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.