Group theory, semidirect products
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From: ar0 on 13 Jun 2010 13:24 Hi, I know the following definition for semi direct products: Let H and N be groups, f : H -> Aut(N) a homomorphism. Then the cartesian product HxN with the multiplication (h,n) * (h', n') := (h*h', f(h'^(-1))(n)*n') (horrible to look at, I know..) forms a group, called semi direct product. Now, as a special case, if G is a group, H an arbitrary subgroup and N a normal subgroup then we always have the homomorphism from H into Aut(N) via conjugation with an element of H. (And if HN = G and H \cap N = {1}, then G is isomorphis to this semidirect product) Now I've got several questions on the relations between the general and specific case: 1.) If the homomorphism f isn't trivial, can one show that there always exists a group G such that H and N are as in the special case and the resulting semi direct products are isomorphic? (I suppose not, because Inn(N) can be a proper subset of Aut(N)) 2.) If someone just writes "H semidirect N" (with that cute symbol mixture of cartesian product + normal in the middle) without any explicit mention of the homomorphism used, what does that person mean? Does he mean I'm supposed to think of a group such that the semidirect product can be thought of as in the special case (with f being the map into the inner automorphisms of N)? I could actually go on and some sort of clarification on this whole matter would really be appreciated. As an example for 2.): I read somewhere that the dihedral group D_n (of order 2n) is isomorphic to C_2 semidirect C_n, where C_n is the cyclic group of order n. As D_n is generated by an n-cycle and an involution, I would identify C_2 with the subgroup generated by the involution and C_n with the subgroup generated by the n-cycle. Now since the homomorphism from C_2 into Aut(C_n) wasn't mentioned explicitly I would guess that it's via conjugation. Is that correct? best regards. -- Sick nature.
From: Arturo Magidin on 13 Jun 2010 16:46 On Jun 13, 12:24 pm, no...(a)nospam.invalid (ar0) wrote: > Hi, > I know the following definition for semi direct products: > Let H and N be groups, f : H -> Aut(N) a homomorphism. > Then the cartesian product HxN with the multiplication > (h,n) * (h', n') := (h*h', f(h'^(-1))(n)*n') > (horrible to look at, I know..) > forms a group, called semi direct product. > > Now, as a special case, if G is a group, H an arbitrary subgroup and N a normal subgroup then > we always have the homomorphism from H into Aut(N) via conjugation with an element > of H. (And if HN = G and H \cap N = {1}, then G is isomorphis to this semidirect product) > > Now I've got several questions on the relations between the general and specific case: > > 1.) If the homomorphism f isn't trivial, can one show that there always exists a group > G such that H and N are as in the special case and the resulting semi direct products > are isomorphic? (I suppose not, because Inn(N) can be a proper subset of Aut(N)) I'm not sure why you talk about "trivial" here with f; the trivial map is the one that maps all of H to 1, so you get the direct product... You mean, is every "external" semidirect product also realizable as an "internal" semidirect product? Of course! Just take G to be the group with underlying set H x N, and mulitplication as given by the action. Then identify H with the subgorup of all (h,1), and N with the subgroup of all (1,n) and you've got your group. I also don't know why you bring up Inn(N). Note that even in the case where N is a normal subgroup of G, the map form Inn(G) to Aut(N) given by restriction does not necessarily land "inside" Inn(N): an inner automorphism of G may induce an outer automorphism of N. For example, consider the case when N is abelian. > 2.) If someone just writes "H semidirect N" (with that cute symbol mixture of cartesian product + normal in the middle) without > any explicit mention of the homomorphism used, what does that person mean? He means that the action is "understood from context". This usually means that there is an obvious action to take. It's a bit of an abuse. > Does he mean I'm supposed to think of a group such that the semidirect product can be thought of as in the > special case (with f being the map into the inner automorphisms of N)? > > I could actually go on and some sort of clarification on this whole matter would really be appreciated. > > As an example for 2.): > > I read somewhere that the dihedral group D_n (of order 2n) is isomorphic to > C_2 semidirect C_n, where C_n is the cyclic group of order n. > As D_n is generated by an n-cycle and an involution, I would identify > C_2 with the subgroup generated by the involution and C_n with the subgroup generated by > the n-cycle. Now since the homomorphism from C_2 into Aut(C_n) wasn't mentioned explicitly I > would guess that it's via conjugation. If you *already* have D_n, then yes; but if what you have is C_n and C_2, it makes no sense to talk about "conjugation" because they are not subgroup of the same group. You are trying to *construct* D_n, not recognize D_n as a semidirect product once you already have it. Now, as it happens, C_n does not have that many automorphisms of exponent 2 (that is, there aren't that many group homomorphisms form C_2 to Aut(C_n). If you think of C_n as the multiplicative group generated by x with exponent taken modulo n, then an automorphism of C_n is completely determined by what it does to x, and must be of the form x|->x^k for some k. If it is of exponent 2, then (x^k)^k = x^{k^2} must be equal to x, so k^2 = 1 (mod n), hence (k-1)(k+1)=0 (mod n). Two obvious choices are k=1 and k=-1; for a few n, there might be a few other values (e.g., if n=15, you could have k=4), but generally speaking, usually the only nontrivial automorphism of exponent 2 is the map that sends every element to its inverse. So *that's* the "obvious" way to define the action of C_2 on C_n, and that is how D_n is defined as a semidirect product. -- Arturo Magidin
From: hagman on 13 Jun 2010 17:09 On 13 Jun., 19:24, no...(a)nospam.invalid (ar0) wrote: > Hi, > I know the following definition for semi direct products: > Let H and N be groups, f : H -> Aut(N) a homomorphism. > Then the cartesian product HxN with the multiplication > (h,n) * (h', n') := (h*h', f(h'^(-1))(n)*n') > (horrible to look at, I know..) > forms a group, called semi direct product. > > Now, as a special case, if G is a group, H an arbitrary subgroup and N a normal subgroup then > we always have the homomorphism from H into Aut(N) via conjugation with an element > of H. (And if HN = G and H \cap N = {1}, then G is isomorphis to this semidirect product) > > Now I've got several questions on the relations between the general and specific case: > > 1.) If the homomorphism f isn't trivial, can one show that there always exists a group > G such that H and N are as in the special case and the resulting semi direct products > are isomorphic? (I suppose not, because Inn(N) can be a proper subset of Aut(N)) The semidirect product *is* such a group. (h,1) * (1,n) *(h^-1,1) = (1,f(h)(n)) Note that Inn(N) != Aut(N) does not matter - Inn(N) is the automorphisms of N obtained by conjugation with elements of N, wheres we're conjugating with elements of H. > > 2.) If someone just writes "H semidirect N" (with that cute symbol mixture of cartesian product + normal in the middle) without > any explicit mention of the homomorphism used, what does that person mean? > Does he mean I'm supposed to think of a group such that the semidirect product can be thought of as in the > special case (with f being the map into the inner automorphisms of N)? No, the operation must usually be stated. In rare cases there is only one (non-trivial) operation, e.g. Z/pZ has exactly one automorphism of order 2, so Z/pZ X| Z/2Z leaves no choice The context (e.g. if the groups are given in a way that suggests one operation) should make it clear but that can't be guaranteed - there are always different levels of "obvious", arent there? If one says "Let N be an abelian group and consider N X| Z/2Z" - what should one make of it? Is there any natuural choice of automotrhism of order 2? Well, from the only special fact mentioned about N (it is abelian), we have that x |-> -x is such an automorphism. But not even hinting the reader towards that is somewhat unfair, esp to beginners. > > I could actually go on and some sort of clarification on this whole matter would really be appreciated. > > As an example for 2.): > > I read somewhere that the dihedral group D_n (of order 2n) is isomorphic to > C_2 semidirect C_n, where C_n is the cyclic group of order n. > As D_n is generated by an n-cycle and an involution, I would identify > C_2 with the subgroup generated by the involution and C_n with the subgroup generated by > the n-cycle. Now since the homomorphism from C_2 into Aut(C_n) wasn't mentioned explicitly I > would guess that it's via conjugation. Conjugation when viewed as subgroups of D_n ("internal" semidirect product). As "external" semidirect product this is one example I mentioned above: C_n is cyclic, hence abelian, hence inversion is an automorphism of order <= 2 (it is of order 1 for n<3). Actually, x |-> x^(-1) is not always the only order 2 automorphims as it just has to be of the form x |-> x^k with k^2 == 1 mod eulerphi(n). Thus e.g. in C_15, f: x |-> x^3 is of order 2. In fact, f is identity on C_3 and inversion on C_5 when writing C_15 = C_3 X C_5. So C_15 X_f C_2 is simply C_3 X D_5, which is not the same as D_15. > Is that correct? If in doubt, give the automorphism used. Better still: *Always* give the automorphism used. > > best regards. > > -- > Sick nature. hagman
From: Arturo Magidin on 13 Jun 2010 17:53 On Jun 13, 4:09 pm, hagman <goo...(a)von-eitzen.de> wrote: > Actually, x |-> x^(-1) is not always the only order 2 automorphims as > it just has to be of the form x |-> x^k with k^2 == 1 mod > eulerphi(n). > Thus e.g. in C_15, f: x |-> x^3 is of order 2. I suspect this is a typo, and I think you meant x |-> x^4; x|->x^3 is an endomorphism, but not an automorphism (x^5 maps to the trivial element). -- Arturo Magidin
From: ar0 on 13 Jun 2010 19:33
Arturo Magidin <magidin(a)member.ams.org> wrote: > I'm not sure why you talk about "trivial" here with f; the trivial map > is the one that maps all of H to 1, so you get the direct product... Yes, that's the case I wanted to exclude. > You mean, is every "external" semidirect product also realizable as an > "internal" semidirect product? Of course! Just take G to be the group > with underlying set H x N, and mulitplication as given by the action. > Then identify H with the subgorup of all (h,1), and N with the > subgroup of all (1,n) and you've got your group. Well, what I meant was: Is every external semidirect product also realizable as an internal semidirect product with the homomorphism being induced by conjugations? >I also don't know why you bring up Inn(N). Because I was a little bit confused. I realized shorlty after posting that those aren't actually inner automorphisms of N induced by conjugations with elements of H. > He means that the action is "understood from context". This usually > means that there is an obvious action to take. It's a bit of an abuse. Hm, ok. > If you *already* have D_n, then yes; but if what you have is C_n and > C_2, it makes no sense to talk about "conjugation" because they are > not subgroup of the same group. You are trying to *construct* D_n, not > recognize D_n as a semidirect product once you already have it. Yeah, the situation is: I have the D_n (as a subgroup of S_n) and conjugation of course is meant to happen inside the S_n. > Now, as it happens, C_n does not have that many automorphisms of > exponent 2 (that is, there aren't that many group homomorphisms form > C_2 to Aut(C_n). If you think of C_n as the multiplicative group > generated by x with exponent taken modulo n, then an automorphism of > C_n is completely determined by what it does to x, and must be of the > form x|->x^k for some k. If it is of exponent 2, then (x^k)^k = > x^{k^2} must be equal to x, so k^2 = 1 (mod n), hence (k-1)(k+1)=0 > (mod n). Two obvious choices are k=1 and k=-1; for a few n, there > might be a few other values (e.g., if n=15, you could have k=4), but > generally speaking, usually the only nontrivial automorphism of > exponent 2 is the map that sends every element to its inverse. So > *that's* the "obvious" way to define the action of C_2 on C_n, and > that is how D_n is defined as a semidirect product. You're right, there aren't that many self inverse automorphisms, and -id is the obvious one to take. I guess I just wanted to hear "yes, you can just leave out the homomorphism" and then one can just (in a canonical way) find some group which has H and N as subgroups just as in the specific case and the homomorphism is induced via conjugation. I wanted to simplify the construct and be able to forget about the homomorphism carried around with it. But I supposed that's not possible. Thanks for the input :) best regards -- Sick nature. |