Talk:Isomorphism theorems/Archive 1

A discussion
I think we're seeing a bit of an over-reaction to the idiotic practice of using "cats" rather than "cat" as the title of an article about cats. This is not an article about a general concept of an isomorphism theorem, of which the reader may be expected to encounter thousands of concrete cases, and another reader may encounter thousands of others that the first reader doesn't run in to. Rather, it is an article about three particular theorems that often are mentioned in the same breath. So "isomorphism theorems" would seem appropriate as the title. Comments? Michael Hardy 01:12 Mar 21, 2003 (UTC)

It would make sense to use Isomorphism theorems as the title -- and it would make sense to use Cats as a title instead of Cat. The singular convention is to make linking easier within text, and that applies in this title as well. Hypothetically:
 * This result is a consequence of the first isomorphism theorem for groups.

Still, it's not a big deal to violate this for a good reason, if there's a redirect.

But as for the reason, there are indeed thousands of different concrete cases for the general concept of isomorphism theorem: three for groups, three for rings, three each for Rmodules for each ring R, etc. There are not just three isomorphism theorems. True, they can all be summarised into just three theorems that begin "Let C be a category such that ...", but of course that's not how we want to state them in this article -- at least not at first.

Thus I would agree with you in some cases, as long as there was a redirect from the singular, but not in this case.

-- Toby 06:27 Apr 1, 2003 (UTC)

Oops .... I had that backwards. I meant: the idiotic practice of using "cats" as a title rather than "cat". I think the singular convention usually is the right thing. But in some cases, e.g., Joint Chiefs of Staff, The Beatles, and Legendre polynomials, the plural is the right way to do it. Michael Hardy 18:34 Apr 1, 2003 (UTC)

I agree with you for the first two examples -- that's a good reason. As for the last example, however -- there are thousands of Legendre polynomials, infinitely many in fact ^_^. -- Toby 05:40 Apr 13, 2003 (UTC)

How many there are, as long as it's more than one, is not the essential point; the essential point is that no one seems interested in any particular Legendre polynomial except because it is one member of that sequence. Michael Hardy 21:55 Apr 13, 2003 (UTC)

Yes, hence my smile. You had implied above that the number was relevant with your talk of "thousands" -- but of course what you have just now is the truly relevant aspect. -- Toby 12:02 May 14, 2003 (UTC)

Are we missing an isomorphism theorem here? My Dummit and Foote Abstract Algebra lists a fourth "Lattice Isomorphism Theorem" which states that the lattice of a quotient group G/H corresponds directly to the lattice of subgroups of G containing H. Also D&F calls the second isomorphism theorem the "Diamond Isomorphism Theorem" so if people know of other sources that use this name maybe we should include it.

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I've seen it published both ways, but isn't it far more conventional to list the second and third in the opposite order of this article? Nnn9245 21:45, 31 May 2006 (UTC)

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It seems like including the commutative diagrammes that go with each theorem would be very useful in visualising what they say. If all I had were these descriptions, and I weren't otherwise familiar with the isomorphism theorems, they would look very difficult. But with the commutative diagrammes, they become almost obvious. Mraj 15:33, 11 October 2006 (UTC)

Pictures!
This article is in desperate need of some more pictures. Each theorem should have a nice diagram to go with it. asmeurer ( talk  |  contribs ) 00:12, 28 October 2012 (UTC)

History
The section "History" seems vague to me. Precise dates should be given. For example W. Burnside, Theory of Groups of Finite Order, 2 éd., 1911, repr. Dover, 2004, p. 41, gives as an exercise : « If H, h are self-conjugate [i. e. normal] sub-groups of G, and if h is contained in H, so that H/h is a self-conjugate sub-group of G/h, shew that the quotient of G/h by H/h is simply isomorphic [i. e. is isomorphic] with G/H. » This is he third isomorphism theorem. When was it stated for the first time ? Marvoir (talk) 17:35, 1 November 2012 (UTC)