Thread: Rigid Motions

Let SM denote the subset of orientation-preserving motions of the
plane. Prove that SM is a normal subgroup of M, and determine its
index in M.


I know if I can show this has an index 2, that means it is normal, but I don't know how to prove this.

you should tell us what "M" is, too. we don't have your text available for ready reference.

my assumption is that M is the set of all rigid motions (isometries) of the euclidean plane. how many orientations can the plane have?

see if you can find a surjection from M to {-1,1} (under the operation of multiplication for the latter set).

is this surjection a group homomorphism? if so, what is its kernel?

Oops. You're right, M is the group of all rigid motions of the plane.
I'm confused about this but I think we can use the determinant map for the rotation matrix, and this will go to 1 if the orientation of the plane is not flipped.

ok, you have a map in mind, det:M --->{1,-1}.

is this surjective?

what does it mean to say det is a homomorphism?

This map is surjective. For det to be a homomorphism, det (AB)=det(A)det(B)

and, is det(AB) = det(A)det(B) a true statement (if it is, you're almost done!)?

what is ker(det)? (remember, ker(det) = {T in M: det(T) = 1}, since 1 is the identity of {-1,1}).

Because detA=detB=1, the statement is true.
So I think the kernel would be all the matrices with det =1

Originally Posted by veronicak5678

Because detA=detB=1, the statement is true.
So I think the kernel would be all the matrices with det =1

no, not ALL transformations have a determinant of 1. for example, if:

T(x,y,z) = (2x,2y,z), det(T) = 4.

I think if the transformation is a reflection, the determinant is -1. So if it is an orthogonal rotation matrix, the determinant is 1.

that is correct, however, you should understand that det(AB) = det(A)det(B) is true for any 2 nxn matrices.

so det is a (group) homomorphism from GL(n,R) to R, no matter what n is (you are dealing with the special case of n = 2, and det(A) = ±1, which is one way of defining M).

so if ker(M) = SM, how does that show SM is normal?

We have proved in my class that the kernel of a group is always normal to it.
How do I find the index of the group?

what is the order of M/SM (which, by the first isomorphism theorem (big important theorem, don't leave home without it!), is isomorphic to det(M) = ____)?

Does M/SM mean the left cosets of SM in M? I tried looking up the first iso theorem, but I don't really understand it...

it does. for a finite group, with a finite normal subgroup: [G:H] = |G/H| = |G|/|H|.

the trouble is, here, we have an infinite group, and an infinite subgroup.

what the first isomorphism theorem says is this: for a group homomorphsim f: G-->G'

1) f(G) is a subgroup of G'
2) ker(f) is a normal subgroup of G
3) the set of cosets of ker(f), is also a group, G/ker(f), and is isomorphic to f(G).

the part that is useful to you is (3). it tells you that the number of cosets of SM in M is the number of possible determinant values elements of M can have,

and this number IS the index of SM in M.

(by the way, since SM is normal, the left cosets and the right cosets of SM in M are the same).

Can you explain how part 3 is telling me that?