Re: The quaternion group as a product
- From: magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin)
- Date: Mon, 19 May 2008 18:55:40 +0000 (UTC)
In article <11669705.1211220360122.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
Adam Burley <ajburley@xxxxxxxxxxxxxx> wrote:
So now let G be a group, H,K subgroups, x e HnK a non-identity
element of order 2 which commutes with every other element. Can we
then make such a product once more?
And further, can we generalise this, by:
i) Allowing other orders than just 2?
ii) Alllowing more than one non-identity element in the intersection?
There are plenty of ways to do this. One is the "central product", if
you want elements of H to commute with elements of K.
Let H and K be any two groups; let
Z(H) = { h in H : hx = xh for all x in H} (the center of H)
Z(K) = { k in K : ky = yk for all y in K}.
Let A be an abelian group, and suppose there are one-to-one
homomorphisms f:A->Z(H) and g:A->Z(K) (i.e., A is a subgroup of both
Z(H) and Z(K)).
The central product of H and K identifying A is the group
G = (H x K)/ { (f(a),g(a)^{-1}) : a in A}
[sometimes you will see the normal subgroup being given as the set of
all pairs (f(a),g(a)) instead; since A is abelian, x|->x^{-1} is an
automorphism of A, so this is obtained by replacing g with the map
obtains by first applying this automorphism and then the old mapping;
the resulting group is isomorphic to this one).
This product has the following characteristicts:
(i) G contains subgroups that are isomorphic to H and to K.
(ii) G is equal to the product of these two subgroups.
(iii) The intersection of these two subgroups is equal to A.
(iv) [H,K] = {1}.
This is not the case with the quaternion subgroup, because H and K do
not commute in the quaternion case. You can define a similar situation
when you have H and K acting on each other in a compatible way, as has
been mentioned earlier, or you can get it more generally if you drop
the requirement that the underlying set of G be some quotient of
H x K. For example, the free product with amalgamation.
Let H, K, C be three groups, and let f:C->H and g:C->K be any two
one-to-one group homomorphisms. The "free product with amalgamation"
of H and K amalgamating C, H*_C K, is given by taking F=H*K to be the
free product of H and K; let N be the least normal subgroup of F that
contains all elements of the form f(c)g(c)^{-1}, with c in C. Then F/N
= H*_C K is a group that:
(i) There are one-to-one group homomorphisms i:H->H*_C K, j:K->H*_C K.
(ii) For all c in C, i(f(c))=j(g(c)), and i(H)/\j(K)=i(f(C)).
(iii) H*_C K = <i(H),j(K)>.
(iv) If G is any group, m:H->G and n:K->G are group homomorphisms
such that m o f = n o g, then there exists a unique group
homomorphism F: H*_C K --> G such that m=Fi and n=Fj.
This works even if C is not central in H and in K.
In particular, any group that you want to create like this will be a
quotient of this group.
--
======================================================================
"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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- The quaternion group as a product
- From: Adam Burley
- Re: The quaternion group as a product
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