Re: Zappa-Szep product in wiki

On 21 jul, 16:11, magi...@xxxxxxxxxxxxxxxxx (Arturo Magidin) wrote:
In article <1185017811.231477.252...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,

Anvita <anvit...@xxxxxxx> wrote:
On 20 jul, 15:14, magi...@xxxxxxxxxxxxxxxxx (Arturo Magidin) wrote:
In article <1184954680.014513.202...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,

Anvita <anvit...@xxxxxxx> wrote:
The definition of the Zappa-Szep product that is given in wikipedia at

http://en.wikipedia.org/wiki/Zappa-Szep_product

includes the property

[...]

b(k, h1*h2) = b(b(k,h2),h1).

However, I suspect that it should instead be

i.e., b(k, h1*h2) = b(b(k,h1),h2).

because, for example, if K normal in G,
we have k h =3D h (k^h)

k*h = h*(k^h)

(this holds regardless of whether K is normal...)

It surely does. But if K is not normal then k^h is in general not in
K which is not what we want.

"We", Kimosabe?


What exactly are you asking?



and so =CE=B2(k, h)=3Dk^h.

k^(h1*h2) = (k^h1)^h2

Again: this holds regardless of whether K is normal: for all elements
a, b, and c in a group

a^(bc) = (bc)^{-1} a (bc) = c^{-1}b^{-1}abc = c^{-1}(b^{-1}ab)c
= c^{-1}(a^b)c
= (a^b)^c.

Hence,
the left-hand side of the above relation is

k^(h1 h2) - (k^h1)^ h2
which equals
b(b(k,h1),h2).
Am I right?

Not entirely. Instead of using the operation x^y = y^{-1}xy, you can
use the operation

beta(k,h) = hkh^{-1} = {}^h k.

This is still not clear to me.

You asked about defining b (beta) via conjugation. There are TWO
conventions for conjugation-by-h. Either x |-> hxh^{-1}, or
x|-> h^{-1}xh.

In a sense, the two are equivalent: it's just a question of left-right
symmetry. You want to use the second convention, which leads to an
equation involving b which is not the one in Wikipedia. But if you
stick to the FIRST convention, then the equation in Wikipedia is the
correct one that codifies that convention, as opposed to the version
you want.

If we define b ( =beta ) this way then,
in order for the relation

k*h=a(k,h)*b(k,h)

To have this equation, you are assuming

It is not *me* who is assuming. I have just taken the assumptions of
the author of the wikipedia article.

that G=HK is an internal
Zappa-Szep product, which requires in addition that H/\K = {1}. So
what we have

Note the use of "we".

is a semi-direct product of K by H. In that situation,
you want to define the maps the ->other<- way, i.e., with the normal
group playing the role of H in the definitions in the Wikipedia page,
and the not-necessarily normal playing the role of K.

You are not saying that in a Zappa-Szep product G=HK the subgroup K
cannot be normal, are you?

to hold, we should define a(k,h)=k*h^2*k^{-1}*h^{-1}.

You do realize that you are transposing the roles of H and K in an
internal product?

I am not transposing the roles of H and K. There are many examples of
products G=HK in which the second factor is normal and the first one
is not.

When G is a semidirect product of two subgroups, the
realization of the Zappa-Szep product assumes that H is normal and K
acts on H, not the other way around;

I disagree. Initially there are no assumptions on whether H or K is
normal. A semidirect product is just a particular case regardless of
which factor is normal or which "conjugation convention" one takes.

the realization in the wikipedia
article (under "Relation to semidirect and direct products") takes
a(k,h)=khk^{-1} (notice the conjugation convention), and
beta(k,h)=k.

What exactly are you trying to do?


Hmmm... I guess I should finish this dialog before getting myself into
further misunderstanding.

Does anyone else in this group see what I am "trying to do"?

Derek Holt, are you reading this?

Anvita

Try starting over and switching your conjugations from x|-> h^{-1}xh
to x|->hxh^{-1}. The equations should come out correctly then.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================

Arturo Magidin
magidin-at-member-ams-org


.

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