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.
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?
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?
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).