Date: Sun, 26 Feb 1995 17:36:35 -0500 (EST)
From: [Permission pending]
Subject: Re: Division algebras and topological groups.
To: rusin@math.niu.edu

[deletia]

Last fall I started wondering about something which I bet many people wonder
when they first see factor groups:  under what circumstances is there an
isomorphism f from G/H to a subgroup K of G, with f(kH) = k for k in K?  And 
outside of the abelian case, what can you say when G is finite?  And I came 
across something pretty nifty in a book called "Noncommutative Algebra" by Farb
and Dennis:  the Schur-Zassenhaus theorem, which says that if |H| is
relatively prime to |G/H| then G is a semidirect product H X| G/H.
Apparently there is a short simple proof using group cohomology.  I don't
know if there is a short proof with straight group theory.  So 
cohomology of groups seems interesting since you can prove things easily.
Your essay on division algebras helped me understand algebras more:  I was
really confused about all those operations running around under one roof.

[sig deleted-- djr]
==============================================================================

Date: Mon, 27 Feb 95 12:39:22 CST
From: rusin (Dave Rusin)
To: [Permission pending]
Subject: Re: Division algebras and topological groups.

[deletia -- djr]

>Last fall I started wondering about something which I bet many people wonder
>when they first see factor groups:  under what circumstances is there an
>isomorphism f from G/H to a subgroup K of G, with f(kH) = k for k in K?  And 
[I'm assuming the  H  is given and you're seeking  K, not the other way round.]
More generally, in many categories it makes sense to ask this question:
given an exact sequence    A --> B --> C  when is it split exact, that is,
when is there is morphism  C --> B  such that the composite  C--> B--> C is
the identity. Observe that it only makes sense to ask this question when 
A --> B --> C --> 0 is exact  (C  is a quotient of B). (Sometimes the question
is phrased thus: when is there a splitting  B--> A ?; in this case it is only
possible if  0 --> A --> B --> C  is exact. In non-abelian categories like
group theory, this is less convenient, since you can't arbitarily form
quotients.)

Of course, the general answer is "sometimes yes, sometimes no". If the category
is the set of modules over a given (commutative, associative, unital) ring,
then _all_ short exact sequences split iff the ring is a field (I think).
You could give sufficient conditions for a splitting (e.g. assume something
is free, or projective), but I doubt an iff condition could be easily 
phrased for most categories.

In the category of groups, your question asks, "Is every normal
subgroup complemented?" (For all H normal in G is there a K with G = HK).  
I would imagine a group G for which this is true for _all_
normal subgroups would tend to be pretty "simple", perhaps in the
sense that all Sylow subgroups are small (or at least close to
elementary-abelian), although of course a group which really is simple
in the technical sense has the property that you specified, vacuously!
For solvable groups, you could show by induction on order that the group
would have to be G = C1.(C2.(C3. ... ))), a sequence of split extensions
by cylic groups.

> the Schur-Zassenhaus theorem, which says that if |H| is
>relatively prime to |G/H| then G is a semidirect product H X| G/H.
>Apparently there is a short simple proof using group cohomology. 

Well, as I recall the simple proof assumes that either  H  or  G/H  is
solvable, which is automatic when they are relatively prime if you believe
that there are no simple groups of odd order; but this fact occupies 200+
pages of proof.

> I don't know if there is a short proof with straight group theory.  So 
>cohomology of groups seems interesting since you can prove things easily.

The critical cohomology fact is that  H*( Q, H)  is annihilated by both
|Q| and |H|, so if these are relatively prime, the cohomlogy group is zero,
and in particular all extensions of  H  by  Q  are split. It's not hard
group theoretically to show that  |H| kills the cohomology, but to show
|Q| kills it is rather trickier (although it's not too hard for  H^1, but
it's  H^2  you need here).

I would agree that cohomology is an interesting part of group theory, but
it isn't long before it takes you off the deep end, so approach this
topic with caution. If you don't mind German, Huppert's group-theory book
treats just the minimum amount of general nonsense needed to get the
few cohomology facts which a finite group theorist would ever really use.
All the other books on group cohomology assume that group cohomology is
worthwhile in its own right, but this is a more dubious proposition.
(I speak here as someone whose main line of work has been in this area.)