From: "Fred E.J. Linton" <fejlinton@usa.net>
To: "categories" <categories@mta.ca>
Subject: Re: Two questions
Date: Fri, 22 Jun 2012 23:09:21 -0400 [thread overview]
Message-ID: <E1SiQAh-0000MH-C7@mlist.mta.ca> (raw)
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.]
[For admin and other information see: http://www.mta.ca/~cat-dist/ ]
next reply other threads:[~2012-06-23 3:09 UTC|newest]
Thread overview: 9+ messages / expand[flat|nested] mbox.gz Atom feed top
2012-06-23 3:09 Fred E.J. Linton [this message]
2012-06-23 15:40 ` Two_questions Joyal, André
[not found] <485qFyBhu8160S03.1340588060@web03.cms.usa.net>
2012-06-25 3:00 ` Two questions Joyal, André
-- strict thread matches above, loose matches on Subject: below --
2012-06-25 1:34 Two_questions Fred E.J. Linton
2012-06-24 16:37 Two_questions Fred E.J. Linton
2012-06-21 13:34 Two questions Michael Barr
2012-06-21 17:07 ` Ronnie Brown
2012-06-21 23:32 ` George Janelidze
[not found] ` <E1ShqME-0001Sf-JM@mlist.mta.ca>
2012-06-22 0:25 ` Peter LeFanu Lumsdaine
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