Symmetric Groups Part 2 

In this particular example, S3 is isomorphic to D3, the dihedral group of order 6. This group gives the symmetries of an equilateral triangle; both in rotation and reflection. So the symmetric groups may not be commutative, but they give us an indication about the symmetries inherent in the group structure.

@@StormCrowAlpha If that would be the real origin of that name, we wouldn't distinguish between dihedral groups and symmetric groups. The fact that S₃ is isomorphic to D₃ is merely coincidental. The truth is, no one knows for sure what's the true origin, because there doesn't seem to be any traces of that (at least not that I would know of). But there are some hints: it might be related to SYMMETRIC POLYNOMIALS which were one of the first area studied by people who introduced groups to mathematics, in the context of finding roots of polynomials. Symmetric polynomials are polynomials in more than one variable, for which it doesn't matter in which order you substitute these variables - the result will be the same. For example, x²·y + x·y² is such a polynomial, because it does the same thing to both `x` and `y`, so it doesn't matter if you substitute, let's say, `7` as `x` and `5` as `y`, the result will be the same. No matter how you PERMUTE these substituents, a symmetric polynomial will be none the wiser, because the result will not change. So these people were using such polynomials as tests for "how much symmetry" does a polynomial have, by checking how many symmetric polynomials can be fooled when they permute the roots of the original polynomial and substitute to the symmetric one. In a way, this was the measure of the "symmetry" of the original polynomial and its solutions. Later on, when they introduced the idea of a group, they called such groups of all possible permutations "symmetric groups" because they represented this "maximal symmetry" of solutions of a polynomial. Dihedral groups (which are groups of symmetries of regular polygons) are SUBGROUPS of symmetric groups (as every other type of groups is, according to Cayley's theorem), because they usually don't include all possible permutations, just some subset of them (because not every permutation is possible for regular polygons' vertices - they must behave like a rigid body, you can't swap just two vertices without breaking the polygon). You can think of symmetric groups as those which have "maximal symmetry possible" (where by "symmetry" is meant how many parts you can move around and still get something that looks similar).

@@bonbonpony Thank you for this insight! Do you have any insight into why alternating groups are called "alternating"?

@@MuffinsAPlenty It might have something to do with even/odd permutations. One can be obtained by swapping (alternating) two elements in pairs, while others cannot. Alternating groups represent those permutations that can (i.e. only the even permutations, excluding the odd ones).

Hey there, I got stuck on the problem, wether the arrows of the trasformation can change anything or not. later then found myself ponder about your problem. So thats my understanding now: your a^2 is wrong, cause your "a" means swapping 1 and 2. doing this twice yields the identity e. if you take a^2 as a separate symbol (not as a composition), you easily get confused. (more to that in the example below) the thing is, that we just use all possible permutations of the set ( n! = 3! = 6 ) and give them a representation 123 123 123 123 123 123 123 312 231 213 321 132 (which is twice a cyclic exchange of 123, followed by a change of 12 -> 21 and another two cyclic exchanges now starting with 213) a b c d e f (arbitrary notation) e b a a^2 (and yours) ab a^2b or a^2b (and yours) if you imagine a equilateral triangle and you write at each edge one of the 12 or 3. then these permutations correspond to the symmetries of the triangle. like a mirror plane through 1 would be m1 and the mirror plane through 2 would be m2 and the same for m3. (three mirror symmetries corresponding to swapping 12, 13, or 23) now there is left a rotation clockwise r and counterclockwise l. (and here two times clockwise f°f or f^2 is the same as one time counterclockwise e, and vice versa. but still they represent different permutations, we just need to have a symbol for that constellation). and the least is just doing nothing e. this is 6 symmeties! here we are with 6 possible permutations which can be represented with a symbol. and that collection of actions represents a group and hence the symmetries of a equilateral triangle

The diagram drawn around 8:43 explains why sigma applied twice is sigma^2. You just apply sigma twice and only keep track of where the initial input (into the first sigma) ends up (after exiting the second sigma). We can use a counting argument to guarantee that we have all of the permutations. A permutation is a one-to-one and onto mapping from a set to itself (meaning that no two numbers can be mapped to the same number as each other, and that every number is mapped to from some number). So if you have n elements, you can put an order on those elements and then just think about how many possibilities you can make. If you are trying to build a one-to-one and onto mapping from {1, 2, 3, 4, ..., n} to itself, you can start with f(1). You have n possible choices for this. Then you consider f(2). Since f(2) has to be different from f(1), you have n-1 choices for f(2). Then you consider f(3). Since this must be different from f(1) and f(2), you have n-2 choice for f(3), etc. Eventually for f(n) you only have 1 choice. So there are n*(n-1)*(n-2)*...*2*1 possibilities of one-to-one and onto functions from {1, 2, 3, 4, ..., n} to itself. This is the same as n!. So there are n! many permutations of a set of size n. When we get to a set of size 3, there are 3! = 3*2*1 = 6 possible permutations. So once we have found 6 distinct permutations, then we are sure we have found all of them.

ok

No.

6

Well, to begin with, every group has itself as a subgroup, as well as the "trivial" group that contains just the identity element (you can call it S₁ if you wish). So that's two. If there are any other subgroups, their order must divide the order of the whole group. The divisors of 6 are 1, 2, 3, 6. We dealt with 1 and 6 already, so there's only 2 and 3 to check. And indeed, S₃ contains two cyclic groups ℤ₂ and ℤ₃ as subgroups. And in fact, S₃ is ℤ₂×ℤ₃, that is, a product of these two cyclic groups, because it consists of two 3-cycles joined with three 2-cycles. And that's pretty much all. Summing up, the complete list of subgroups of S₃ is just this: { S₃, ℤ₃, ℤ₂, S₁ }. And there's just 4 of them, not 6.