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Groups are naturally "the symmetries of an object". To me, the group axioms are just a way of codifying what the symmetries of an object can be so we can study it abstractly.

However, this heuristic breaks down in the case of many abelian groups. Abelian groups more often arise as a "receptacle for addition". What I mean is that they are more intuitively counting combinations of some generating elements. See for example: solution spaces to linear equations, the underlying group of rings and modules, (co)homology groups. They rarely act naturally on anything except themselves, which seems like a copout.

It bugs me that these two intuitions are so far from each other, even though the underlying axioms differ by a single, deceptively-mild assumption. Is there a way to reconcile these two perspectives? Do abelian groups satisfy the group axioms by accident?

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Oh, I agree entirely. That case is potentially more interesting, because of how fruitful it can be to apply one perspective to the other (see: non-commutative geometry). However, this seemed like the simpler question. – Greg Muller Oct 26 '09 at 4:22
I reminds me of a thing I often think about: why abelian varieties and affine algebraic groups feel so different. – Evgeny Shinder Dec 18 '09 at 21:27
Historically, groups and abelian groups came up in very different contexts. On the one hand, Lagrange, Ruffini, and eventually Galois used permutation groups to study solubility of polynomial equations by radicals. On the other hand, abelian groups seem to have been introduced in (or at least motivated by) Abel's work on uniformisation of complex tori. – R. van Dobben de Bruyn Jun 11 at 19:18

11 Answers 11

Here's how I think about it: if the automorphism group of an object is abelian, this means something very strong about the object. It sort of means you can affect its structure in two different ways, in any order, and they won't affect each other. This to me hints that maybe the object consists of separate "pieces" that can only be affected independently.

The first example like this which comes to mind is a direct (commutative ring) product of finite fields of different characteristics. Each field has a cyclic automorphism group, making it very "simple", and the fields can't map into each other, meaning they can only be affected "independently."

So I'd posit that if you want to think of abelian groups as symmetry groups, you can imagine them as the symmetry groups of objects that "break apart" into "simple" pieces which can't interact with each other (intentionally vague, since I don't want to commit to a particular category).

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From a representation theory perspective, abelian groups are distinguished by the richness of their invertible representations, since (in the locally compact case), the groups are completely determined by one dimensional complex characters. For nonabelian groups, the one dimensional characters only determine the abelianization, and one must find actions on larger vector spaces to recover the full group structure. Abelian groups admit a straightforward Fourier transform, which is a change of basis between functions on a group and functions on the dual (i.e., character) group. For a nonabelian group, the situation becomes substantially more complicated, involving representations of quantum doubles.

From a homotopy theory perspective, groups can be delooped once, but abelian groups can be delooped infinitely many times. This is why you can have Eilenberg-MacLane spaces K(G,n) for n > 1 if and only if G is abelian. Since these spaces represent cohomology with G coefficients, this is why higher degree cohomology only exists for G abelian.

There are intermediate levels of loopiness, corresponding to commutativity up to homotopy plus some intermediate levels of coherence of choices of homotopies. The quantum double of a group G, for example, lives on the two-fold loop space G/G. A discrete two-fold loop space H is automatically an abelian group, since \pi2 of the two-fold delooping is H, and higher homotopy groups are abelian (this is one form of the Eckmann-Hilton argument). As Qiaochu mentioned, the 2-morphisms in a 2-groupoid with one object and one 1-morphism form a natural two-fold "loop set" which therefore is an abelian group.

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Scott, can you make precise and/or provide references for your statement that "the quantum double lives on the two-fold loop space"? – Mariano Suárez-Alvarez Dec 18 '09 at 20:38
By "lives on" I mean sheaves on G/G have a convolution operation from the E[2] structure of G/G, and one can construct the quantum double as sections of a sheaf on G/G together with an E[2] multiplication map using the convolution. I don't have a precise reference, since I worked it out while trying to understand some writing by Majid about transmutation. In the finite case, there is some information in Willerton: 0503.5266. I think Ben-Zvi-Francis-Nadler may have some rough sketches in more general contexts. – S. Carnahan Dec 22 '09 at 19:47

I also like to think of abelian groups as quite different from (general) groups. To me an abelian group should be thought of as being more like a module than a group. Indeed, abelian groups are the same thing as Z-modules and the category of abelian groups is an abelian category (which has many nice properties not shared by the category of groups). The prototypical examples of abelian categories are categories of modules over a ring. In the case of abelian groups the ring is Z (the ring of integers).

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I'd like to think that the answer to this question shouldn't require the definition of a ring. In fact it suggests another question: why is it so important that addition in a ring be commutative? – Qiaochu Yuan Oct 26 '09 at 5:03
One reason is that the distributive property actually implies it: 0=(1-1)(a+b)=a+b-a-b, hence a+b=b+a. Thus commutativity of addition is actually superfluous as a ring axiom. – Andrew Critch Oct 26 '09 at 5:31
I think this is a great answer. (Of course, as a ring theorist, I certainly have a bias!) In response to Qiaochu's question of why a ring should be commutative, one way to think of a ring is a multiplicative monoid with the bonus structure of an abelian group. Now taking a "quotient" of a monoid can be very hard. But if we add (ha ha) the extra abelian group structure, this makes it much easier to "mod things out," by forming ideals (or even one-sided ideals, if we only want to create a module). – Manny Reyes Oct 26 '09 at 22:50
In fact, the commutation of addition is a consequence of the axiom of distributivity only in the case of unital rings. – Gérard Lang Jun 12 at 9:49

I think there is something to this idea that abelian groups are only groups "by accident". Here is at least a construction of the category of commutative monoids which does not mention the monoidal operation directly and cannot be easily modified to yield a definition of the category of all monoids.

Let Fin denote the category of finite sets and all maps, and let Span denote the category of spans in Fin: its objects are the objects of Fin, and a morphism from X to Y is a diagram X ← Z → Y in Fin. The identity span on X is X ← X → X, and composition of spans is by pullback. This is properly a 2-category, so make it into a 1-category by replacing each Hom-category by its set of isomorphism classes of objects (discarding the automorphisms).

This category has finite products—they are given by disjoint union of the underlying sets. Let Comm be the category of finite-product-preserving functors from Span to Set. Then Comm is equivalent to the category of commutative monoids.

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It almost seems that the "non" in non-abelian is a positive quality.

I remember my advisor saying that "nonlinear" isn't a hypothesis but the absence of a hypothesis. In that sense "nonlinear" means "not necessarily linear." But you can turn nonlinearity into a positive assertion by looking at things that are nonlinear in a specific way, such as strict convexity, that excludes linearity.

Along the same lines, maybe the key is to define non-abelian groups in a positive way, to focus on what these groups have that abelian groups lack. Or to do this for some subset of non-abelian groups.

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One way to do this is to define non-abelian as "has nontrivial inner automorphisms." – Qiaochu Yuan Oct 26 '09 at 16:49

Here's a totally different suggestion, which is why I'm writing it separately. As has already been remarked, Ab has the really nice property that it's enriched over itself and monoidal closed. We don't have this property in Grp because Hom(A, B) doesn't have a natural notion of composition if A != B. One can, however, construct a category with the elements of Hom(A, B) as objects and invertible functors fixing A and B as morphisms, and this category is a groupoid. In other words, we should think of Grp as enriched over Gpd - even better, we should think of Grp as living in Gpd, which is enriched over itself and Cartesian closed. So one reason Grp feels different from Ab is that Grp is neither Cartesian nor monoidal closed.

I don't know whether Gpd feels more similar to Ab than Grp does, though.

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I'm somewhat confused by your definition of Gpd. Are the objects all morphisms from FIXED groups A and B, or are they morphisms between any two groups? In either case, what are the morphisms? – Greg Muller Oct 26 '09 at 19:06
My understanding is this: let C = {A, B} be the subcategory of Grp consisting of two fixed groups A and B. Then Hom(A, B) can be thought of as the groupoid whose objects are morphisms from A to B and whose morphisms are invertible functors from C to C such that F(A) = A, F(B) = B. – Qiaochu Yuan Oct 26 '09 at 19:15
Thats a bit of an odd fellow. So, if I am correct, then if there are no non-trivial morphisms from B to A, then Hom(A,B) is the action groupoid of tautological action of the symmetric group of the set Hom(A,B)? – Greg Muller Oct 26 '09 at 19:24
The Gpd-enriched Hom(A, B) in Grp is obtained by just regarding A and B as categories (groupoids) with one element and taking functors as objects and natural transformations as morphisms. If you work this out then the objects are group homomorphisms A -> B and the automorphisms of an object f are the elements of B which commute with the image of f (more generally a morphism f -> g is an element b such that g = b f b^{-1}). – Reid Barton Oct 27 '09 at 0:26
My mistake, then. I'm going more off of intuition than anything else. – Qiaochu Yuan Oct 27 '09 at 0:49

A group is a category with a single object and all morphisms invertible; an abelian group is a monoidal category with a single object and all morphisms invertible.

This can be seen via the trick that if a set (in this case the morphisms of the category) has two multiplications that commute with each other (in this case composition of morphisms and tensor product of morphisms) and have the same unit (the unique identity morphism in the category), then they agree and are commutative.

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The way I think about it, the difference between abelian and non abelian groups is that the former are much better understood. As John Cook says in his answer, abelianness is an extra condition that narrows down the possibilities.
Agreed that abelian groups act as "a receptacle for addition", but additions, or translations, are symmetries too!

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Here's sort of an "opposite" question: Is it possible to naturally and intuitively go from the idea of "groups as group presentations/quotients of the free group" to the idea of groups as symmetry groups? Certainly the reason for why abelian groups are counting things arises immediately in the first context... but of course this isn't a particularly nice place to work, since all sorts of things are undecidable.

Oh, right, it's totally possible to go from group presentations to symmetry groups, although not particularly nicely. There is, of course, a standard construction given a group presentation of a guy on which the group acts... the problem is, we either have to work with a silly definition of automorphism which nobody really uses in real life, or introduce some rather gross and unintuitive combinatorial gadgets that complicate things, or accept that there are a bunch of extra automorphisms that the presentation doesn't account for, and which in particular will generally (always?) make the automorphism group non-abelian... Is there a good way to get around this problem?

ETA: Now that I think about it, though, this isn't even necessarily the core issue with the "co-question." It's pretty easy to think about the free abelian group: It's just how we count apples and oranges if we don't want 3 apples and 3 oranges to make 6 fruits. (If you don't want to deal with negative oranges, fine; it's how you, a stock speculator, keep track of your investments [counting short sells] if you -- for good reason -- don't want to mix up your shares of Reynolds Firearms and Roslin Pharmaceuticals.) From there, it's an easy abstraction to think, well, if I lay out my apples and oranges on the table, then the order they're in doesn't matter. And then you (now you're a mathematician and not a grocer) might think, oh, but what if order does matter, which gets you to the free group. But the conceptual leap from there to general finitely presented groups is non-obvious to me. Maybe I should go back to stacking these cantaloupes...

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To nitpick a bit, if order doesn't matter you're talking about the free abelian monoid and if it does you're talking about the free monoid. You can't take away apples or oranges that aren't there. – Michael Lugo Oct 26 '09 at 16:39
Yeah, but you can get to a group by pretending there are negative apples oranges, and everything works out fine -- not nearly as basic as counting, but it's not a mathematical issue. – Harrison Brown Oct 26 '09 at 16:56
"the problem is, we either have to work with a silly definition of automorphism which nobody really uses in real life..." The Cayley graph is the 1-skeleton of the universal cover of a presentation complex for your group, and the group is realized as covering transformations. What's silly about that? – HJRW Dec 18 '09 at 20:28

I don't think this directly answers the question, but I thought the following perspective is nice: Groups and abelian groups are actually group objects in different categories! Namely:

  • A group is a group object in Set
  • An abelian group is a group object in Mon, the category of monoids

This follows from the observation that the inverse map $\iota:G \to G$ is a monoid homomorphism iff $G$ is abelian, plus the following consequence of the Eckmann-Hilton argument alluded to in some other answers (or see the Wikipedia article on group object):

  • A monoid is a monoid object in Set
  • An abelian monoid is a monoid object in Mon

Edit: The rest of what I wrote doesn't hold; see Todd's comments below for the right statement.

One could do the same for Hopf algebras versus commutative Hopf algebras (over a fixed field $k$):

  • A Hopf algebra is a group object in $k$-Mod
  • A commutative Hopf algebra is a group object in $k$-Alg

Since Hopf algebras are the function algebras of quantum groups, while commutative Hopf algebras are the function algebras of groups, I'm tempted to say on a whim that

abelian groups : groups :: groups : quantum groups

i.e. the difference between abelian groups and groups is analogous to the difference between groups and quantum groups.

I don't know enough about quantum groups or Hopf algebras to say whether this is a significant (or even true) analogy. I'd be happy to hear more about it!

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Group objects make sense in cartesian monoidal categories, i.e., monoidal categories where the tensor product is a cartesian product. The point is that you need projections and diagonal maps to internalize the group axioms. For instance, the category of cocommutative coalgebras over $k$ has cartesian product given by $\otimes_k$. But $\otimes_k$ for $k$-Mod doesn't work like that. So I think one should rather say that cocommutative Hopf algebras are group objects in cocommutative coalgebras, and commutative Hopf algebras are cogroups in commutative $k$-algebras. – Todd Trimble Jun 11 at 20:38
Isn't a Hopf algebra a group object in $(k-\mathbf{Mod})^{\mathrm{op}}$ (as opposed to $k-\mathbf{Mod}$)? – Qfwfq Jun 11 at 20:47
Oh wait, I guess I wanted to say what Todd Trimble said in the last sentence and I even messed up – Qfwfq Jun 11 at 20:48
Thanks for the clarication! I've struck out the erroneous parts of my answer. Is it still true that a Hopf algebra is commutative iff it's antipode is an algebra homomorphism? I've come across books where the antipode is defined to be an algebra homomorphism - surely that can only hold for commutative Hopf algebras. – L.Z. Wong Jun 11 at 21:02
Do those books really say that, and not that it's an antihomomorphism? Actually you do not need to assume that, only that it's a linear map which obeys the usual antipode equation, and it turns out that being an antihomomorphism (at both the algebra and coalgebra level) is a consequence. A derivation of this fact is given here: Anyway, I think your surmise (L.Z.) is correct. – Todd Trimble Jun 11 at 22:21

Look at Cayley's embedding Theorem which realises every group as a group of permutations. Permutations are functions (from a set to itself . . .). So an abelian group is a set of commuting permutations. For functions, commuting is a serious restriction. And so abelian groups have special properties.

Look at Kronecker-Weber theorem in Algebraic Number Theory. The group of field automorphisms being abelinn means the automorphism functions should commute with each other. So the field has to be very special kind (contained ina field generated by roots of the equation $x^n-1=0$.)

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