Is there a classification theorem for "finitely generated abelian groups with a distinguished element"? If it helps, you can restrict this to the cases where the order of each element divides the order of the distinguished element.

My idea would be to try to use this for a classification theorem for rings with a finitely generated additive group, where the distinguished element is 1.

  • $\begingroup$ Well, it would in particular be a finitely generated abelian group, so the usual structure theorem would apply; the distinguished element would, presumably, show up in the definition of homomorphisms (which would be required to respect this zero-ary operation) and of substructure (same reason, you require the substructure to be closed under the zero-ary operation). But you would still have the decomposition into direct sums of cyclic groups... just that those cyclic summands need not be "subgroups-with-distinguished-element". $\endgroup$ – Arturo Magidin Sep 7 '10 at 19:59
  • $\begingroup$ well I think those cyclic summands also have distinguished elements. For example $(\mathbb{Z}^2,(1,2))$ is the direct sum of $\mathbb{Z},1$ and $(\mathbb{Z},2)$. Then every such finitely generated group can be expressed uniquely as a direct sum of summands of the form $(\mathbb{Z},n)$ with $n\ge 0$ and $(\mathbb{Z}/p^n,p^k)$ with $p$ prime and $k\le n$. $\endgroup$ – HenrikRüping Sep 7 '10 at 20:15
  • $\begingroup$ @Henrik: but then the direct summands are not subgroups-with-distinguished-element. While you can map $(\mathbb{Z},1)$ to $(\mathbb{Z}^2,(1,2))$ by sending $1$ to $(1,2)$ (and must map it so), you cannot map $(\mathbb{Z},2)$ to $(\mathbb{Z},(1,2))$: you would have to map the distinguished element $2$ to $(1,2)$, and that means that the image of $1$ would have to satisfy $f(1)+f(1) = (1,2)$ in $\mathbb{Z}^2$. No such element exists, so the second direct summand you have is not isomorphic to any subgroup-with-distinguished-element of $(\mathbb{Z}^2,(2,1))$. This seems like a likely headache. $\endgroup$ – Arturo Magidin Sep 7 '10 at 20:28

You say that we may assume that the order of every element divides the order of the distinguished element. That's the easy case. If the distinguished element generates a finite cyclic group $C$ of order $n$ and every element has order dividing $n$, then in fact $C$ is a direct summand. ("For the ring $\mathbb Z/n$ the free module of rank one is injective.")

  • $\begingroup$ Oh, I see, you are thinking about rings. But the $1$ will be the least of your worries, even if you stick to commutative rings. Even when the order of $1$ is a prime $p$ you will be seeking to classify in particular all $0$-dimensional local rings over the field with $p$ elements, which sounds hard over any field. $\endgroup$ – Tom Goodwillie Sep 7 '10 at 21:59
  • $\begingroup$ Every possible order divides infinity. If the distinguished element has infinite order, is it necessarily a member of a cyclic direct summand? I googled "the free module of rank one is injective.", and it didn't find it (not even on this page). Where would I find a proof that C is a direct summand? $\endgroup$ – user5810 Sep 7 '10 at 23:30
  • 1
    $\begingroup$ An $R$-module $I$ is injective if for modules $M\subset N$ the restriction map $Hom(N,I)\to Hom(M,I)$ is always surjective). This is true for all pairs $(N,M)$ if it is true in the case when $N=R$. In the case when $R=\mathbb Z/n$ ($n\ge 1$) it is easy to see that this condition holds for $I=R$. This observation is the basis for the usual proof(s) of the classification of finite abelian groups, isn't it? Choose a cyclic subgroup of maximal size ad split it off. Induct. $\endgroup$ – Tom Goodwillie Sep 8 '10 at 0:34
  • $\begingroup$ And if $1$ has infinite order in a ring then let $n$ be the largest integer such that there exists $x$ in the ring with $nx=1$. If $n\ge 2$ then $n^2x^2=1$ gives a contradiction. $\endgroup$ – Tom Goodwillie Sep 8 '10 at 0:39
  • $\begingroup$ That skips all the work, which is showing that there is a largest such integer. However, what it gives is only that if 1 is in a direct summand, then 1 generates the direct summand. Does 1 always generate a direct summand? $\endgroup$ – user5810 Sep 8 '10 at 2:05

This is only a partial answer assuming the "if this help" (interpreted in a certain way, hoping it is the intended one).

Let G be a finite abelian group and g in G such that the order of each element of G divides the order of g. (I assume this means that the order of g is finite.)

There exist n_1,...,n_r such that n_i divides n_(i+1) and and isomorphism f from G to C_(n_1) + ... + C_(n_r) such that f(g) = (0,...,0,1).

If the element does not have maximal order the situation is certainly more complex.

  • $\begingroup$ Sorry, the first answer was not there when I started mine, thus the duplication. Feel free to delete it. $\endgroup$ – user9072 Sep 7 '10 at 22:06
  • $\begingroup$ I don't think I will, but ... How would I delete an answer? $\endgroup$ – user5810 Sep 8 '10 at 5:04

I have some general comments.

Summary: such a general theory would be much harder than that for abelian groups (and in fact, would contain the classification of finitely generated abelian groups as a special case), and you would lose a lot of the "good" properties of the category of abelian groups along the way. Many constructions (such as the coproduct) would be different from what we are used to. The advantage is that you can fit into the general context of Universal Algebra, so all of the familiar "abstract nonsense" theory will apply (suitably interpreted). On the other hand, if you restrict to the case you are actually interested in, in which the order of each element divides the order of the distinguished element, then you end up with a much simpler theory, but it falls outside the usual confines of Universal Algebra. You can write down an easy description, but it's not going to be well-behaved relative to homomorphisms.

Some details. To consider the general theory from the point of view of Universal Algebra, you are looking at algebras of type (2,1,0,0), where the binary operation is denoted $+$ (with the usual infix notation), the unary operation is "$-$", one of the zero-ary operations is denoted "$0$", and the second unary operation is denoted "$1$". The operations are required to satisfy the obvious identities: $(a+b)+c = a+(b+c)$, $a+b=b+a$, $a+0=a$, $a+(-a)=0$, and no identity is required of the second unary operation $1$. Homomorphisms are required to respect addition, inversion, map $0$ to $0$, and map $1$ to $1$. The class of abelian groups (as usual) is a variety of algebras of this type, characterized by the identity $1=0$. So you have an embedding from abelian groups to "pointed abelian groups", and this embedding has left adjoint which corresponds to taking the quotient modulo the subgroup generated by $1$. You also have a "forgetful" functor from "pointed abelian groups" to abelian groups, obtained by forgetting about the pointed element;

This class, however, is no longer an abelian category. I'm pretty sure of the following, but I could have made some mistakes: The product of two pointed abelian groups $(A,1)$ and $(B,1')$ is, as Henry suggested in the comments, the abelian group $A\times B$ with distinguished element $(1,1')$, and the structure projections are the canonical projections. The free pointed abelian group on a set $X$ will be the direct sum of $|X|+1$ copies of $\mathbb{Z}$, with the generator of the extra copy being the distinguished element. The coproduct of $(A,1)$ and $(B,1')$ will be the quotient of $A\oplus B\oplus\mathbb{Z}$ modulo the subgroup generated by $(1,0,1)$ and $(0,1',1)$, and the distinguished element is the image of $(0,0,1)$. The canonical immersions go are the canonical maps into the direct sum followed by the quotient map.

This category includes the entire theory of abelian groups as the class of those groups in which $0=1$; so any classification theory will be much harder than the one for abelian groups. For example, the $1$ generator pointed abelian groups: every $1$-generator pointed abelian group is really a $2$-generator abelian group: it is generated by the given generator plus the distinguished element $1$. So suppose $x$ is the given generator. If $\langle 1\rangle\subseteq\langle x\rangle$, then you get an isomorphism type of the form $(C,C')$, where $C$ is a cyclic group and $C'$ is a subgroup of $C$ ($C$ corresponds to the cyclic group generated by $x$, and $C'$ to the subgroup generated by $1$). But these do not exhaust all possibilities. You also get groups in which $\langle x\rangle\cap\langle 1\rangle = \{0\}$; if the orders are coprime, you are back to the previous case, but if the orders are not coprime then the description is a bit more complicated. And then the case where they intersect nontrivially but neither contains the other, and so on. As you can see, things get complicated even in this simplest of cases...

Two pointed abelian groups that are isomorphic as pointed abelian groups will be isomorphic as abelian groups, but you can have to pointed abelian groups whose forgetful image is isomorphic, but that are not isomorphic as pointed abelian groups; for example, $(\mathbb{Z},1)$ and $(\mathbb{Z},2)$ are not isomorphic as pointed abelian groups.

The advantage is that you have the entire weaponry from Universal Algebra at your disposal, as well as two reasonably well-behaved functors to abelian groups: the forgetful functor, and the adjoint of the inclusion of the variety of abelian groups to the variety of pointed abelian groups. The disadvantage is that it is unfamiliar, and more complicated than the usual theory of abelian groups. (Think about how things get more difficult going form abelian groups to not-necessarily-abelian groups).

Different take: For the application you want, however, you don't need such a general theory. In your application, the order of every element divides the order of the distinguished element. The disadvantage is that this cannot be expressed as an identity, so you do not have a variety of algebras. You do not have the nice constructions such as coproducts, products, free objects, etc. a priori. On the other hand, I think that the classification is pretty straightforward: every such group can be expressed as a product of cyclic groups, $C_{n_1}\times\cdots\times C_{n_k}$, with $n_1|n_2|\cdots|n_k$, and with distinguished element a specific generator of the largest cyclic factor in the case where $n_k>0$, and an arbitrary nonzero element of the last cyclic factor if $n_k=0$ and a torsion element plus a nonzero element of the last cyclic factor if $n_k=0$. However, this decomposition is not decomposing the objects into something which is a direct product of pointed abelian groups (the direct summands other than the largest ones do not contain the distinguished element). The proof in the finite case can be done the usual way; for an abelian $p$-group $A$, if $x$ is an element of maximal order then $A$ is isomorphic to $A/\langle x\rangle \oplus \langle x\rangle$, so you can start by picking the distinguished element; otherwise, consider the $p$-parts and think of them as quotients of the original pointed group, with distinguished element the image of the distinguished element; then put them together in the usual way, and the distinguished element will correspond to a generator of the largest cyclic factor. In the infinite case, it gets a bit more complicated, but it should go through: take $A/A^{tor}$, with distinguished element the image of the distinguished element of $A$; this is a direct sum of cyclic groups of infinite order as an abelian group, and the subgroup generated by the distinguished element is a submodule of rank $1$. So if I remember my modules over PIDs correctly, there is a basis for $A/A^{tor}$ of the form $x_1,\ldots,x_r$, in which $1$ is a multiple of $x_r$; then deal with the torsion part as you would with a normal finite abelian group. Edit: No, the infinite case does not go through like this; see comment. You can get it so that it will be of the form $(a,(0,\ldots,0,d))$, where $a$ is in the torsion part and $(0,\ldots,0,d)$ is an element of the torsion free part with $d>0$, but you may not be able to "get rid" of the $a$. This is problematical.

The advantage is that there's the description of all such groups. There are some obvious properties, such as: you can only map from $(A,1)$ to $(B,1)$ if the order of the distinguished element of $A$ is a multiple of the order of the distinguished element of $B$; the map must send the distinguished generator of the largest cyclic factor of the expression for $A$ to the distinguished generator of the largest cyclic factor of $B$. Among the disadvantages is that it is not a "decomposition" in the sense of expressing the pointed group in terms of "smaller" pointed groups, and so it may not lead to any sensible "simplification", just a "description", which is probably not what you want. A more sensible simplification would require you to identify all "product irreducible" pointed abelian groups, and this might be pretty difficult.

  • $\begingroup$ 3 has infinite order in Z, which is certainly maximal, but Z/<3> + <3> has an element of order 3 and Z doesn't, so Z is not isomorphic to Z/<3> + <3>. Could you give or point me to a proof of what you meant to say? Also, a "description" is what I want, since I'm just wondering what elements I would have to worry about being 1 in an invariant factor decomposition. $\endgroup$ – user5810 Sep 8 '10 at 0:13
  • $\begingroup$ Oh, sorry; I'm misreporting/misremembering a theorem from Lang. It holds for finitely generated abelian $p$-groups, not for finitely groups. I'll fix it. $\endgroup$ – Arturo Magidin Sep 8 '10 at 0:28
  • $\begingroup$ ...and it seems that the last part of the penultimate paragraph, when your distinguished element is of infinite order, may not be quite right either. As I note in another comment, I don't see how to arrange it if we had $A=C_2\times C_0$ with distinguished element $(1,2)$ (where $C_2$ is cyclic group of order $2$, $C_0$ the infinite cyclic group). The quotient is cyclic of order $4$, so we cannot write it as $C_2\times C_0$ with distinguished element contained in the summand $C_0$. $\endgroup$ – Arturo Magidin Sep 8 '10 at 4:54

As far as I understand, the classification problem for finite abelian groups with a fixed element is not such a 'slight' extension. (Without any simplifying assumptions, such as divisibility of order.) It is probably an overkill for your situation, but it actually already came up on mathoverflow at least once: see the answer and comments to this question, so I thought I'd mention it:

The answer and comments in the link lead to the following conjecture (which hopefully should not be too hard). First, we may assume without losing generality that G is an abelian p-group. Its isomorphism type is given by a partition.

Conjecture. If g∈G is a non-zero element, the isomorphism type of (g,G) is given by a bipartition.

Justification. There is an analogy between the theory of finite abelian groups and the theory of Jordan form (both classify modules over PIDs). Under this analogy, p-groups correspond to vector spaces equipped with a nilpotent operator. (And indeed, both sides of the analogy are classified by partitions.) The problem at hand is thus analogous to classifying finite dimensional vector spaces equipped with a nilpotent operator and a non-zero vector. Such triples are known to be classified by bipartitions: see Achar, Henderson: Orbit closures in an enhanced nilpotent cone.

  • $\begingroup$ Yeah, when I asked this I thought it would be much simpler than it turned out to be. However, how do we "assume without losing generality that G is an abelian p-group"? The unsolved case is when the distinguished element has infinite order. $\endgroup$ – user5810 Sep 8 '10 at 19:01
  • $\begingroup$ That's why I said 'finite abelian groups'. However, the case of an element of infinite order easily reduces to the case of finite group (unless I am hallucinating). Indeed, suppose our group is $G\times T$, where G is free and T is finite. Suppose the image of $g\in G\times T$ in G is divisible by k (but not by anything larger). Then it is clear that classification problem is the same as classifying the image of g in $T/(k\cdot T)$, which is a finite group $\endgroup$ – t3suji Sep 8 '10 at 20:56
  • $\begingroup$ It is not clear to me how it is enough to classify the image of g in T/(k*T). $\endgroup$ – user5810 Sep 9 '10 at 0:42
  • $\begingroup$ Suppose you have a group H that is isomorphic to $G\times T$. Let us fix such an isomorphism $\phi$, then an element $h\in H$ would correspond to a pair $(g,t)\in G\times T$. Now, what is $(g,t)$ defined up to? If we change $\phi$ by an automorphism of $G\times T$, the following things happen: $g$ gets acted upon by an automorphism of $G$, $t$ is acted upon by an automorphism of $T$, and an element of $k\cdot T$ is added to $t$ (where $k$ is the max. divisibility of $g$). All this follows because we know what automorphisms of $G\times T$ look like. Any such change of the pair $(g,t)$ occurs. $\endgroup$ – t3suji Sep 9 '10 at 2:22

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