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Below, I present an outline of a proof of the first isomorphism theorem for groups. This is how I usually think of the first isomorphism theorem for ______, but groups will get the points across. My question is whether there is any textbook which takes this basic approach.

Let $\pi: C \to L$ be a surjection. I like to think of the elements $l$ of $L$ as labels for the preimages $\pi^{-1}(l)$. So I call any surjection a "labeling function", and the codomain a "set of labels".

If $f: C \to R$ is function such that $\pi(x) = \pi(y) \implies f(x) = f(y)$ I say that "$f$ respects the labeling of $\pi$".

Theorem 1: $f$ factors uniquely through $\pi$.

For example if $C$ is the set of all candy bars in a store, $L$ is the set of all "types" of candy bars (mars, trix, heath, ...), $R$ is $\mathbb{R}$ and $f$ assigns to each candy bar its price, then it is plain that $f$ respects $\pi$'s labeling, and so I might as well have just assigned a price to each type of candy bar, rather than pricing each individual candy bar.

Theorem 2: If $\pi_1$ and $\pi_2$ respect each other's labeling, then the induced maps between their codomains are inverses. Hence the codomains are isomorphic.

For example if $\pi_1$ gives the English label of a candy bar and $\pi_2$ gives the Spanish label, then there is an obvious bijection induced between the set of English and Spanish labels (In fact, this is the principle which allows us to learn other languages).

Theorem 3 If $C$, $L$, and $R$ are groups, and $\pi$ and $f$ are homomorphisms, then the induced map from $L$ to $R$ is also a homomorphism.

This is just a couple lines of manipulation.

Theorem 4 If two surjective group homomorphisms respect each others labeling, then their codomains are isomorphic.

This is really a corollary of theorem 2 and 3

Observation: for a group homomorphism $\phi$, $\phi(a) = \phi(b) \Longleftrightarrow \phi(ab^{-1}) = 1$, i.e. $ab^{-1} \in Ker(\phi)$. Thus "$f$ respects $\pi$'s labeling" can be rephrased "$Ker(\pi) \subset Ker(f)$".

So a rephrasing of theorem 4 is

Theorem 4':Any two group surjective group homomoprhisms with the same kernel have isomorphic codomains

or observing that any homomorphism is a surjection onto its image, you can say

Theorem 4'': Any two group homomorphism with the same kernel have isomorphic images.

This is the first isomorphism theorem right? I have not mentioned normal subgroups,cosets, or factor groups. Those only come in as a construction to pick one representative of the collection of all images of homomorphisms having the same kernel, or when you want to start characterizing which subgroups of a group can be the kernel of some homomorphism.

I think that this very basic outline gets lost on students when they see the isomorphism theorem for the first time. Most textbooks defines normal subgroups, cosets, puts a group structure on the cosets, all without any kind of real motivation, and at the end the first isomorphism theorem seems kind of magical, and not at all natural. I know I thought it was magic when I first learned it. But the first isomorphism theorem does not seem surprising at all when you follow my sequence of theorems 1 through 4''.

So I ask: Is this approach taken by any textbook out there?

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One does not learn foreign languages by learning bijections: languages are seldom in bijection in any sensible way! –  Mariano Suárez-Alvarez Oct 13 '10 at 1:37
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I like this presentation and agree that this is more or less how I see these theorems but I don't recall actually seeing them presented this way in a book or in class. Perhaps people are concerned that too many intermediate results will confuse students? While we're on the topic of irrelevantly pedantic comments, Trix is a cereal -- the candy bar is Twix. –  Noah Stein Oct 13 '10 at 17:59
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I should've hoped a Moderator would know better: guys, can we take the chatter to Meta tea.mathoverflow.net/discussion/707/… please? Please only add comments here if you have something similar to what Noah has to say in the first half of his comment (namely, about the actual question asked above by Steven Gubkin). –  Willie Wong Oct 13 '10 at 18:17
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Also, should this question be Wiki? Technically the final question asked has a definite answer (either yes or no, assuming we don't run into problems with the incompleteness theorem), but just that would be rather unhelpful. Thus I suspect this is liable to become a list of textbooks, and hence a Community Wiki would be better? –  Willie Wong Oct 13 '10 at 18:19
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@Mariano: All I meant is that if you are stranded on an island with someone who speaks a foreign language, and you want to establish some basic form of communication, then you will have to start by pointing at things and saying their name. In order to build the dictionary of words you have to look at their preimages under the mapping which takes objects to words. There is no way to build an accurate dictionary from the words alone. Learning a language in school you use the dictionary which someone else has written - they have already done this hard work hundreds of years ago. –  Steven Gubkin Oct 13 '10 at 19:30
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2 Answers 2

In my experience, not exactly. A close match is Chapter 1 in "Algebras, Lattices, Varieties" by McKenzie, McNulty, and Taylor. There may be other texts which take a universal algebraic approach to abstract algebra, but others will have to inform you about such. Pedagogically, I suggest showing a couple other examples before springing the theorems on an unsuspecting student.

Gerhard "Ask Me About System Design" Paseman, 2010.10.13

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Slightly off topic. Gerhard: I notice that you have 7 separate accounts with a total of 1000+ rep in them. Normally I'd just ask in the Meta thread to have your many accounts merged, but since you also particpate there fairly regularly, I figure you may actually prefer having them separate. If that assumption is false, you can ask the moderators to merge your accounts here: tea.mathoverflow.net/discussion/605/merge-two-user-ids –  Willie Wong Oct 13 '10 at 19:15
    
Your assumption is not only good, it is correct. I am actually using the different accounts for different purposes at the moment. Later I may ask for a merge, but not now. Thanks for reminding me of the option. –  Gerhard Paseman Oct 13 '10 at 21:06
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You might try pages 47-55 of Burris and Sankappanavar (pages 63--71 of the pdf document).

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Very nice reference. Thanks! –  Andres Caicedo Dec 10 '10 at 2:18
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