I just figured out that this will be 1 of 4 possible orbits. All 4 orbits (namely (1,2),(1,3), and (2,4) are all orbits of x under conjugation) satisfy the conjugation rule above.Now is this just 1 of many possible orbits of x, or is it the only one?
No, (23)(14) is indistinguishable from (14)(32); the orbit only has 3 elements. The stabilizer thus has 8. The stabilizer is a subgroup, and you have an odd element (12) which clearly stabilizes (12)(34). Note that if y is in the stabilizer of x, then yxy' = x, where y' is the inverse of y. But this is just the statment yx = xy, so it's easier (perhaps) to just look for those elements which commute with x, and we can tell by inspection that (12) does so. So we know that the stabilizer consists of a subgroup of order 4, let's call it H, that consists of even elements, together with (12)H. So our job is reduced to finding the 4 even elements that commute with x. Well e and x clearly commute with x, and if we can find one more even element y, then xy will be our 4th and final even element. If H contains an element y with order 4, then since 4-cycles are the only elements of order 4, it contains a 4-cycles, but those are odd, so H must contain no element of order 4, hence y is of order 2. The only even permutations of order two are the products of a pair of disjoint transpositions (same cycle structure as x), so we quickly see that H consists of e plus the orbit of x, and from this, H U (12)H gives us the stabilizer.Oxymoron said:!!! If what I wrote is correct, then I see now why you gave the hint to look up "cycle type". Because it was simply a matter of finding all permutations of
[tex] (x_1 x_2)(x_3 x_4) [/tex]
Obviously (12)(34), (13)(24), (14)(32), and (23)(14) are members of the orbit set. But say (21)(43) is not because it is indistinguishable from (12)(34). The same goes for every other.
[itex]S_4[/itex] has 24 elements. The orbit [itex]G_O[/itex] has 4 elements. By Lagrange's theorem, the stabilizer set [itex]G_X[/itex] has 6 elements.
Damn, I missed that! So the cardinality of the orbit is 3 and so the stabilizer subgroup consists of 24!/3 = 8 elements. ok.No, (23)(14) is indistinguishable from (14)(32); the orbit only has 3 elements.
Is the cycle (12) odd (as you say) because its length is even?The stabilizer is a subgroup, and you have an odd element (12) which clearly stabilizes (12)(34). Note that if y is in the stabilizer of x, then yxy' = x, where y' is the inverse of y. But this is just the statment yx = xy, so it's easier (perhaps) to just look for those elements which commute with x, and we can tell by inspection that (12) does so
The stabilizer is a subgroup of order 4. But it has 8 elements?So we know that the stabilizer consists of a subgroup of order 4, let's call it H, that consists of even elements, together with (12)H.
Okay, Im not 100% sure what even and odd elements are. These elements are cycles like (12) right? So is a cycle of even length, odd? and vice-versa.So our job is reduced to finding the 4 even elements that commute with x. Well e and x clearly commute with x, and if we can find one more even element y, then xy will be our 4th and final even element. If H contains an element y with order 4, then since 4-cycles are the only elements of order 4, it contains a 4-cycles, but those are odd, so H must contain no element of order 4, hence y is of order 2. The only even permutations of order two are the products of a pair of disjoint transpositions (same cycle structure as x), so we quickly see that H consists of e plus the orbit of x, and from this, H U (12)H gives us the stabilizer.
Of course! (1,2) is odd because it is the product of 1 transposition...itself! Silly me, I always forget that 1 is odd! :)(12) is odd since it is the product of an odd number of transpositions, (12)(34) and (123) are even: (123)=(12)(23).
...because a cycle with n = even, can always be written as a product of n+1 (odd) disjoint transpositions. Since this 'decomposition' into disjoint transpositions is unique in parity, a cycle with n=even elements, can ONLY be written as a product of an odd number of disjoint transpositions. ?a simple cycle (123,..,n) is odd iff it has an even number of things in the bracket, annoyingly.
I still dont quite understand this last bit of reasoning. :(the stabilizer contains (12), if it contained (1234) or any other 4-cycle, then these two thigs generate S_4 and the stabilizer would not have 8 elements.
As someone else mentioned, any permutation is odd if it is the product of an odd number of transpositions. It can be proven that no matter how you write out a permutation as a product of transpositions, if you can write it out as an even number of transpositions, then every other way of writing it out will have evenly many transpositions. As a trivial example, the permutation (123) is even since it can be written (13)(12), or as (13)(12)(12)(12) [the (12)'s on the end just cancel, hence the example is trivial]. When it comes to n-cycles, the permutation is even if and only if n is odd.Oxymoron said:Is the cycle (12) odd (as you say) because its length is even?
You have to read the entire sentence. I said the stabilizer consits of H together with (12)H. H is a subgroup of order 4. (12)H is not a subgroup, but is a subset of cardinality 4, and which is disjoint from H.The stabilizer is a subgroup of order 4. But it has 8 elements?
Yes, but like I said above, the notion applies to any permutation, not just cycles.Okay, Im not 100% sure what even and odd elements are. These elements are cycles like (12) right? So is a cycle of even length, odd? and vice-versa.
Remember, the stabilizer is H U (12)H. H is our even "half" of the stabilizer, and (12)H will be the odd half. If we can find the even half, then multiplying all those elements by (12) will give us the odd half, and we'll have the whole. So let's just find the even half of the stabilizer, H. Well since the stabilizer has order 8, H has order 4. Any group of order for is either cyclic (thus contains an element of order 4) or is ismorphic to [itex]\mathbb{Z}_2 \times \mathbb{Z}_2[/itex]. Well, if it contains an element of order 4, the only permutations of that order are 4 cycles, so it would contain a 4-cycle. But we're trying to determine H, the even half of the stabilizer, and a 4-cycles is not even, so it can't be in H, so we know that H is not cyclic. The even elements of [itex]S_4[/itex] are the 8 3-cycles, the 3 disjoint-transposition-pairs, and identity. All the 3-cycles have order 3, so clearly we're left with the remainig 4 elements, those being e and the disjoint-transposition-pairs.Since e = (1) and x = (12)(34) commute with x (ie. xy=yx) then these are 2 members of the stabilizer subgroup. Now you say that the stabilizer subgroup cannot contain a cycle of order 4 because this will be odd. Briefly, why cant the stabilizer subgroup contain a cycle of order 4?
We're looking for even stabilizers (because once we've found those, having already found one odd stabilizer in (12), we know all the odd stabilizers. (****) is not even, and (***) has order 3.We have found e and x as stabilizers so far. Another stabilizer y will be of order 2. We are still looking for even permutations, so y will be in this form (**)(**). Why not (****)? or (***)?
That's not the stabilizer, that's just H. Take the union of H with (12)H, so you get:Then the stabilizer subgroup consists of {e, x, (13)(24), and (14)(23)}. But I thought there should be 8 elements in the stabilizer subgroup? This is only 4.
No, the center of a group is the subgroup of elements that commutes with all other elements. The stabilizer of an element x of a set X under an action of G is a subset of G consisting of those g in G such that the corresponding bijection from X to X, g', sends x to itself, i.e. g'(x) = x.Is the stabilizer also called the center?
No, it can always be written as a product of n+1 transpositions, but they wouldn't be disjoint. A product of disjoint transpositions can't even be a cycle (except if you have one transposition, then it's a 2-cycle).Oxymoron said:...because a cycle with n = even, can always be written as a product of n+1 (odd) disjoint transpositions.
Neither do I, because it's wrong. (12) and (1234) do generate all of [itex]S_4[/itex] I believe, but it is not that (12) and any 4-cycle together generate [itex]S_4[/itex].I still dont quite understand this last bit of reasoning. :(