# Group Action

1. Jun 16, 2005

### Oxymoron

I need some help on solving these sorts of problems in Group theory.

Question

Consider the group action of $S_4$ on itself by conjugation. Determine the orbit and the stabilizer of

$$x = (12)(34)$$

2. Jun 16, 2005

### Oxymoron

Attemped solution

In this question we are considering the conjugation action. That is

$$\phi\sigma\phi^{-1}$$

The orbit of $x$ is a set of elements of $S_4$ to which $(12)(34)$ can be permuted by other elements of $S_4$. I will denote the orbit as

$$Gx = \{gx : g\in S_4\}$$

First thing I tried was to see if $(1,2)$ was an orbit

$$(1,2)(1,2)(3,4) = (3,4)$$ ...[1]

$$(1,2)(3,4)(1,2) = (3,4)$$ ...[2]

That was a lucky guess! I would like to know if there is a more formal way of finding the orbit rather than just guessing??

This is true because (computing [1] and [2] from right to left...). [1]...

1 2 3 4
1 2 4 3
2 1 4 3
1 2 4 3 = (3,4)

and [2]...

1 2 3 4
2 1 3 4
2 1 4 3
1 2 4 3 = (3,4)

Hence this is an orbit of $x=(1,2)(3,4)$.

Now is this just 1 of many possible orbits of x, or is it the only one?

Stabilizers next...

Last edited: Jun 16, 2005
3. Jun 16, 2005

### Oxymoron

I just wanted to note that in my previous post [1] and [2] were formulated by the conjugation action $\phi\sigma\phi^{-1} = I$ or in other words $\phi\sigma = \phi$.

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.

But why was my guess of (1,2) still lucky? As in, why are all four orbits transpositions? How am I meant to find the orbit if I dont want to guess?

The stabilizer of x is going to be a set of elements of $S_4$. The stabilizer set will have 6 elements by Lagrange's theorem...

$$|G_O| = \frac{|S_4|}{|G_X|} = \frac{4!}{4} = 6$$

One stabilizer is the trivial one: $(1)$. Obviously this is a stabilzer because it does nothing to any element of $S_4$ by conjugation.

But what are the other 5, and how do I find them?

4. Jun 16, 2005

### Muzza

You've confused a few things. The orbit of x is indeed the set {gx; g in G}, but you have to remember that "gx" does NOT necessarily mean the normal group product of g and x (indeed, a group can act on an arbitrary set, not just itself, so that the the normal group product of g and x need not even be defined).

In this case, where the action is conjugation and G = S_4, the orbit of x is {g.x; g in G} = {g * x * g^-1; g in S_4}. (Here, . denotes the "action" multiplication and * ordinary multiplication in S_n). Hence, the elements of the orbit you've determined are all wrong. I also wish you wouldn't say things like "(12) is an orbit" ;) Surely "(12) is in the orbit of x" makes more sense (the orbit being a set and all...).

There is a very easy way of determining orbits in S_n, look up "cycle type". You could of course also simply go through all elements g in S_4 and compute g * x * g^-1, but that doesn't seem like much fun.

5. Jun 16, 2005

### matt grime

Can i also add to "the orbit being a set and all" the extra words "and a unique one ie given x there is exactly one set Orb(x), one orbit"

6. Jun 16, 2005

### Oxymoron

Ok so when I say "orbit" I should be referring to a set of particular cycles?

You said cycle type. Are the cycles that constitute the orbit of $(12)(34)$ of the same cycle type? As in $(x_1 x_2)(x_3 x_4)$ ?

If so, I believe that (12)(34), (13)(24), (14)(32), and (23)(14) are the required members of the orbit.

Since by acting on the group member $x=(12)(34)$ by these cycles $\phi_1,\phi_2,\phi_3,\phi_4$ above, by $g.x$.

Am I getting the orbit set mixed up with the stabilizer set? How do I go about finding the stabilizer set now? They are just cycles which do nothing to $x$ under conjugation right? - they "fix" $x$ under conjugation.

Last edited: Jun 16, 2005
7. Jun 16, 2005

### Oxymoron

!!! 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

$$(x_1 x_2)(x_3 x_4)$$

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.

$S_4$ has 24 elements. The orbit $G_O$ has 4 elements. By Lagrange's theorem, the stabilizer set $G_X$ has 6 elements.

8. Jun 16, 2005

### AKG

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.

You can do this problem without having to actually multiply any two permutations.

Last edited: Jun 16, 2005
9. Jun 17, 2005

### Oxymoron

Damn, I missed that! So the cardinality of the orbit is 3 and so the stabilizer subgroup consists of 24!/3 = 8 elements. ok.

Is the cycle (12) odd (as you say) because its length is even?

The stabilizer is a subgroup of order 4. But it has 8 elements?

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.

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 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 (***)?

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.

10. Jun 17, 2005

### matt grime

all permtuations can be written as a product of transpositions (things like (12) that swap 2 objects). this is not unique, but the parity (oddness or evenness) is unique and independent of the choice of transpositions.

(12) is odd since it is the product of an odd number of transpositions, (12)(34) and (123) are even: (123)=(12)(23).

a simple cycle (123,..,n) is odd iff it has an even number of things in the bracket, annoyingly.

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.

11. Jun 17, 2005

### Oxymoron

Thankyou Matt, you have answered some of my questions.

Of course! (1,2) is odd because it is the product of 1 transposition...itself! Silly me, I always forget that 1 is odd! :)

...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. ?

I still dont quite understand this last bit of reasoning. :(

12. Jun 17, 2005

### Oxymoron

Is the stabilizer also called the center?

13. Jun 17, 2005

### Muzza

The center Z(G) of a group G is the set $$\{x \in G; xy = yx \mbox{ for all } y \in G\}$$ (i.e. the set of all elements which commute with all other elements). It's a normal subgroup of G.

The stabilizer of an element $$a \in G$$, when the action happens to be conjugation, is called the centralizer of a. It can be written as $$\{x \in G; xax^{-1} = a\} = \{x \in G; xa = ax\}$$, i.e the centralizer of a is the set of all elements which commute with a (which would explain why its name was chosen to be so similar to "center").

14. Jun 17, 2005

### AKG

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.
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.
Yes, but like I said above, the notion applies to any permutation, not just cycles.
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 $\mathbb{Z}_2 \times \mathbb{Z}_2$. 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 $S_4$ 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.
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.
That's not the stabilizer, that's just H. Take the union of H with (12)H, so you get:

{e, x, (13)(24), (14)(23)} U (12){e, x, (13)(24), (14)(23)}
={e, x, (13)(24), (14)(23)} U {(12), (34), (1324), (1423)}
={e, x, (13)(24), (14)(23), (12), (34), (1324), (1423)}
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.

In this particular case, G = X, and g' is defined by:

$$g'(x) = gxg^{-1}$$

In this case, the stabilizer of x are those elements G which commute with x (but they need not commute with all elements). Again, only in this particular example, if you took the intersection of all the stabilizers, you'd get the center of G, but there's no necessary connection whatsoever between the center of a group, and the stabilizer of an element under the action of a group.

15. Jun 17, 2005

### AKG

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).
Neither do I, because it's wrong. (12) and (1234) do generate all of $S_4$ I believe, but it is not that (12) and any 4-cycle together generate $S_4$.