# Countably generated abelian groups

The following theorem is one that many of us use practically daily.

Theorem 1 If ${A}$ is a finitely-generated abelian group, then ${A}$ can be written as a direct sum

$\displaystyle A \cong {\mathbb Z}^r \oplus \bigoplus_{p\text{ prime}} \bigoplus_{n=1}^\infty ({\mathbb Z}/p^n{\mathbb Z})^{e_{p,n}}$

in a unique way — that is, the exponents ${r}$ and ${e_{p,n}}$ are uniquely determined by ${A}$ (and all but finitely many of them are zero).

This theorem is really two important theorems in one. The first is a decomposition theorem, which lets us decompose our objects (finitely generated abelian groups) in a simple way (a direct sum of cyclic groups). The second is a uniqueness theorem, which tells us when two of our objects are the same (in this case, precisely when the exponents ${r}$ and ${e_{p,n}}$ are the same).

In particular, this theorem tells us all we’d ever need to know about finite abelian groups. When we go further, though, the situation becomes far more difficult. Here are a few radically different examples of infinite abelian groups (all of which are countable, even!):

• ${{\mathbb Z}}$, which we’ve already taken care of.
• ${{\mathbb Q}}$, the rational numbers.
• ${{\mathbb Q}/{\mathbb Z}}$.
• ${{\mathbb Z}_{(p)}}$, the set of rational numbers with denominators prime to ${p}$.
• ${{\mathbb Z}/p^\infty = {\mathbb Z}_{(p)}/{\mathbb Z} = \varinjlim {\mathbb Z}/p^n}$ along the maps that are multiplication by the obvious powers of ${p}$.

We’d like to generalize the above classification theorem to deal with infinite abelian groups. This is surprisingly difficult for high cardinalities, and as far as I can tell in my limited research, there’s no real hope for such a theorem for uncountable abelian groups. For countable abelian groups which are assumed to be torsion, however, there’s a rather nice theorem, due to Ulm in the 30’s [4], and in this talk we’ll attempt to prove it. More generally, we’ll exhibit a couple convenient ways of decomposing abelian groups.

Most of this is from Kaplansky’s little red book [2], which is a great and short read for people of all mathematical backgrounds. I’m going to omit the word abelian’ a lot. Additionally, all of this applies mutatis mutandis for countably generated modules over a PID.

1. Torsion and torsion-free

The first, and most familiar, way to decompose abelian groups is the following. The torsion subgroup of ${A}$ is the subgroup ${T(A)}$ consisting of all torsion elements, that is, those of finite order. It’s easy to see that ${A/T(A)}$ is torsion-free, and so we get an extension

$\displaystyle 0 \rightarrow A/T(A) \rightarrow A \rightarrow T(A) \rightarrow 0.$

This isn’t a perfect decomposition because the extension doesn’t always split. On the other hand, it’s both unique and natural: ${A}$ only has one torsion subgroup, and group homomorphisms preserve torsion. So classifying abelian groups mostly reduces to classifying torsion and torsion-free groups. As I’ll discuss later, torsion-free groups are pretty difficult, so we’ll mostly discuss torsion groups.

2. Primary components

Localization gives us the next decomposition method. If ${A}$ is torsion, then ${A_{(p)} = A \otimes {\mathbb Z}_{(p)}}$ is exactly the subgroup of elements whose order is a power of ${p}$, which we call the ${p}$primary component of ${A}$. We have a direct sum decomposition

$\displaystyle A = \bigoplus_p A_{(p)}.$

The easiest way to prove this is, actually, is using the internal direct sum we all learned about in our first-year algebra classes. Each ${A_{(p)}}$ is a subgroup of ${A}$; their pairwise intersections are trivial; and every element of ${A}$ is a sum of ${p}$-primary elements for various ${p}$.

This is a first indication of what goes wrong for non-torsion groups. Though we can study them through their ${p}$-localizations, these might not be subgroups, we certainly won’t have a direct sum decomposition, and often ${p}$-localizing won’t get us anything new (for example, ${{\mathbb Q}_{(p)} = {\mathbb Q}}$).

3. Divisible and reduced

A group is divisible if for every ${x \in A}$ and ${n \in {\mathbb Z}}$, there exists a ${y \in A}$ with ${ny = x}$.

Note that such a ${y}$ might not be unique, which keeps us from using the notation ${x/n}$.

A group ${A}$ is injective if the following equivalent conditions hold:

• Whenever ${A}$ is a subgroup of another group, it is a direct summand.
• If ${X \hookrightarrow Y}$ is an injection of groups and ${X \rightarrow A}$ is a map, then this map extends to a map ${Y \rightarrow A}$.
• The functor ${\mathrm{Hom}(-,A)}$ on the category of abelian groups preserves exact sequences.

Proposition 2 An abelian group is divisible if and only if it’s injective.

Proof: If ${A}$ is injective, then for any ${x \in A}$, the map ${{\mathbb Z} \rightarrow A}$ given by ${1 \mapsto X}$ extends along ${{\mathbb Z} \stackrel{n}{\rightarrow} {\mathbb Z}}$, which precisely gives an element ${y}$ with ${ny = x}$.

Conversely, if ${A}$ is divisible, ${f:X \rightarrow A}$ is a map, and ${X \hookrightarrow Y}$ is an injection, let ${z \in Y - X}$. If no multiple of ${z}$ is in ${X}$, then we can extend ${f}$ to ${X \oplus \langle z\rangle}$ by sending ${z}$ to any element. On the other hand, if ${nz = z' \in X}$, then choose a ${y \in A}$ with ${ny = f(z')}$ and define ${f(z) = y}$. Thus in both cases, ${f}$ extends to ${X + \langle z \rangle}$. By Zorn’s lemma, we can extend our map all the way to ${Y}$. $\Box$

Proposition 3 Every abelian group has a unique maximal divisible subgroup, which is also a direct summand.

Proof: Look at the subset of all elements of the group satisfying the divisibility property, and check that the sum of two such elements is again a divisible element. Thus this set is a subgroup, and it’s clear it’s the unique maximal divisible subgroup. Since divisible groups are injective, this is in fact a direct summand. $\Box$

A group with no divisible elements is reduced.

So we’ve got a direct sum decomposition into a reduced and a divisible group, and fortunately, we can take care of the divisible part immediately.

Theorem 4 Any divisible group is uniquely a direct sum of copies of ${{\mathbb Q}}$ and ${{\mathbb Z}/p^\infty}$ (for various ${p}$).

Proof: Let ${D}$ be a divisible group with torsion subgroup ${T}$ and torsion-free quotient ${F}$. Then ${T}$ is also divisible, so the torsion exact sequence splits, and we have ${D = T \oplus F}$. ${F}$ is torsion-free, so division by ${n}$ is unique: if ${ny = ny'}$, then ${n(y-y') = 0}$, so ${y = y'}$. Thus ${F}$ is a ${{\mathbb Q}}$-vector space, and has a unique dimension. We will speak no more of ${F}$.

To deal with ${T}$, split it into its primary components. If ${x \in T_{(p)}}$ is ${p}$-torsion, we can find a ${{\mathbb Z}/p^\infty}$ summand containing ${x}$, and an easy Zorn’s Lemma argument (using the fact that any ${p}$-primary torsion group has nonempty ${p}$-torsion) gives you an expression of ${T}$ as a direct sum of copies of ${{\mathbb Z}/p^\infty}$. The cardinality of the direct sum is just the dimension of the ${p}$-torsion of ${T_{(p)}}$ as an ${\mathbb{F}_p}$-vector space. We will speak no more of ${T}$.

We will speak no more of ${D}$. $\Box$

One popular divisible group not mentioned in the above theorem is ${{\mathbb Q}/{\mathbb Z}}$. You can check that this is just ${\bigoplus_p {\mathbb Z}/p^\infty}$!

So we’re reduced to considering reduced groups.

4. Ulm’s theorem

Let’s consider the case of a finite primary abelian group. One way to get at the number of summands of order ${p^k}$ is by noting that each such summand contains ${p}$ elements of height ${p^{k-1}}$, and no elements of height ${p^k}$ (again, the height of zero is ${\infty}$). Thus, if we let ${G_k}$ be the subgroup ${p^{k-1}G}$, the dimension of the ${\mathbb{F}_p}$-vector space ${G_k/G_{k+1}}$ is precisely the number of ${{\mathbb Z}/p^k}$ summands.

But now let ${G}$ be the countable group ${{\mathbb Z}/p \oplus {\mathbb Z}/p^2 \oplus \dotsb}$, and ${H = G \oplus G}$. Both ${G_k/G_{k+1}}$ and ${H_k/H_{k+1}}$ are countably infinite-dimensional ${\mathbb{F}_p}$-vector spaces (in fact, ${G_k \cong G}$ and ${H_k \cong H}$, so it always suffices to take ${k = 0}$). But the groups are not isomorphic — if we look at the ${p}$-torsion height zero elements, then ${G}$ has only an ${\mathbb{F}_p}$‘s worth, while ${H}$ has an ${\mathbb{F}_p^2}$‘s worth.

The order of an element in a torsion group is a well-understood way of studying that element. Dually, in a reduced ${p}$-group, we could ask for something like the ${p}$-adic valuation — how many times can you divide an element by ${p}$? The right way to define this is, as it turns out, to value it on the ordinals. We inductively define

• ${G_0 = G}$,
• ${G_{\alpha + 1} = pG_\alpha}$,
• for ${\beta}$ a limit ordinal, ${G_\beta = \bigcap_{\alpha < \beta} G_\alpha}$,
• the height ${h(x)}$ (for ${x}$ not divisible) is the largest ${\alpha}$ such that ${x \in G_\alpha}$,
• and ${h(0)}$ is ${\infty}$‘, or at any rate an ordinal greater than any of the heights of nonzero elements.

Since our groups have underlying sets (groups with underlying proper classes will be dealt with in the next lecture), this filtration stabilizes at some ordinal ${\lambda}$, called the length of ${G}$; we note that ${G_\lambda}$ is precisely the divisible summand of ${G}$. It’s possible, though, for ${G}$ to have elements of infinite height even if it’s reduced, an easy example being the group

$\displaystyle \langle x_0,x_1,x_2,\dotsc | px_0, p^kx_k - x_0\rangle.$

The following properties hold:

• ${h(x + y) \ge \min(h(x), h(y))}$, with equality if the heights of ${x}$ and ${y}$ are unequal.
• ${h(px) > h(x)}$.

So height behaves like a transfinite ${p}$-adic valuation. We’re now ready to make the following definition.

The ${\alpha}$th Ulm invariant of ${G}$ is

$\displaystyle U_G(\alpha) = \dim_{\mathbb{F}_p} \mathrm{Ann}_p(G_\alpha)/\mathrm{Ann}_p(G_{\alpha + 1}).$

Theorem 5 (Ulm) Two countable reduced primary groups are isomorphic iff they have the same Ulm invariants.

Let ${G}$ and ${H}$ be two such groups. Our approach will be to use induction on the elements of ${G}$, which is why this only works for ${G}$ countable. Our eventual isomorphism will obviously preserve height, so for the induction hypothesis, suppose we have finite subgroups ${S \subseteq G}$, ${T \subseteq H}$, and an isomorphism ${\phi:S \rightarrow T}$ which preserves heights as computed in ${G}$ and ${H}$. Then given an element ${x \in G}$, we’ll extend ${\phi}$ to an isomorphism ${\langle S, x\rangle \rightarrow \langle T, y\rangle}$ for some ${y = \phi(x)}$. The equality of the Ulm invariants of ${G}$ and ${H}$ allows us to always find a ${y}$ that satisfies the same constraints on ${x}$ given by height and relationship to the subgroup ${S}$. We begin with the following technical definition, construction, and lemma.

Let ${S \subseteq G}$ be a subgroup. An element ${x \in G}$ is proper with respect to ${S}$ if ${x}$ has maximal height in the coset ${x + S}$.

If ${S}$ is finite, such elements always exist. The ultrametric inequality lets us compute heights in ${x + S}$ based on heights in ${S}$ and the height of a proper element.

Given a subgroup ${S}$, let ${S_\alpha^*}$ be the set of elements ${x \in S \cap G_\alpha}$ with ${px \in G_{\alpha + 2}}$. Obviously, this holds for all of ${S \cap G_{\alpha + 1}}$, so ${S_\alpha^*/S \cap G_{\alpha + 1}}$ could be thought of as the elements in ${S}$ of height at least ${\alpha}$ for which multiplication by ${p}$ doesn’t work as expected with respect to height. Given ${x \in S_\alpha^*}$, ${px = py}$ for some ${y \in G_{\alpha + 1}}$, and this ${y}$ is determined up to ${\mathrm{Ann}_pG_{\alpha + 1}}$. Thus ${x - y}$ is well-defined in ${P_\alpha / P_{\alpha + 1}}$. If ${x \in S \cap G_{\alpha + 1}}$, we can take ${y = x}$, and at last we get a map

$\displaystyle \theta:S_\alpha^*/(S \cap G_{\alpha + 1}) \hookrightarrow P_\alpha/P_{\alpha+1}.$

In particular, ${\dim_{\mathbb{F}_p} S_\alpha^*/(S \cap G_{\alpha + 1}) \le U_G(\alpha)}$, and the following lemma tells us when this is strict.

Lemma 6 An element ${v \in P_\alpha}$ is proper of height ${\alpha}$ with respect to ${S}$ iff ${\overline{v}}$ isn’t in the image of ${\theta}$.

Proof: (${\Rightarrow}$) If ${v}$ has height ${\alpha}$ and is in ${\mathrm{im}(\theta)}$, then ${v \equiv x - y}$ mod ${P_{\alpha + 1}}$ with ${x \in S_\alpha^*}$ and ${y \in G_{\alpha + 1}}$. Thus ${h(v - x) = h(y) \ge \alpha + 1}$, so ${v}$ is not proper.

(${\Leftarrow}$) If ${\overline{v}}$ is outside ${\mathrm{im}(\theta)}$, then ${v \not\in P_{\alpha + 1}}$, so ${h(v) = \alpha}$. If there exists ${s \in S}$ with ${h(s-v) > \alpha}$, then ${s-v = pt}$ for some ${t}$ of height at least ${\alpha}$, and ${p^2t = p(s-v) = ps}$ has height at least ${\alpha + 2}$. Thus ${s \in S_\alpha^*}$ and ${\theta(s) = s-pt = v}$, whichi s a contradiction. Thus ${v}$ is proper. $\Box$

Proof: As outlined above, suppose for induction that there are finite subgroups ${S \subseteq G}$, ${T \subseteq H}$, and a height-preserving isomorphism ${\phi:S \rightarrow T}$, and let ${x \not\in S}$ have height ${\alpha}$. Since ${G}$ is torsion, we can pick ${x}$ so that ${px \in S}$; additionally, we normalize ${x}$ to be proper in its coset and with ${h(px)}$ maximal among proper ${x}$ in this coset. (Since ${S}$ is finite, these normalizations present no difficulty). Let ${z = \phi(px)}$. It suffices to find a ${y \in H}$ which has height ${\alpha}$, is proper with respect to ${T}$, and satisfies ${py = z}$. There are two cases.

The easy’ case is where ${h(px) = \alpha + 1}$; we don’t even need the statement about Ulm invariants to do this, and we can take ${y}$ to be any height-${\alpha}$ element of ${H}$ with ${py = z}$. First we show that ${y \not\in T}$; if ${y}$ is in ${T}$, then ${y = \phi(w)}$ for some ${w \in S}$ of height ${\alpha}$, and using properness of ${x}$ and the ultrametric inequality, we get ${h(x-w) = \alpha}$. But ${px - pw = 0}$, which contradicts maximality of ${h(px)}$. Thus ${y\not\in T}$.

If ${y}$ is not proper, there is some ${\phi(s) = t \in T}$ with ${h(y + t) \ge \alpha + 1}$. But then

$\displaystyle h(px) = \alpha + 1 < h(py + pt) = h(z + pt) = h(px + ps),$

again contradicting maximality of ${h(px)}$. This proves this case.

The second case is that ${h(px) = \beta > \alpha + 1}$. Write ${px = pv}$ with ${h(v) \ge \alpha + 1}$. Then ${x - v}$ is ${p}$-torsion, height ${\alpha}$, and proper with respect to ${S}$ since for ${s \in S}$,

$\displaystyle h(x - v + s) = h(x+s) \le h(x) = h(x - v),$

using the ultrametric inequality twice. By the previous lemma, ${\dim S_\alpha^*/(S \cap G_{\alpha + 1}) < U_G(\alpha)}$. But ${\phi}$ establishes an isomorphism ${S_\alpha^*/(S \cap G_{\alpha + 1}) \cong T_\alpha^*/(T \cap H_{\alpha + 1})}$; the Ulm invariants of ${G}$ and ${H}$ are the same by hypothesis; thus by the lemma again, there’s a ${y_1 \in H}$ which is ${p}$-torsion, height ${\alpha}$, and proper with respect to ${T}$. If we let ${y_2}$ be some element of height ${\ge \alpha + 1}$ with ${py_2 = z}$, then ${y = y_1 + y_2}$ has height ${\alpha}$, satisfies ${py = z}$, and is proper with respect to ${T}$ by the argument beginning the paragraph. $\Box$

5. Some applications

Thus given a general countable torsion abelian group, we decompose it into ${p}$-primary components, and look for each ${p}$ at the rank of the divisible summand as a free module over ${{\mathbb Z}/p^\infty}$ and the Ulm invariants. This gives us a cardinal and an ordinal-indexed sequence of cardinals which are complete invariants of the group. One easy thing to note is that these are all additive under direct sums. The following corollaries are then immediate:

Corollary 7 If ${G}$ and ${H}$ are countable torsion groups and ${G \oplus G \cong H \oplus H}$, then ${G \cong H}$.

Corollary 8 If ${G}$ and ${H}$ are countable torsion groups and ${F}$ is a finitely generated group such that ${F \oplus G \cong F \oplus H}$, then ${G \cong H}$.

Corollary 9 If ${G}$ and ${H}$ are countable torsion groups with each one isomorphic to a direct summand of the other one, then ${G \cong H}$.

For an example of why we can’t put subgroup’ in place of direct summand,’ take ${G = ({\mathbb Z}/p^2)^\omega}$, ${H = {\mathbb Z}/p \oplus ({\mathbb Z}/p^2)^\omega}$.

We can also use this observation to characterize the direct sums of cyclic groups, thus generalizing the classification of finitely generated abelian groups. A cyclic ${p}$-group has one-dimensional ${p}$-torsion and thus only a single nonzero Ulm invariant ${U_{{\mathbb Z}/p^k}(k-1) = 1}$.

Theorem 10 A countable torsion group is a direct sum of cyclic groups iff it has no elements of infinite height.

Corollary 11 A subgroup of a countable torsion direct sum of cyclic groups is a direct sum of cyclic groups.

Kaplansky proved the corollary with the countability hypothesis dropped. On the other hand, the theorem itself is false with for uncountable groups.

Let ${T}$ be the torsion subgroup of ${\prod_k {\mathbb Z}/p^k}$. Obviously, ${T}$ has no elements of infinite height. The maps ${{\mathbb Z}/p^k \stackrel{p}{\rightarrow} {\mathbb Z}/p^{k+1}}$ induce an isomorphism ${T \rightarrow pT}$, so all the (finite) Ulm invariants are the same, and it’s easy to see that they’re all 1. Thus, if ${T}$ were a direct sum of cyclic groups, it would be ${\bigoplus_k {\mathbb Z}/p^k}$, but it’s not — for instance, it’s not even countable. (There’s a distinct element ${x_A \in T}$ for each ${A \subseteq {\mathbb N}}$ given by ${(x_A)_k = p^{k-1} \in {\mathbb Z}/p^k}$ for ${k \in A}$, ${0}$ for ${k \not\in A}$; a standard cardinal-arithmetic argument shows ${|T| = 2^{\aleph_0}}$.) Moreover, the quotient of ${T}$ by this direct sum is divisible!

As I said, this whole theory carries through for countably generated modules over a PID. One PID of great interest is ${k[T]}$ for ${k}$ a field — a module over this is a ${k}$-vector space equipped with a linear transformation. We could ask the following question: if ${f}$ and ${g}$ are endomorphisms of a vector space ${V}$ over ${k}$, is there an automorphism ${\tau}$ such that ${\tau f \tau^{-1} = g}$? Ulm invariants allow you to answer this when ${V}$ is countable-dimensional and ${f}$ and ${g}$ are locally algebraic, meaning that any ${v \in V}$ is in the kernel of some polynomial in the endomorphism.

It feels unfair to conclude without a few words on torsion-free groups. The first invariant of a torsion-free group is its rank, which is the largest number of linearly independent elements in the group. In 1937, Baer classified torsion-free groups of rank 1, and from then on the problem gets rapidly hopeless. Current research seems to center around the following approach: rank ${n}$ torsion-free groups correspond to subgroups of ${{\mathbb Q}^n}$ containing a basis for ${{\mathbb Q}^n}$, which is apparently best thought of as a topological space with its Borel ${\sigma}$-algebra, and isomorphism is an equivalence relation on this standard Borel space’ ${X}$, which one can prove is a Borel subset of ${X \times X}$. There’s a partial ordering on such `Borel equivalence relations’ corresponding roughly to complexity. A recent result of Simon Thomas [3] shows that the Borel equivalence relations corresponding to the classification of torsion-free groups of rank ${n}$ get strictly more complex as ${n}$ increases, which gives you some idea as to where things are in this subject. For more, see [1].

6. Bibliography

[1] Coskey, Samuel. The classification of torsion-free abelian groups of finite rank up to isomorphism and up to quasi-isomorphism. Trans. Amer. Math. Soc. 364 (2012), no. 1, 175Ð-194.

[2] Kaplansky, Irving. Infinite Abelian Groups. Publications in Mathematics, Issue 2, University of Michigan Press (1969).

[3] Thomas, Simon. The classification problem for torsion-free abelian groups of finite rank. J. Amer. Math. Soc. 16 (2003), no. 1, 233Ð-258.

[4] Ulm, Helmut. Zur Theorie der abzŠhlbar-unendlichen Abelschen Gruppen. Math. Ann. 107 (1933), 774Ð-803.