Re: Two_questions

["Joyal, André" <joyal.andre@uqam.ca>]

Thank you Fred for your nice description of Sammy's proof
that every subgroup is an equaliser (is a regular subgroup).
I would like to suggest a small variant which might be simpler.
Sammy's proof can be decomposed in three steps:

(1) the pullback of a regular subgroup is regular;

(2) Every subgroup of a group G is the stabiliser of a point in some G-set;

(3) If E! is the full permutation group of a set E,
then the stabiliser S(p) of a point p in E is a regular subgroup of E!.

The key argument lies in step (3). Let E' be the set obtained from E by
adding copy p' of p. There are two embeddings u,u':E-->E',
the first u is the inclusion of E in E', and the second u' is defined 
by putting u'(p)=p' and u'(x)=x for x different than p.
This leads to a pair of homomorphisms h,h':E!-->E'! the equaliser of which
is the stabiliser S(p) of p in E!.

This last argument seems to differ from the argument you have presented.
 
Am I making an error?

Best, 
André





-------- Message d'origine--------
De: Fred E.J. Linton [mailto:fejlinton@usa.net]
Date: ven. 22/06/2012 23:09
À: categories
Objet : categories: Re: Two questions
 

On Fri, 22 Jun 2012 01:32:10 +0200, George Janelidze wrote, inter alia:

> 6. Concerning your second question: What you saw (with a special argument
> for 2) is on the same Page 21 of the above-mentioned Eilenberg--Moore
paper.
> The argument was used to prove that every epimorphism of groups is
> surjective with no mention of regular monomorphisms, but in fact they prove
> that, for every homomorphism
> 
> j : H --> G, there exist two homomorphisms
> 
> k, l : G --> P with k(g) = L(g) only when g is in j(H).
> 
> This also appears as Exercise 5 of Section 5 of Chapter I in Mac Lane's
> "Categories for the Working Mathematician", with very precise hints.

What the origins of the argument Mac Lane outlines here are, I don't know. 
I do seem to recall that I first saw more or less such an argument in a 
graduate course Sammy gave at Columbia during the "golden era" 1958-1963,  
and that Sammy himself, probably at a Bowdoin NSF summer semester on 
categories, revealed his deft trick by which to convert the argument for 
the "index greater than 2" case to a uniform argument ignoring the 
subgroup's index.

Let me record that argument here.

Given are a group G, a subgroup H of G, and an element a in G \ H.

P is to be the group of permutations of the underlying set of the left 
G-set got by forming the coproduct 

G/H + 1

of the principal left G-set G/H of left cosets xH of H in G (x ? G) 
with a (trivial) terminal left G-set 1 = {*} (assume * ? G/H):

P = perm(|G/H| ? {*}) = (|G/H| ? {*})! .

One permutation in particular is to be singled out for attention: the 
transposition t ? P interchanging * with the coset H itself.

Let me use r: G --> P for the result of composing the regular 
representation of G by left action (g, xH) |-> (gx)H on the cosets of H 
with the obvious injection of (|G/H|)! into (|G/H| ? {*})! -- thus:

[r(g)](xH) = (gx)H  ,
[r(g)](*) = * .

And let me write s: G --> P for the result of conjugating by t the various 
assorted values of r -- thus:

[s(g)](xH) = t([r(g)](t(xH))) ,
[s(g)](*) = t([r(g)](t(*))) = t([r(g)](H)) = t(gH) .

The end-game strategy is now this:

(i) for h ? H: r(h) = s(h) ; yet
(ii) for g = a, [r(a)](*) = * but [s(a)](*) = aH (whence r ?  s).

Details: Bear in mind that if, for x in G and h in H, we have (hx)H = H, 
we must have hx in H, whence also x in H so that xH = H. Then for (i):

(a) if xH ? H -- [s(h)](xH) = t([r(h)](xH)) = t((hx)H) = (hx)H = [r(h)](xH)
;
(b) if xH = H -- [s(h)](H) = t([r(h)](*)) = t(*) = H = [r(h)](H) ;
(c) and at * -- [s(h)](*) = t([r(h)](H)) = t(H) = * = [r(h)](*) .

And bear in mind also that aH ? H because a ? H; thus:

for (ii) -- [s(a)](*) = t([r(a)](H)) = t(aH) = aH ? * = [r(a)](*) .

By the way, if the subgroup H had index 3 or more in G, one need not 
require recourse to any external element * as above -- one could let any 
coset other than H and aH (when there are such) play the role of *, and, 
with but a few additional wrinkles (unless I am mistaken (which is always  
possible :-) )), I believe that is essentially how the proof George cites  
from Mac Lane's exercise works.

Cheers, -- Fred

[NB: I'm writing ? for an element symbol, ? for a crossed-out 
element symbol, ? for a union symbol, and ? for a crossed-out equal 
sign. With luck these HTML glyph constructs will simply display as the 
glyphs they're meant to represent; and if not, they're no more painful to  
decipher than their TeX counterparts, which HTML can't display as glyphs.]




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