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I added a condition that B and D are normal subgroups of A and C, using the usual triangle notation. I'd be surprised if this is not a necessary condition for the conclusion to hold. Please correct if original formulation was actually correct.

Perhaps the article should explicitly state in words which subgroups are normal subgroups of others. For example, one of the conclusions of the lemma is that is a normal subgroup of (which is currently implicit in the notation because of the quotient notation). Saying this in words provides somebody learning group theory a way to look up the meaning, of say normal subgroup, elsewhere in wikipedia, which notation is not as convenient for.

Although it's rather obvious, the article could emphasize that the butterfly showing the Hasse diagram is upside down in the sense that smaller subgroups are higher. DRLB (talk) 15:42, 9 April 2008 (UTC)[reply]

Characters

In my opinion one of the following formulaes from the two isomorphic groups is wrong

   (A\cap C)B/(A\cap D)B is isomorphic to (A\cap C)D/(B\cap C)D. 

I think the first one must be changed to

   B(A\cap C)/B(A\cap D)

in order to coincide with the graph. I am not capable of changing the symbols so I ask you,(the next capable reader of this post) to do so. Thanks. 129.132.211.6 (talk) 13:24, 4 April 2013 (UTC)[reply]

Maybe the graph is wrong? The External Link's formulation of the lemma seems to coincide with what is stated in the text, not the graph (switching factors on both sides, which is equivalent AFAICS). --Roentgenium111 (talk) 22:04, 26 May 2013 (UTC)[reply]

Cute, but

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Despite the cuteness of the butterfly, it actually obscures the mathematical content. In Cohn's _Classic Algebra_ the proof is much more obvious visually by drawing the parallelograms properly which illustrate the application of the second isomorphism theorem. Also, drawing the lattice upside down as it's done here (joins at bottom) than it's usually drawn (joins at the top) is a bit confusing as well. 86.127.138.67 (talk) 04:38, 17 April 2015 (UTC)[reply]

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