Talk:Five lemma

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Please add the apostrophe for the small letters where it's needed(e.g. you have c where you should have c apostrophe). Thanks!

FWIW, the diagram in the subject page does not display correctly on my web browser. I am using Konqueror in Linux. I don't know if there is some way to make it appear correctly on all web browers or not.

Nevermind. Apparently It was being worked on as I wrote that! -- Ram-Man

:-) I need to bump that discussion on TeX-Wiki, it'll look way bettern with that. -- Tarquin

I always hated this lemma. It appeared in every second proof and all of them became so un-obvious ... If you give me the TeX-code I can make an image from it. -- JeLuF

Is this lemma also true for non-commutative groups? AxelBoldt 18:05 Oct 29, 2002 (UTC)

I believe so; IIRC, there is at least a very similar lemma called the five lemma which uses the same diagram and exact sequences; I just can't remember now (and am not quite in the head space to derive it) whether all the morphisms are of the same class as that shown here (i.e., epi- mono- and iso- respectively). I'll check tonite. Cheers Chas zzz brown 00:11 Oct 30, 2002 (UTC)

Wait, I'll check right now :). Recall that for any group homomorphism f and element a, f(a ^-1) = (f(a)) ^-1; then the steps of the proof that relate to the underlying binary operation commutes could be rewritten in multiplicative notation (without invoking the commuting assumption a*b = b*a) as:

• (Given the steps in the first part of the proof, ...)
  • Then t(n(c)) = p(h(c)) = t(c'), so t(c' * n(c ^-1)) = t(c)*t(n(c ^-1)) = t(c)*(t(n(c))) ^-1 = t(c)*(t(c)) ^-1 = 1.
  • By exactness, c' * n(c ^-1) = c' * (n(c)) ^-1 must be in the image of s; let b' be an element of the inverse image of c'.
  • Since m is surjective, we can find b in B such that b = m(b').
  • By commutativity, n(g(b)) = s(m(b)) = c' * (n(c)) ^-1.
  • Since n is a homomorphism, n(g(b) * c) = n(g(b)) * n(c) = c' * (n(c)) ^-1 * n(c) = c'.
  • Therefore, n is surjective.

Voila. The second proof of the other four lemma doesn't seem to make rely on any statements which would change for a non-commutative operation, so (assuming these proofs are correct) they would hold for groups as well (just exchange 0s for 1s in multiplicative notation).

Would it then be proper to say "This proof applies to all objects in the category of groups"? Chas zzz brown 00:45 Oct 30, 2002 (UTC)

You would probably say something like: "The Five lemma is also valid in the category of groups, and the same proof applies." AxelBoldt 17:32 Oct 30, 2002 (UTC)

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NOTE: This Lemma is FALSE for the category of groups. Consider the exact sequences 1->Z/4-> D_8->Z/2->1 and 1->Z/4->Q_8->Z/2->1 (where Q_8 is the quaternion group of order 8 and D_8 is the dihedral group of order 8). We have isomorphisms everywhere but D_8 and Q_8 (1 -> 1 is both epi and mono). Thus this FAILS in the Category of groups! --Polfbroekstraat —Preceding unsigned comment added by 128.12.72.244 (talk) 15:47, 15 October 2007 (UTC)

But your proposed diagram is not a commutative diagram, which is required for the lemma!. We need (at a minimum) a morphism n : D_8 -> Q_8 which projects D_8 onto a copy of Z_4 (the same one produced by the composition of m and s in the diagram in the article).

Assume D_8 = <x,y : x^4, y^2, x^y = x^-1>. The only normal subgroup of D_8 of index 4 is <x^2>; but D_8/<x^2> is not isomorphic to Z_4; it is instead isomorphic to D_4 (the Klein group). So, the required morphism does not exist; and so the diagram cannot commute. Cheers - Chas zzz brown 21:39, 16 October 2007 (UTC)

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This is one of my favourite lemmas! :-) -- Tarquin

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Um, it is said that the proof does not use commutativity and yet it certainly seems to. Jcreed 14:25, 4 May 2006 (UTC) Wait, never mind, that's diagram commutativity, not multiplication commutativity. Jcreed 14:26, 4 May 2006 (UTC) Good point. I have made the commutativity clearer. --IkamusumeFan (talk) 08:34, 14 November 2014 (UTC)