Re: Zappa-Szep product in wiki

In article <1185108579.188245.59410@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Anvita <anvita21@xxxxxxx> wrote:

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


Okay, lest I contribute to futher misunderstanding, how about starting
over?

Given a group G and two subgroups HK, if H/\K={1} and G=HK, then we
say G is the Zappa-Szep product of H and K.

If we assume that G is the Zappa-Szep product of H and K, then this
allows us to define to mappings, a:K x H -> H and b:K x H -> K,
(set-theoretic, not necessarily group morphisms) defined as follows:

Since HK is a group, we must have HK=KH as sets. Thus, given h in
H and k in K, there must exist h' in H and k' in K such that
kh=h'k'. We define a(k,h) = h' and b(k,h)=k'.

Because H/\K = {1}, this is well defined, since h'k' = h''k'' implies
that (h'')^{-1}h' = k''(k')^{-1}, so this element lies in the
intersection.

This definition implies that a and b have certain properties. Your
question was about a particular property of b, namely property (v)
below. You couched it in terms of conjugation, but I think that this
is the wrong way to approach it. Rather, we should use the definition
above.

To warm up, I look at property (iv) first.

(iv) a(k1*k2,h) = a(k1,a(k2,h)).

Well, a(k1*k2,h) is the unique h' in H such that
(k1*k2)*h = h'k.

Okay, start with (k1*k2)h. We can first rewrite this as

(k1*k2)h = k1*(k2*h)
= k1*(a(k2,h)*b(k2,h))
= [k1*a(k2,h)] * b(k2,h)
= [a(k1,a(k2,h))*b(k1,a(k2,h))]b(k2,h)
= a(k1,a(k2,h))*[b(k1,a(k2,h))*b(k2,h)],
where b(k1,a(k2,h))*b(k2,h) lies in K, and a(k1,a(k2,h)) lies in H.

So we have

a(k1*k2,h) = a(k1,a(k2,h))
b(k1*k2,h) = b(k1,a(k2,h))b(k2,h)

the latter being listed as the last property on the wikipedia page.

Moving on:

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

Your question was whether this should instead be

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

Your justification was in terms of conjugation in the case when K is
normal, which I think may have led to my saying the wrong thing,
focusing on conjugation. Let us look at the definition directly.

Proceeding as above, b(k,h1*h2) is the unique k' in K such that
k*(h1*h2) = h*k' for some h in H. Since

k*(h1*h2) = (k*h1)*h2
= [a(k,h1)*b(k,h1)]*h2
= a(k,h1)*[b(k,h1)*h2].
= a(k,h1)*[a(b(k,h1),h2)*b(b(k,h1),h2)]
= [a(k,h1)*a(b(k,h1),h2)]*b(b(k,h1),h2).

Which means you were correct. We have

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

So... Go ahead and fix it, and sorry for wasting your time by talking
about other conjugation conventions.


--
======================================================================
"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

.