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From: ar0 on 13 Jun 2010 19:44
> No, the operation must usually be stated.
> [...]
> not even hinting the reader towards that is somewhat unfair, esp
> to beginners.
Ok, that's clear enough. I guess I should pay more attention to context and think
about the various possibilies for non-trivial homomorphisms into Aut(N).


> If in doubt, give the automorphism used.
> Better still: *Always* give the automorphism used.
I would wholeheartedly concur. But I think, as you said,
the levels of "obviousness" are differing for a student
who is exposed to the subject for the first time and
a professor who knows the constructions inside out.

For other comments: see my other reply.

> hagman

Thanks for the reply.
best regards.

--
Sick nature.
From: Arturo Magidin on 13 Jun 2010 20:08
On Jun 13, 6:33 pm, no...(a)nospam.invalid (ar0) wrote:
> Arturo Magidin <magi...(a)member.ams.org> wrote:

> > 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?

And the answer is "yes": once you have constructed the "external"
semidirect product G = N x|_f H (f is the map H-->Aut(N)), you can
view *it* is an internal semidirect product in which the subgroup of
all (1,n) correspond to N, the subgroup of all (h,1) correspond to H.

What is the action by conjugation of an element (1,h) on the subgroup
we have denoted N, according to the rule of mulitplication we placed
on G? You had

(h,n) * (h', n') := (h*h', f(h'^(-1))(n)*n')

Now, the inverse of (h,1) in G is (h^{-1},1). So if you conjugate
(1,n) by (h,1), you get:

(h,1)(1,n)(h^{-1},1) = (h*1 , f(1^{-1})(1)*n ) * (h^{-1} ,1)
= (h, n) * (h^{-1},1)
= (h*h^{-1} , f(h)(n)*1)
= (1, f(h)(n) )

so the action "by conjugation" inside G is exactly the action induced
by the map f:H-->Aut(N). Thus, you have realized the original exterior
semidirect product as an internal one in the group you constructed.


> > 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.

As I said, and as Hagman said, it is a bit of an abuse. In most cases,
experience or context makes it "clear" which one is meant. But not
always, alas. So sometimes such an expression leads to an inexact
result.

> > 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.

Right; but in this case you *have* D_n. What if you didn't know what
D_n was, though? One way to "tell" someone what D_n means is to
describe it as a subgroup of S_n; another is to describe it as a
semidirect product of C_n by C_2. The latter can be done without any
reference to S_n, or to a regular n-gon; it's a purely abstract
construction.

While for certain values of n there are several non-isomorphic
semidirect products of C_n by C_2 (depending on the map f:C_2--
>Aut(C_n)), there is a choice that works for every n which is to map
C_2 to the subgroup corresponding to the inversion map x|->x^{-1}.
This will work for every n, and thus gives you a construction that
works for every n. So, from context, it makes sense (after a non-
trivial amount of thought and work, mind you) that "the" semidirect
product being refered to here is the only you obtain by using that
action uniformly (that is, for all values of n).

> 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.

Alas, no, that is not the case.

> 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.

No, since there are nonisomorphic semidirect products with the same N
and the same H.

--
Arturo Magidin
From: Derek Holt on 14 Jun 2010 04:32
On 13 June, 21:46, Arturo Magidin <magi...(a)member.ams.org> wrote:
> 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.

I would agree that inversion is the obvious way to define an action of
C_2 on C_n. But I am not so sure about "generally speaking, usually
the only nontrivial automorphism of exponent 2 is the map that sends
every element to its inverse".

Let n = product n_i from i=i to k, with n_i = p_i^{e_i}, e_i>0.

be the prime power factorization of n. Then Aut(C_n) is the direct
product of the Aut(C_{n_i}).

For p odd, Aut(C_{n_i}) is cyclic and has a unique element of order 2.

For p=2 and n_2 > 2, Aut(C_{n_i}) has 3 elements of order 2:
inversion, x->x^{2^{n_2-1}+1}, and x->x^{2^{n_2-1}-1}.

So C_n has either t2^{k-1}-1 automorphisms of order 2 where t=1, 2 or
4,, and these all give rise to non-isomorphic semidirect products.

Derek Holt.


From: Arturo Magidin on 14 Jun 2010 15:01
On Jun 14, 3:32 am, Derek Holt <ma...(a)warwick.ac.uk> wrote:
> On 13 June, 21:46, Arturo Magidin <magi...(a)member.ams.org> wrote:

On my wrongheaded statement that:

> > 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.

[...]

> I would agree that inversion is the obvious way to define an action of
> C_2 on C_n. But I am not so sure about "generally speaking, usually
> the only nontrivial automorphism of exponent 2 is the map that sends
> every element to its inverse".
>
> Let n = product n_i from i=i to k, with n_i = p_i^{e_i}, e_i>0.
>
> be the prime power factorization of n. Then Aut(C_n) is the direct
> product of the Aut(C_{n_i}).
>
> For p odd, Aut(C_{n_i}) is cyclic and has a unique element of order 2.
>
> For p=2 and n_2 > 2, Aut(C_{n_i}) has 3 elements of order 2:
>   inversion, x->x^{2^{n_2-1}+1}, and x->x^{2^{n_2-1}-1}.
>
> So C_n has either t2^{k-1}-1 automorphisms of order 2 where t=1, 2 or
> 4,, and these all give rise to non-isomorphic semidirect products.

In the back of my mind, I knew that you only needed to have k
congruent to 1 or -1 for each prime power dividing n, and so long as
at least one was congruent to -1 you would get a nontrivial
automorphism; nonetheless, I go ahead and shoot my mouth off...
Thanks.

Perhaps better is to say that (looking at the automorphism x|-> x^k
for <x> = C_n) choosing k=-1 is the only choice that works for all
values of n, so in a general discussion about D_n expressed as a
semidirect product of C_n by C_2, this is the only value of k that
will define a nonabelian semidirect product for every value of n.

--
Arturo Magidin
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