arbitrarily. They map to $A',B'$, say. Now given $Y\in\Bbb R^2$ (for the generic case: not on the line $AB$) there are exactly two points $X_1,X_2$ such that $ABX_i$ is congruent to $A'B'Y$ (found by intersecting circles as you described). Also, there is only one second point $Z$ such that $A'B'Z$ is congruent to $A'B'Y$. Now $f$ must map $X_1$ and $X_2$ to $Y$ or $Z$ and cannot map both to $Z$.

To treat all points at once, pick a triangle $ABC\in\Bbb R^2$. This maps under the given isometry to a congruent triangle $A'B'C'$. Now the reflection at the perpendicular bisector of $A,A'$ maps $A'\mapsto A$ (and so $A'B'C'\mapsto AB''C''$). The perpendicular bisector of $B,B''$ passes through $A$ (why?) and maps $AB''C''\mapsto ABC'''$. By the circle-ntersection argument, we have either $C=C'''$, or $AB$ is the perpendicular bisector of $C,C'''$ in which case another reflection maps $ABC'''\mapsto ABC$. At any rate, a sequence of two or three reflections, i.e., an isometry acts as inverse of $f$ at least as far as $A,B,C$ are concerned. But any other point is uniquely determined by its distances to $A,B,C$ (again, by the intersecting-circles argument) so that the compsed map must be the identity of all of $\Bbb R^2$. A map with a (two-sided) inverse is bijective.

• Am I to assume that $Y$ falls on the intersection of the circles around $A'$ and $B'$? Oct 6, 2015 at 22:24