Symmetry group of a line segment in Euclidean spaces

I am self studying the chapter of Symmetry Groups from Gallian's Abstract Algebra. There I encountered the following paragraph

" It is important to realize that the symmetry group of an object depends not only on the object but also on the space in which we view it. For example, the symmetry group of a line segment in $$R^1$$ has order 2, the symmetry group of a line segment considered as a set of points in $$R^2$$ has order 4, and the symmetry group of a line segments viewed as a set of points in $$R^3$$ is of infinite order."

I guess in $$R^1$$ , it is the group generated by the isometry of rotation by 180 degree.

I noticed that translation is an isometry of infinite order. Does that play a role in the symmetry group considered in $$R^3$$ ? What about $$R^2$$ ?How exactly are the groups determined?

Please please help me understand the underlying concepts. A lot of thanks for your time!!

1 Answer

I think Gallian is talking about the symmetries of $${\bf R}^n$$ that fix the line segment. So for $$n=2$$, you get a symmetry that flips the plane around the line containing the line segment, and the symmetry that flips the plane around the perpendicular bisector of the line segment (and their composition, and the identity, making four). For $$n=3$$, you can rotate the space through any angle $$\theta$$, $$0\le\theta<2\pi$$, around an axis containing the line segment, so infinite (in fact, uncountably infinite).

• @GerryMyerson..With a little bit of imagination , your answer has helped me make sense of the paragraph. Thank you so much. Apr 21, 2020 at 11:11
• You have the option of "accepting" my answer, by clicking in the check mark next to it. Apr 21, 2020 at 11:21