From mboxrd@z Thu Jan 1 00:00:00 1970 X-Msuck: nntp://news.gmane.io/gmane.science.mathematics.categories/7348 Path: news.gmane.org!not-for-mail From: =?iso-8859-1?Q?Joyal=2C_Andr=E9?= Newsgroups: gmane.science.mathematics.categories Subject: Re: Two_questions Date: Sat, 23 Jun 2012 11:40:13 -0400 Message-ID: References: Reply-To: =?iso-8859-1?Q?Joyal=2C_Andr=E9?= NNTP-Posting-Host: plane.gmane.org Mime-Version: 1.0 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable X-Trace: dough.gmane.org 1340500698 17481 80.91.229.3 (24 Jun 2012 01:18:18 GMT) X-Complaints-To: usenet@dough.gmane.org NNTP-Posting-Date: Sun, 24 Jun 2012 01:18:18 +0000 (UTC) To: "Fred E.J. Linton" , "categories" Original-X-From: majordomo@mlist.mta.ca Sun Jun 24 03:18:14 2012 Return-path: Envelope-to: gsmc-categories@m.gmane.org Original-Received: from smtpx.mta.ca ([138.73.1.80]) by plane.gmane.org with esmtp (Exim 4.69) (envelope-from ) id 1SibSv-0000BH-Lg for gsmc-categories@m.gmane.org; Sun, 24 Jun 2012 03:18:13 +0200 Original-Received: from mlist.mta.ca ([138.73.1.63]:50700) by smtpx.mta.ca with esmtp (Exim 4.77) (envelope-from ) id 1SibS0-0000O8-5G; Sat, 23 Jun 2012 22:17:16 -0300 Original-Received: from majordomo by mlist.mta.ca with local (Exim 4.71) (envelope-from ) id 1SibS1-0001iA-PZ for categories-list@mlist.mta.ca; Sat, 23 Jun 2012 22:17:17 -0300 Thread-Topic: categories: Re: Two questions Thread-Index: Ac1RQiayuNV2b9YyRxaLcUuS3Ad0OAABWVof Precedence: bulk Xref: news.gmane.org gmane.science.mathematics.categories:7348 Archived-At: 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=20 by putting u'(p)=3Dp' and u'(x)=3Dx 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. =20 Am I making an error? Best,=20 Andr=E9 -------- Message d'origine-------- De: Fred E.J. Linton [mailto:fejlinton@usa.net] Date: ven. 22/06/2012 23:09 =C0: categories Objet : categories: Re: Two questions =20 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 >=20 > j : H --> G, there exist two homomorphisms >=20 > k, l : G --> P with k(g) =3D L(g) only when g is in j(H). >=20 > 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.=20 I do seem to recall that I first saw more or less such an argument in a=20 graduate course Sammy gave at Columbia during the "golden era" = 1958-1963, =20 and that Sammy himself, probably at a Bowdoin NSF summer semester on=20 categories, revealed his deft trick by which to convert the argument for = the "index greater than 2" case to a uniform argument ignoring the=20 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=20 G-set got by forming the coproduct=20 G/H + 1 of the principal left G-set G/H of left cosets xH of H in G (x ? G)=20 with a (trivial) terminal left G-set 1 =3D {*} (assume * ? G/H): P =3D perm(|G/H| ? {*}) =3D (|G/H| ? {*})! . One permutation in particular is to be singled out for attention: the=20 transposition t ? P interchanging * with the coset H itself. Let me use r: G --> P for the result of composing the regular=20 representation of G by left action (g, xH) |-> (gx)H on the cosets of H=20 with the obvious injection of (|G/H|)! into (|G/H| ? {*})! -- thus: [r(g)](xH) =3D (gx)H , [r(g)](*) =3D * . And let me write s: G --> P for the result of conjugating by t the = various=20 assorted values of r -- thus: [s(g)](xH) =3D t([r(g)](t(xH))) , [s(g)](*) =3D t([r(g)](t(*))) =3D t([r(g)](H)) =3D t(gH) . The end-game strategy is now this: (i) for h ? H: r(h) =3D s(h) ; yet (ii) for g =3D a, [r(a)](*) =3D * but [s(a)](*) =3D aH (whence r ? s). Details: Bear in mind that if, for x in G and h in H, we have (hx)H =3D = H,=20 we must have hx in H, whence also x in H so that xH =3D H. Then for (i): (a) if xH ? H -- [s(h)](xH) =3D t([r(h)](xH)) =3D t((hx)H) =3D (hx)H =3D = [r(h)](xH) ; (b) if xH =3D H -- [s(h)](H) =3D t([r(h)](*)) =3D t(*) =3D H =3D = [r(h)](H) ; (c) and at * -- [s(h)](*) =3D t([r(h)](H)) =3D t(H) =3D * =3D [r(h)](*) = . And bear in mind also that aH ? H because a ? H; thus: for (ii) -- [s(a)](*) =3D t([r(a)](H)) =3D t(aH) =3D aH ? * =3D = [r(a)](*) . By the way, if the subgroup H had index 3 or more in G, one need not=20 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 =20 possible :-) )), I believe that is essentially how the proof George = cites =20 from Mac Lane's exercise works. Cheers, -- Fred [NB: I'm writing ? for an element symbol, ? for a crossed-out=20 element symbol, ? for a union symbol, and ? for a crossed-out equal=20 sign. With luck these HTML glyph constructs will simply display as the=20 glyphs they're meant to represent; and if not, they're no more painful = to =20 decipher than their TeX counterparts, which HTML can't display as = glyphs.] [For admin and other information see: http://www.mta.ca/~cat-dist/ ]