Wreath Products (Run away!)
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Wreath Products (Run away!)
Is there anybody out there who can help me understand these better that is not a website?
Yeah, I know I'm a drunk Irish cowboy ninja...
Re: Wreath Products (Run away!)
What's confusing about them? Do you understand semidirect products in general? It would be helpful if you asked more specific questions.

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Re: Wreath Products (Run away!)
It is my understanding that semidirect products are essentially ntuples of the first set cross the second set. Although this is the first day I encountered them both the semidirect products and wreath products), I think my real flaw in understanding is the decomposition of the group of permutations of a finite set into disjoint sets. How does this happen and what other information is required to form the specific decomposition?
The reason I posted this question was for 2 purposes:
(1) I am a Junior undergrad at the University of Wyoming studying Inverse Galois Theory, though I have no business being in there since I'm concurrently in grad Abstract Algebra. I took it for fun and got confused today (not the first time) on wreath products (although I got the gist). I just joined the forum today and thought, Hey, why not ask around?
(2) To probe the math brains who regularly attend the forum and see if I might like to frequent the forums myself. At first it looked like a lot of calc stuff when it came to math but I have been pleasantly surprised.
Thanks!
The reason I posted this question was for 2 purposes:
(1) I am a Junior undergrad at the University of Wyoming studying Inverse Galois Theory, though I have no business being in there since I'm concurrently in grad Abstract Algebra. I took it for fun and got confused today (not the first time) on wreath products (although I got the gist). I just joined the forum today and thought, Hey, why not ask around?
(2) To probe the math brains who regularly attend the forum and see if I might like to frequent the forums myself. At first it looked like a lot of calc stuff when it came to math but I have been pleasantly surprised.
Thanks!
Yeah, I know I'm a drunk Irish cowboy ninja...
Re: Wreath Products (Run away!)
hashbrowns wrote:It is my understanding that semidirect products are essentially ntuples of the first set cross the second set.
No, that's wreath products, and you should say "group."
hashbrowns wrote:I think my real flaw in understanding is the decomposition of the group of permutations of a finite set into disjoint sets.
I don't understand what decomposition you're referring to. It sounds like you need to brush up on your definitions a little.

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 Joined: Thu Feb 04, 2010 12:41 am UTC
 Location: Laramie, Wyoming
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Re: Wreath Products (Run away!)
So if what I stated WAS a wreath product, then what is a generic semidirect product? I mean, I read the wiki definition and all, but if you gave me the two subgroups to make the group via semidirect product, I would be at a loss.
as for the decomp, let's see if i can get the imath to work...
here's what we got in our notes
let [imath]\Omega[/imath] be a finite set, [imath]Sym(\Omega)[/imath] a group of permutations of [imath]\Omega[/imath]
Def: A permutation representation of [imath]G[/imath] on [imath]\Omega[/imath] is a homomorphism from [imath]G \rightarrow Sym(\Omega)[/imath]
Def: [imath]\rho : G \rightarrow Sym(\Omega)[/imath] is transitive (or irreducible) if [imath]]\rho(G)[/imath] has a single orbit on [imath]\Omega[/imath]
My question lies above with the decomposition of [imath]\Omega[/imath] into its Gorbits. Then I think the definition of wreath product, [imath]S_e \wr S_m[/imath] would make more sense.
as for the decomp, let's see if i can get the imath to work...
here's what we got in our notes
let [imath]\Omega[/imath] be a finite set, [imath]Sym(\Omega)[/imath] a group of permutations of [imath]\Omega[/imath]
Def: A permutation representation of [imath]G[/imath] on [imath]\Omega[/imath] is a homomorphism from [imath]G \rightarrow Sym(\Omega)[/imath]
Def: [imath]\rho : G \rightarrow Sym(\Omega)[/imath] is transitive (or irreducible) if [imath]]\rho(G)[/imath] has a single orbit on [imath]\Omega[/imath]
My question lies above with the decomposition of [imath]\Omega[/imath] into its Gorbits. Then I think the definition of wreath product, [imath]S_e \wr S_m[/imath] would make more sense.
Yeah, I know I'm a drunk Irish cowboy ninja...
Re: Wreath Products (Run away!)
About semidirect products:
Let G be the dihedral group [imath]D_8[/imath], and let [imath]A=\langle a\rangle[/imath] be a subgroup of order 4 and [imath]b[/imath] an involution (element of order 2) outside of [imath]A[/imath]. Clearly [imath]A[/imath] is a normal subgroup and [imath]B=\langle b\rangle[/imath] is a subgroup, we have [imath]A \cap B=1[/imath], and [imath]AB=G[/imath]. These are almost the criteria for a group to be a direct product  if only [imath]B[/imath] we normal also, we would be done  but clear [imath]G[/imath] is not the direct product of [imath]A[/imath] and [imath]B[/imath], since it is nonabelian.
What we have here is a semidirect product. Here, the one subgroup is normal, but the other subgroup isn't, because it acts on the first one. In a direct product [imath]H=X\times Y[/imath], the two groups [imath]X[/imath] and [imath]Y[/imath] are normal and intersect trivially. Since [imath][X,Y]\leq X\cap Y=1[/imath], we see that [imath]X[/imath] and [imath]Y[/imath] must commute. So in any direct product, the conjugation action on one factor induced by elements of the other (remember both a normal, so there is a conjugation action) is trivial.
Now let us return to our dihedral example. Here, [imath]b[/imath] inverts [imath]a[/imath], and so the action of [imath]B[/imath] on [imath]A[/imath] (by conjugation) is nontrivial. (This is why [imath]B[/imath] is not normal in [imath]G[/imath].) This is a semidirect product; i.e., a semidirect product is a group with two trivially intersecting subgroups, one of which is normal, that generate the group.
A semidirect product, denoted [imath]G=A\rtimes B[/imath], gives you groups [imath]A[/imath] and [imath]B[/imath], and a way of associating, to each element of [imath]B[/imath], a conjugation map on [imath]A[/imath]. This last piece of data is really a homomorphism [imath]B\to \mathrm{Aut}(A)[/imath]. Conversely, if you are given [imath]A[/imath], [imath]B[/imath], and a map [imath]B\to\mathrm{Aut}(A)[/imath], you can make a group realizing these data. This construction is given in detail in any book, and I don't think it would help here.
Let G be the dihedral group [imath]D_8[/imath], and let [imath]A=\langle a\rangle[/imath] be a subgroup of order 4 and [imath]b[/imath] an involution (element of order 2) outside of [imath]A[/imath]. Clearly [imath]A[/imath] is a normal subgroup and [imath]B=\langle b\rangle[/imath] is a subgroup, we have [imath]A \cap B=1[/imath], and [imath]AB=G[/imath]. These are almost the criteria for a group to be a direct product  if only [imath]B[/imath] we normal also, we would be done  but clear [imath]G[/imath] is not the direct product of [imath]A[/imath] and [imath]B[/imath], since it is nonabelian.
What we have here is a semidirect product. Here, the one subgroup is normal, but the other subgroup isn't, because it acts on the first one. In a direct product [imath]H=X\times Y[/imath], the two groups [imath]X[/imath] and [imath]Y[/imath] are normal and intersect trivially. Since [imath][X,Y]\leq X\cap Y=1[/imath], we see that [imath]X[/imath] and [imath]Y[/imath] must commute. So in any direct product, the conjugation action on one factor induced by elements of the other (remember both a normal, so there is a conjugation action) is trivial.
Now let us return to our dihedral example. Here, [imath]b[/imath] inverts [imath]a[/imath], and so the action of [imath]B[/imath] on [imath]A[/imath] (by conjugation) is nontrivial. (This is why [imath]B[/imath] is not normal in [imath]G[/imath].) This is a semidirect product; i.e., a semidirect product is a group with two trivially intersecting subgroups, one of which is normal, that generate the group.
A semidirect product, denoted [imath]G=A\rtimes B[/imath], gives you groups [imath]A[/imath] and [imath]B[/imath], and a way of associating, to each element of [imath]B[/imath], a conjugation map on [imath]A[/imath]. This last piece of data is really a homomorphism [imath]B\to \mathrm{Aut}(A)[/imath]. Conversely, if you are given [imath]A[/imath], [imath]B[/imath], and a map [imath]B\to\mathrm{Aut}(A)[/imath], you can make a group realizing these data. This construction is given in detail in any book, and I don't think it would help here.
Re: Wreath Products (Run away!)
 MartianInvader
 Posts: 808
 Joined: Sat Oct 27, 2007 5:51 pm UTC
Re: Wreath Products (Run away!)
A common mistake is to think there's such a thing as "the semidirect product of two groups". The trick is that you need more information.
To make a semidirect product, I don't just have to hand you a pair of groups. I also have to specify an action of the second group on the first (where the action is by group automorphisms). People like to talk about "The semidirect product of G with H", but really there's some implied action going on.
The underlying set of a semidirect product is just the product set. So if I've got G and H, the elements look like (g,h). But how do I multiply (g,h) by (g',h')? Well, visually, I want to turn g h g' h' into something that looks like g g' h h', so I can multiply the components together using the group multiplication from G and H. In a normal direct product, that's what I do.
But in a semidirect product, I have to pay a price for switching the g' with the h. The price is that h acts on g' as it goes past, using whatever group action is specified . So I get that (g,h)(g',h') = (gh(g'), hh'), where h(g') is what you get by taking the specified group action of H on G, and having h act on g'.
For example, let's take G = H = Z. I can't just say "the semidirect product of Z with itself", I also need to specify an action of Z on itself, where acting by each individual element induces an automorphism. Let's choose the action n(m) = (1)^n m, ie acting by n just multiplies by 1 n times. I'll let you check that this is, in fact, a group action. (bonus points to whoever identifies what group this is)
The group we get then has elements of the form (a,b), where a and b are integers, BUT (a,b)(c,d) isn't always (a+c, b+d). It's (a+c, b+d) if b is even, and it's (ac, b+d) if b is odd. You can check for yourself that this "multiplication" does give us a genuine, bonafide group.
I hope that helps you with semidirect products. I don't have time to explain wreath products right now, but I can come back and describe them a bit later.
To make a semidirect product, I don't just have to hand you a pair of groups. I also have to specify an action of the second group on the first (where the action is by group automorphisms). People like to talk about "The semidirect product of G with H", but really there's some implied action going on.
The underlying set of a semidirect product is just the product set. So if I've got G and H, the elements look like (g,h). But how do I multiply (g,h) by (g',h')? Well, visually, I want to turn g h g' h' into something that looks like g g' h h', so I can multiply the components together using the group multiplication from G and H. In a normal direct product, that's what I do.
But in a semidirect product, I have to pay a price for switching the g' with the h. The price is that h acts on g' as it goes past, using whatever group action is specified . So I get that (g,h)(g',h') = (gh(g'), hh'), where h(g') is what you get by taking the specified group action of H on G, and having h act on g'.
For example, let's take G = H = Z. I can't just say "the semidirect product of Z with itself", I also need to specify an action of Z on itself, where acting by each individual element induces an automorphism. Let's choose the action n(m) = (1)^n m, ie acting by n just multiplies by 1 n times. I'll let you check that this is, in fact, a group action. (bonus points to whoever identifies what group this is)
The group we get then has elements of the form (a,b), where a and b are integers, BUT (a,b)(c,d) isn't always (a+c, b+d). It's (a+c, b+d) if b is even, and it's (ac, b+d) if b is odd. You can check for yourself that this "multiplication" does give us a genuine, bonafide group.
I hope that helps you with semidirect products. I don't have time to explain wreath products right now, but I can come back and describe them a bit later.
Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!

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Re: Wreath Products (Run away!)
those both help, thanks a lot.
Yeah, I know I'm a drunk Irish cowboy ninja...
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