Talk:Five lemma

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 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)

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 = . The only normal subgroup of D_8 of index 4 is ; but D_8/ 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)

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)