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Old 04-08-2014, 09:06 AM   #1
cowmoo32
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Zell, Mash, anyone else good with higher math...

Is there an intuitive way to picture the stabilizer and orbit of a group? I can easily find them, but don't really understand what they're doing. Like for the kernel and image, I knew how to find them, but didn't really understand what they were until I saw this:
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Old 04-08-2014, 09:13 AM   #2
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Going to sidebar this thread for a moment to compliment your stellar photography. Enjoyed looking through that. Carry on.
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Old 04-08-2014, 10:19 AM   #3
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Pretty sure I have it figured out but if you guys can think of a better metaphor, feel free. S is a list of things, say, colored balls. G is a machine that matches the balls together and contains every possible combination of colors. The stabilizer for a given color is the method by which G outputs the same color you feed in. And the orbit will be the method that produces every possible combination for the color you feed in, including the stabilizer.

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Going to sidebar this thread for a moment to compliment your stellar photography. Enjoyed looking through that. Carry on.
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Old 04-08-2014, 10:31 AM   #4
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Old 04-08-2014, 10:51 AM   #5
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If you were talking about planets, I could help you.
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Old 04-08-2014, 10:54 AM   #6
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Is there an intuitive way to picture the stabilizer and orbit of a group? I can easily find them, but don't really understand what they're doing. Like for the kernel and image, I knew how to find them, but didn't really understand what they were until I saw this:
da fuq are you talking about?
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Old 04-08-2014, 10:57 AM   #7
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If you were talking about planets, I could help you.
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Trust me. I am probably better at math than anyone on the boards right now. I know how to do it right and I get perfect results every time because I practice the discipline of math and I don't skip steps.
lol

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da fuq are you talking about?
Permutation groups niqqa

edit: Where the fvck is Zell, I know he knows this stuff.
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Old 04-08-2014, 11:04 AM   #8
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Permutation groups niqqa
What class is this for? Glad I never had to deal with this crap.
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Old 04-08-2014, 11:07 AM   #9
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I have a confession to make, most of the higher level math I know (group theory stuff) is from physics classes, I haven't taken many formal higher level math courses, and even those were a long time ago and I haven't used them much since. I can't help you here
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Old 04-08-2014, 11:08 AM   #10
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lol


Permutation groups niqqa

edit: Where the fvck is Zell, I know he knows this stuff.
I'll read iiiiit, I r been busy lately. THIS THREAD HAS BEEN NOTED
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Old 04-08-2014, 11:10 AM   #11
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Since I have a little time on my hands I'll try to help you out a little. Let me start with a question and then go through some proofs that will hopefully help you to understand.

How many rotational symmetries does a cube have? This question can be answered in a number of ways. Perhaps the one that most readily occurs to people is this: each vertex can end up in one of eight places; once youíve decided where to put it, there are three places you can put one of its neighbours; once youíve decided where to put that, the rotation is determined, so the total number of rotations is 8\times 3=24.

Hereís another proof. Take one of the faces. It can go to one of six other faces, and once youíve decided which face it will go to, one of the vertices on the face has four places it can go, and once youíve decided that youíve fixed the rotation. So the total number of rotations is 6\times 4=24.

And hereís another. Take one of the midpoints of the twelve edges. There are twelve places it can end up, and once youíve decided where to put it, there are two choices for how you send the two endpoints of the original edge to the endpoints of the new edge. So the total number of rotations is 12\times 2=24.

And hereís yet another. Take a point half way between the centre of one of the faces and one of the vertices. (If the cube is the set of all points (x,y,z) such that x, y and z all lie in the closed interval [-1,1], then we could, for instance, take the point (1/2,1/2,1).) There are 24 places that that point can end up ó the 24 points that are half way between the centre of some face and one of the vertices of that face. There is only one rotational symmetry that takes the original point to any given one of those 24 points. Therefore, the number of rotations of the cube is 24\times 1=24.

If you understood those arguments, then you basically understand the orbit-stabilizer theorem, even if you think you donít. To try to convince you of that, Iíll first show how to convert one of the proofs into a proof that explicitly uses the orbit-stabilizer theorem. Iíll then discuss how much you have to remember in order to remember the proof of the orbit-stabilizer theorem. As usual, I shall aim to show that the answer is ďvery little indeedĒ. However, routine proofs in abstract algebra have a somewhat different flavour from routine proofs in set theory and analysis, and as this one is a good representative example, it bears discussing in some detail. (Later I may try the same with the first isomorphism theorem, which you should be meeting fairly soon.)

To illustrate how to count symmetries using the orbit-stabilizer theorem, let me convert the 6\times 4 proof that used faces. The obvious way to capture the idea of ďwhich face a particular face goes toĒ is to let the symmetry group of the cube act on the faces of the cube. Then the set of places that a particular face can go to is nothing other than the orbit of that face. How does the group act on the faces? Well, any rotation of the cube will permute the faces, and that permutation is the one that corresponds to the rotation.

In the earlier version of the argument, we said that our chosen face can go to any one of six other faces. The fancy way of saying that is that the orbit of the chosen face is the set of all faces, which has size 6.

And in the earlier version of the argument, we said that once we had decided which face our chosen face would go to, there were four ways of lining up the vertices. This doesnít quite correspond to a statement about stabilizers, but let us note for now that if the chosen face maps to itself, then we have four choices. So the stabilizer of our chosen face has size 4. Multiplying those two numbers together gives us 24.

The content of the orbit-stabilizer theorem is basically that if a group G acts on a set X, then once you know how many ways there are of sending an element x of X to itself, you also know how many ways there are of sending it to any other element in its orbit. In the faces-of-the-cube example, knowing that there are four ways of rotating the cube so that a face F ends up where it was before tells us that there are also four ways of sending it to some other face F'. And that is because ďall faces look basically the sameĒ from the point of view of the rotation group. In general, ďall points of the orbit of x look basically the sameĒ from the point of view of the transformations performed by the elements of G.

Iím going to give a few different proofs of the orbit-stabilizer theorem, not because they are fundamentally different, but because they have certain stylistic features that are worth commenting on. Also, it may be interesting to see how much variety is possible even when presenting a proof, even when the underlying argument is the same.

First, Iíll remind you of the statement we are trying to prove.

Theorem. Let G be a group that acts on a set X, let x be an element of X, let O_x be the orbit of x and let S_x be the stabilizer of x. Then |O_x||S_x|=|G|.

In words, the size of the group is the size of the orbit times the size of the stabilizer.

Proof 1. For this proof I want to try to capture as closely as possible our intuition in the cube example that the set of rotations that takes one face to another is ďjust like the set of rotations that fix that face, but over at another placeĒ. However, Iíll argue in general.

What am I trying to show in the abstract? I want to show that if y belongs to the orbit of x, then the set of g\in G such that gx=y is the same size as S_x, which is the set of g\in G such that gx=x. (Here Iím slipping into the condensed notation and writing gx instead of \phi(g,x) or \phi(g)(x). But itís important to bear in mind that the ďproductĒ gx is not the group operation, since itís only g that belongs to a group. Rather, it is what one might call the group action operation, which has some nice group-like properties such as (gh)x=g(hx) and ex=x.)

What is the most natural way to show that two sets have the same size? Itís to produce a bijection between them. (It isnít the only way. Another is to calculate their sizes, possibly by completely different methods, and show that you get the same answer in both cases. But that is not an appropriate method here, since weíre not given enough information to calculate anything.)

Let me write S_{xy} for the set of g\in G that take x to y. If Iím given an element of S_x, how do I turn it into an element of S_{xy}? Well, in the faces-of-cube example, what I might do is this. I fix, once and for all, a rotation \rho that takes my initial face to the given face. And then, given a rotation that fixes the initial face, I compose it with \rho to get a rotation that takes the initial face to the given face.

Letís try that with S_x and S_{xy}. Since y belongs to the orbit of x (by assumption), there is some h such that hx=y. Now let me define a map from S_x to S_{xy} by mapping g to hg. Note that hg takes x first to x (since g\in S_x) and then to y (since h\in S_{xy}). Thus, hg really does belong to S_{xy}.

However, is the map weíve just defined a bijection? Letís see if we can prove that itís an injection and a surjection. No, letís not do that. Letís try to find an inverse. Given an element u of S_{xy}, how might we convert it into an element of S_x? Thereís only one thing we can even dream of trying at this point. Since u takes x to y and h also takes x to y, h^{-1}u takes x to x, as we want. So the map u\mapsto h^{-1}u takes S_{xy} to S_x. Does it invert the previous map? Yes of course it does.

Weíve shown that for each y\in O_x there are precisely |S_x| elements of G that take x to y. But every element of G takes x to something in the orbit, so |G|=|O_x||S_x|, as we were trying to prove. \square

In case you found that off-puttingly long, here it is stripped of all the accompanying chat.

The real Proof 1. Let y be an arbitrary element of O_x, and let S_{xy}=\{g\in G:gx=y\}. Pick h\in S_{xy} and define a map \phi:S_x\to S_{xy} by \phi:g\to hg. The map \psi:S_{xy}\to S_x defined by \psi:u\to h^{-1}u inverts \phi, so |S_x|=|S_{xy}| for every y\in O_x. But the sets S_{xy} with y\in O_x form a partition of G. It follows that |G|=|S_x||O_x|. \square

I cheated slightly there by not checking carefully that \phi really does take S_x to S_{xy} and that \psi really does take S_{xy} to S_x. But those facts are easy enough to be left to the reader to do in his/her head.

That proof is so short that one might wonder why another proof could possibly be desirable. Well, itís mostly an aesthetic point, at least as far as this proof is concerned, but thereís something a bit ugly about choosing that h. Why? Because I didnít say how to do it. So, for example, the way I chose h for one y might be completely unrelated to the way I chose it for another y. Wouldnít it be nicer if there were some natural way of defining h in terms of y?

Well, yes it would, but itís not that easy to do. Just think of the cube example. Suppose I take a face F and I ask you to define, for each other face F', a rotation of the cube that takes F to F'. I want you to do it in a nice systematic way. How might you do it? Well, if F'=F then probably the most natural thing to do is take the identity. How about if F' is one of the neighbouring faces to F? Perhaps the most natural thing there is to take a 90-degree rotation about an axis through the centre of the cube thatís parallel to the edge where F and F' meet. But then what about the face opposite F? There are four ways of rotating the cube so that F lands up at the opposite face, and they are of two kinds: two of them are half turns about axes that go through the midpoints of two opposite faces, and the other two are half turns about axes that go through the midpoints of two opposite edges. (If you draw a diagram, I hope youíll be able to see what Iím talking about here. Itís one of those things that may be easier to work out for yourself.) Anyhow, for each of these two types of half turn, there is absolutely nothing to choose between the two half turns of that type. So there just isnít a natural way to choose a rotation for each face. As mathematicians might say, there isnít a canonical choice.

So is that the end of the story? Not quite. Thereís a useful principle that sometimes applies in this kind of situation. It says this.

If you canít make a canonical choice, then make all choices at once.
This meme was embedded in my brain as a result of a nice Mathoverflow question, though I suppose I was aware of it less consciously before that. Letís see how it plays out with our proof.

If I donít choose just one h\in S_{xy}, then what do I do instead? I choose all h that belong to S_{xy}.

A problem then arises when I try to define a bijection from S_x to S_{xy}. Never mind, though. Letís just define a map from S_{xy}\times S_x to S_{xy} by sending (h,g) to hg. Note that this is like what we did before, but weíre doing it for every h and not just one chosen h. And whatís nice about it is that we are now doing precisely the same thing for every y.

Letís call our map \phi. Obviously, \phi is not a bijection, but that doesnít matter. Weíre trying to show that itís a bit like a putting together of |S_{xy}| bijections. So what weíd like to show is that every element of S_{xy} has exactly |S_{xy}| preimages under \phi. That will tell us that |S_{xy}||S_x|=|S_{xy}|^2, which will imply that |S_{xy}|=|S_x|.

So letís take u\in S_x. How many preimages has it got? Well, a preimage of u is a pair (h,g) such that hg=u. So I want to show that there are precisely |S_{xy}| solutions of the equation hg=u with h\in S_{xy} and g\in S_x. But thatís easy: for each h\in S_{xy} there is precisely one possible g, namely h^{-1}u, which does indeed belong to S_x.

Let me again give the condensed version, just to convince you that what Iíve written doesnít add up to a long proof. In fact, I donít even need to bother to define the function \phi.

Proof 2. Let y be an arbitrary element of O_x and let S_{xy}=\{g\in G:gx=y\}. For every u,h\in S_{xy} there is exactly one g\in S_x such that hg=u, namely h^{-1}u. It follows that every u\in S_{xy} can be written in |S_{xy}| ways as hg with h\in S_{xy} and g\in S_x. Also, hg\in S_{xy} for every h\in S_{xy} and every g\in S_x. It follows that |S_{xy}||S_x|=|S_{xy}|^2. Therefore, |S_x|=|S_{xy}|. But the sets S_{xy} with y\in O_x form a partition of G. It follows that |G|=|S_x||O_x|. \square

Here is a variant of the above argument that I like better.

Proof 3. Let y be an arbitrary element of O_x and let S_{xy}=\{g\in G:gx=y\}. Let u\in S_{xy} and let us see in how many ways we can write u=hg with h\in S_{xy} and g\in S_x. Well, for each h\in S_{xy}, there is exactly one g\in S_x such that hg=u, namely h^{-1}u. So we can write u=hg in |S_{xy}| ways. But also, for each g\in S_x there is exactly one h\in S_{xy} such that hg=u, namely ug^{-1}. This shows that we can write u=hg in |S_x| ways. Therefore, |S_x|=|S_{xy}|. But the sets S_{xy} with y\in O_x form a partition of G. It follows that |G|=|S_x||O_x|. \square

Or do I like it better? Itís starting to look, with its non-canonical choice of u\in S_{xy}, a bit too like the first argument.

I still donít feel as though Iíve arrived at the neatest and most symmetrical argument, and Iíve also not satisfied another urge, which is to prove that |G|=|O_x||S_x| by finding a nice bijection between G and O_x\times S_x. There are good reasons for the second problem, as Iíve already discussed, but I might be able to satisfy my urge by finding not a one-to-one correspondence, but something a bit more general and multivalued.

So letís take a point (y,g)\in O_x\times S_x. What can we do with it? We can use it to define the element gy of X, but that doesnít seem very exciting or helpful. What weíd really like is to define an element of G, which earlier we did by fixing some h\in S_{xy} and taking hg. But that felt too non-canonical, so instead we preferred to take all h.

If we take all h, then thatís saying that we can get from x to y in several different ways. So we seem to get a map from G\times S_x to G, which is very simple: it takes (h,g) to hg. But whatís the point of it? And where does O_x come in?

One way of making O_x come in is to go one step further and let hg act on x to create hgx=hx. So now weíve got a map from G\times S_x to O_x, the map that takes (h,g) to hx. How many preimages does an element y have?

Iím not going to answer that question (though I could) because once weíve got to the point of mapping G\times S_x to O_x using the map above, we see that g isnít really playing a role, so why not just focus on the map from G to O_x that takes g (which I was calling h above) to gx? If y\in O_x, then how many preimages does y have?

I wonít bother to answer that question either: itís the same as asking how big the set S_{xy} is, and weíve established already that that is |S_x|, so it wouldnít be a particularly new proof that resulted. However, one remark thatís worth making is that the set S_{xy} is a left coset of S_x. Probably the proof you were given in lectures was this.

Proof 4. We shall show that there is a bijection between O_x and the set of left cosets of S_x. To do this, we map the left coset gS_x to gx. We must show that this is well-defined. But if gS_x=hS_x, then h^{-1}gS_x=S_x, and therefore, since S_x is a subgroup, h^{-1}g\in S_x. From that it follows that h^{-1}gx=x, so gx=hx.

Since the number of left cosets of S_x is |G|/|S_x|, we are done. \square

If the phrase ďwell-definedĒ worries you in the above argument, then I recommend a post I once wrote about what it means.

Hereís a different way of writing the above proof, where we think more about equivalence relations than about partitions.

Proof 5. Here are two equivalence relations on G. For the first, I define g\sim_1h if gx=hx. For the second, I define g\sim_2h if h^{-1}g\in S_x. Note that this is the same as saying that g^{-1}h\in S_x and also the same as saying that gS_x=hS_x.

Now gx=hx if and only if h^{-1}gx=x if and only if h^{-1}g\in S_x. So the two equivalence relations are the same. The number of equivalence classes for the first relation is obviously |O_x| and the size of each equivalence class for the second relation is obviously |S_x|, so the result is proved. \square

Let me make one final remark about the orbit-stabilizer theorem. Why, one might ask, is it a useful result? A rather general answer is that it gives us a relationship between three quantities, namely |G|, |S_x| and |O_x|, that allows us to determine any one of them from the other two. Situations crop up quite frequently in group theory where it is not very easy to see what one of these quantities is directly, but quite easy to calculate the other two. The orbit-stabilizer theorem then gives us the hard one too. For example, in the case of counting rotational symmetries of a cube, it isnít easy to think of all the rotations unless you partition them in some nice way, such as looking at what they do to a particular vertex or face, which, as we saw before, amounts to counting orbits and stabilizers.

Itís not always |G| thatís the tough one to get a handle on. For example, in question 10 of Examples Sheet 3 itís the size of the orbit that isnít obvious. (In general with the bunch of questions around there, I recommend thinking to yourself, ďWhat is the orbit-stabilizer theorem giving me?Ē)
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Old 04-08-2014, 11:14 AM   #12
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What class is this for? Glad I never had to deal with this crap.
Modern algebra, finishing up a math minor.

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I have a confession to make, most of the higher level math I know (group theory stuff) is from physics classes, I haven't taken many formal higher level math courses, and even those were a long time ago and I haven't used them much since. I can't help you here
I hope I never deal with it again in this type of setting. It's interesting because all of modern cryptography has a base in it, and I know things like SL(2) and GL(2) pop up in physics, but I hate writing proofs with a passion. Apply it and I'm fine, but math for math's sake is awful.
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Old 04-08-2014, 11:15 AM   #13
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Well, I don't think you're gonna get a better answer than that
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Old 04-08-2014, 11:15 AM   #14
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lol


Permutation groups niqqa

edit: Where the fvck is Zell, I know he knows this stuff.
LOL gotta laugh at that one. Maybe I'll just stick to my kinematics/mechanics that actually have practical applications.

BTW 217 doesn't know anything about this. He copied this blog post:

http://gowers.wordpress.com/2011/11/...rem/#more-3720
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Old 04-08-2014, 11:20 AM   #15
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LOL gotta laugh at that one. Maybe I'll just stick to my kinematics/mechanics that actually have practical applications.

BTW 217 doesn't know anything about this. He copied this blog post:

http://gowers.wordpress.com/2011/11/...rem/#more-3720
niqqa i wrote that blog post
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Old 04-08-2014, 11:20 AM   #16
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prove it
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Old 04-08-2014, 11:24 AM   #17
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Well, I don't think you're gonna get a better answer than that
If you can't explain it to your grandmother then you don't know it well enough, so there has to be a better answer. Anyway, I've found some stuff about the Rubik's Cube Group that helped elucidate the idea. My idea was close, but I missed it a little. The stabilizer lives in G so my picture isn't quite right, but it's good enough for me. All I need is a picture and I can remember the formula.

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LOL gotta laugh at that one. Maybe I'll just stick to my kinematics/mechanics that actually have practical applications.
Like I sad before, this stuff underlies all of modern cryptography. It has applications but they're more abstract than classical physics.
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Old 04-08-2014, 11:29 AM   #18
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If you can't explain it to your grandmother then you don't know it well enough, so there has to be a better answer. Anyway, I've found some stuff about the Rubik's Cube Group that helped elucidate the idea. My idea was close, but I missed it a little. The stabilizer lives in G so my picture isn't quite right, but it's good enough for me. All I need is a picture and I can remember the formula.
Tell that to the guys who write mathematics Wikipedia articles. They're the least useful descriptions ever.

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Originally Posted by cowmoo32 View Post
Like I sad before, this stuff underlies all of modern cryptography. It has applications but they're more abstract than classical physics.
I'm sure it's got applications. Most math does. This just isn't particularly practical for many people to learn.
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Last edited by WDE46; 04-08-2014 at 11:29 AM.
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Old 04-08-2014, 11:40 AM   #19
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Tell that to the guys who write mathematics Wikipedia articles. They're the least useful descriptions ever.
https://simple.wikipedia.org seriously, it's great.
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Old 04-08-2014, 11:48 AM   #20
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niqqa i wrote that blog post
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