Let H be the group of the symmetries of a square. Denote by R reflection about the y-axis and by ρ the rotation by 90 degrees around the origin.
a)Give a representation of each of the 8 elements of H: e, ρ, ρ², ρ³, R, ρR, ρ²R, ρ³R as a permutation of {1,2,3,4}. We can therefore view H as a subgroup of S(4).
b) Construct all the left cosets of H in S(4).
c)Compute the right coset H(1,2) and compare it to the left coset (1,2)H. Are they equal?
I missed the class where all of this type of stuff was discussed, and am so lost. I've read through my book but it is just not very helpful and doesn't offer many examples. I would sooooo appreciate any help on this problem
a) what you want to do here is somehow assign the numbers 1,2,3 and 4 to geometric objects that are "symmetrical" in the square. there are two ways to do this: one can label the sides, or the vertices (corners). normally, the vertices are chosen (you get an equivalent representation in either method).
so how does a rotation of 90 degrees (we'll assume counter-clockwise) affect a square with the vertices labelled 1,2,3,4 (also in a counter-clockwise fashion)?
it sends 1-->2 2-->3 3-->4 4-->1. this is the permutation of S4, (1 2 3 4) (a 4-cycle). so we have ρ <--> (1 2 3 4), which means that ρ^2 = ρoρ = (1 3)(2 4):
1-->3
2-->4
3-->1
4-->2
i leave it to you to figure out which permutation is ρ^3. since ρ^4 = 1, this will be ρ^-1.
if we assume that the vertex "1" is in the 4th quadrant, with the origin at the center of the square (so it's "symmetrically placed"), then R can be represented by the permutation:
1-->4
2-->3
3-->2
4-->1, which can be written in cycle notation as: (1 4)(2 3). convince yourself that any reflection must be either a double 2-cycle, or a 2-cycle consisting of switching one pair of "diagonally opposite corners", and that these reflections are exactly the elements of H with "R" in them.
b) explicitly computing these would be quite a chore: S4 has 24 elements, and H has 8, for a total of 192 computations to see which cosets are the same. there must be a better way.
the key is to see that the index of H in S4 is just 3, so all we need to do is find 3 different cosets, and that will do it. and right off the bat, H is one of them, which means we only need to find 2 more. well, if we find some permutation σ that's not in H, then σH is a second coset.
to find the third, just find an element of S4 not listed in any of the other two cosets (this takes the number of calculations down from 192 to a mere 8).
c) this is just a matter of computation. for example, we know that (1 2 3 4) is in H, so (1 2 3 4)(1 2) is in H(1 2). (1 2 3 4)( 1 2) =
1-->2-->3
2-->1-->2
3-->3-->4
4-->4-->1, (first we switched 1 and 2, and then "cycled" 1,2,3 and 4), which is the permutation (1 3 4) in S4 (this makes sense that it is a 3-cycle, because 2 stays fixed).
the identity is: the "do nothing" action. which is also equivalent to rotating the square 360 degrees. see how that works?
ρ is a rotation. imagine turning the square 90 degrees (one way or the other. in math, rotations counter-clockwise, are considered "positive"). so the vertex that was in the 4th quadrant, now is in the 1st quadrant, the vertex in the 1st is now in the 2nd, the vertex in the 3rd quadrant is now in the 4th quadrant. like this:
2-1.......1-4
3-4 --->2-3 (you'll have to imagine these are squares). R does this:
2-1.......1-2
3-4 --->4-3 (here, the y-axis runs down the middle of the square).
so you have these two types of things, physical ways you can move a square so that the "ending positon" of the "moved square" looks like its "starting position", and a corresponding re-arrangement of the vertices.
you can also visualize the elements {e,ρ,ρ^2,ρ^3,R,ρR,ρ^2R,ρ^3R} as a set of 8 2x2 matrices. in this case, ρ =
[0 -1]
[ 1 0]
while the matrix of R is:
[-1 0]
[ 0 1], and e is the matrix I =
[1 0]
[0 1].
each of these matrices corresponds to a "rigid motion" of the plane, that maps a square to a square. and we see that these motions map sides to sides, and corners to corners. so either by numbering the sides, or numbering the corners, we see that each of these motions takes a set of four things, to the same set of four things, but perhaps in a different arrangement (in the originial arrangement, side 1 might be "up", but after a rotation of 180 degrees, side 1 is now "down").
in other words, each of these 8 operations, 4 rotations, and 4 reflections, is a bijection on the set of corners (or sides). and bijections on 4 things is just what the symmetry group S4 is MADE of. so we have a subset of S4, of order 8, which being a group, is a subgroup.