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Do translations form a normal subgroup of the group of rigid motions in a general Euclidean plane with no underlying field?

This is a question that has puzzled me for the past few days. In the wikipedia article for normal subgroups it is stated that the translation group is a normal subgroup of the Euclidean group. When the Euclidean plane is over an arbitrary field, it is straightforward to verify that conjugation by rotations or reflections results in translation only, showing that the translation group is a normal subgroup.

I've been unsuccessful in proving this over a Euclidean plane that does not have an underlying arbitrary field. Suppose instead that my definition of translation is that a rigid motion $T$ is a translation if for any two points $A$, $B$ in the plane, we have that segments $AT(A)\cong BT(B)$, (here $\cong$ is the undefined notion of congruence). So I can't merely say that translations are defined by translating every point in the plane by some fixed translation vector.

I have been able to show that this definition of a translation is congruent to saying $A, B, T(A), T(B)$ form a parallelogram (in the normal Euclidean sense) when $A, B, T(A)$ are not collinear. So when there is no underlying coordinate system, how can one show that translations still form a normal subgroup? I've been having a heck of time getting a grasp on the objects without a coordinate system to pin them down, as I don't think I can write rotations and translations in terms of rotation matrices and vector addition. Thanks for any input.

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Your definition of a translation is invariant under conjugation, so... – Qiaochu Yuan Mar 28 '11 at 14:43
up vote 3 down vote accepted

I'm not sure exactly what definition of Euclidean geometry you are using, but I'll assume a rigid motion takes parallelograms to parallelograms (using Klein's viewpoint that you don't define geometric objects that aren't preserved under rigid motions).

In that case the collection of T such that A, T(A), T(B), B are parallelograms for every pair of points A, B forms a normal subset:

If G is any rigid motion, then G(A), G(B) are two points, so G(A), T(G(A)), T(G(B)), G(B) is a parallelogram by definition of T. Apply the inverse of G to get A, T^G(A), T^G(B), B is a parallelogram, and so T^G is also a translation.

If you've already shown the composition of translations is a translation, a translation is invertible, and its inverse is a translation, then this shows you have a normal subgroup. I think those proofs require actually knowing what axiom system you are using.

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Thanks Jack, I already know that the set of translations is an abelian subgroup, so this should work nicely. Do you mind explaining what T^G means? I'm not familiar with that notation. Does it mean the same as $T\circ G^{-1}$? – yunone Mar 28 '11 at 15:02
$T^G$ means "$T$ conjugated by $G$". $T^G = G^{-1}\circ T\circ G$. – Arturo Magidin Mar 28 '11 at 15:03
@Arturo, thank you for the clarification. – yunone Mar 28 '11 at 15:07

I think a proof should be able to proceed along the following lines:

Assuming that a rigid motion of the plane is defined as a map that maps all line segments to congruent line segments, a rigid motion can be characterized as the unique rigid motion that maps a certain oriented line segment $R$ to a certain congruent oriented line segment $S$ (where an oriented line segment is just a line segment with an ordering of the end points). Given some oriented line segment $R$, there is a bijective map between the group of rigid motions and the set of congruent oriented line segments $S$ they map $R$ to. (This is not the case in higher dimensions, where rotations can leave line segments invariant.)

Given oriented line segments $R$ and $S$ characterizing a rigid motion, the rigid motion can be applied to any point $p$ by constructing the point $q$ that forms a triangle with $S$ that is congruent to the triangle $p$ forms with $R$. (Here we need the orientation of the line segments to distinguish the two possible points.)

Now let any rigid motion M, any translation T and any two points $A$, $B$ be given, and let $R$ be the oriented line segment from $A$ to $B$. This is mapped to some oriented line segment $S$ by $M$, with end points $A'$ and $B'$. Applying the translation $T$ to $S$ yields a parallelogram $A'B'T(A')T(B')$. If you now apply $M^{-1}$, which is the unique rigid motion that maps the oriented line segment $S$ to the oriented line segment $R$, the corresponding construction of congruent triangles shows that $M^{-1}(A')M^{-1}(B')M^{-1}(T(A'))M^{-1}(T(B'))=ABM^{-1}(T(A'))M^{-1}(T(B'))$ is again a parallelogram.

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Thanks for the response, Joriki. – yunone Mar 28 '11 at 15:07
@yunone: You're welcome :-) – joriki Mar 28 '11 at 15:09

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