Call an abelian group $G = (G,+)$ $m$-torsion for some natural number $m$ if one has $m \cdot x = 0$ for all $x \in G$. A subgroup $H$ of $G$ is said to be complemented if one can write $G = H \oplus K$ for some other subgroup $K$ of $G$. I was able to establish the following fact:

Proposition. If $G$ is a finite abelian $m$-torsion group, and $H$ is an index $k$ subgroup of $G$, then $H$ contains a complemented subgroup $H'$ of $G$ of index at most $C(m,k)$, for some quantity $C(m,k)$ that depends only on $m$ and $k$.

For instance, if $G = ({\bf Z}/4{\bf Z})^n$ for some large $n$, then $G$ is $4$-torsion, and the index $2$ subgroup $H := \{ (x_1,\dots,x_n) \in G: 2(x_1+\dots+x_n)=0\}$ is not complemented, but it contains the complemented index $4$ subgroup $H := \{ (x_1,\dots,x_n) \in G: x_1+\dots+x_n=0\}$. The restriction to $G$ being finite is probably not essential, but I include it here for simplicity.

I am not satisfied with my proof because it relies a fair bit on the classification of finite abelian groups. Here is my proof:

  1. By Pontryagin duality, it suffices to establish the dual claim that every order $k$ subgroup $K$ of $G$ is contained in a complemented subgroup $K'$ of order at most $C(m,k)$.
  2. By induction on $k$, we may assume without loss of generality that $K$ is simple. Indeed, if $K$ contained a non-trivial normal subgroup $N$, then by induction $N$ is contained in a complemented subgroup $N'$ of bounded size, and after quotienting out by this subgroup $N'$, $K$ will also be contained in a complemented subgroup of bounded size. From this it is easy to place $K$ itself in a complemented subgroup of bounded size.
  3. By Cauchy's theorem, $N$ is now of some prime order $p$, which must divide $m$. By the Schur-Zassenhaus theorem, we may now pass to the Sylow $p$-group and assume that $G$ is a $p$-group.
  4. By the classification of finite abelian groups, one can write $G$ as the product of $O_m(1)$ many factors $({\bf Z}/p^i {\bf Z})^{n_i}$, where $i$ is bounded but $n_i$ need not be. By projecting $N$ to each of these factors separately, and then taking Cartesian products of any complemented subgroups located, we may reduce to the case of a single factor $G = ({\bf Z}/p^i {\bf Z})^{n_i}$.
  5. If we let $e$ be a generator of $N$, one can now perform an explicit computation in coordinates to locate a root $p^{i-1} e' = e$ of $e$. Taking $N'$ to be the group generated by $e'$, one can check by hand that $N'$ is complemented and of order $p^i$, giving the claim.

My (somewhat vague) question is whether there is a more conceptual or high-level proof of this proposition that does not rely so much on the classification of finite abelian groups, or working in explicit coordinates. In a somewhat different direction, I am also curious as to what the correct dependence of the constant $C(m,k)$ on $m,k$ should be; the argument I gave above is quite inefficient in this regard.

  • 2
    $\begingroup$ This kind of result usually makes sense and should be true in the context of modules over PIDs (changing "finite index" to "the quotient has finite length"), so ac.commutative-algebra would make sense. Or maybe modules over artinian rings. $\endgroup$
    – YCor
    Dec 3, 2023 at 16:49
  • $\begingroup$ One way of proving the classification of finite Abelian groups is to restrict to the $p$-group case, and show an element of maximal order generates a complemented subgroup. Perhaps that approach would work here too? $\endgroup$
    – Steve D
    Dec 3, 2023 at 16:53
  • $\begingroup$ Are there cyclic p-groups $P_1, P_2, ... ,P_k$ such that $G = P_1 \oplus P_2 \oplus ... \oplus P_k$ and $H = (P_1 \cap H) \oplus (P_2 \cap H) \oplus ... \oplus (P_k \cap H)$ ? Then $H'$ can be choosen as the direct sum of the $P_j$ with $P_j \subset H$ . $\endgroup$
    – jjcale
    Dec 6, 2023 at 19:34
  • $\begingroup$ Example where the $P_k$ don't exist : $G = {\bf Z}/8{\bf Z} \times {\bf Z}/2{\bf Z}$, $H = <(2,1)>$ . $\endgroup$
    – jjcale
    Dec 10, 2023 at 13:45

2 Answers 2


Here is a pretty elementary argument for the best possible $C(m,k)$. It suffices to consider the case when $G$ is a $p$-group. If $m=p^a$ and $k=p^b$ then $C(m,k)=p^{ab}$. That is:

Claim 1: Suppose that $G$ is an abelian group with exponent $p^a$, i.e. such that for every $g\in G$ we have $g^{p^a}=1$. Then every finite subgroup $H\subset G$ is contained in some complementable subgroup $H'\subset G$ such that $|H'|$ divides $|H|^a$.

The key is the fact that when $|H|=p$ then there is a complementable and cyclic subgroup containing $H$. I'll prove this at the end.

Proof of the claim, assuming this fact:

We induct on $|H|$. Choose a subgroup $C\subset H$ of order $p$. Let $C'\subset G$ be a complementable and cyclic subgroup such that $C\subset C'$. So $|C'|$ divides $p^a=|C|^a$. Write $G=C'\times G_0$. The subgroup $C'H\subset G$ (i.e. the subgroup generated by $C$ and $H$) contains $C'=C'\times 1\subset C'\times G_0$, so that $$C'H=C'\times H_0 $$ for some subgroup $H_0\subset G_0$. By inductive hypothesis $H_0$ is contained in a complementable subgroup $H_0'\subset G_0$ such that $|H'_0|$ divides $|H_0|^a$. Now define $$ H'=C'\times H_0'. $$ Then $|H'|$ divides $(|C||H_0|)^a$. But $|H_0|=|C'H|/|C'|=|H|/|C'\cap H|$, so that $|C||H_0|$ divides $|H|$.

It remains to prove

Claim 2: If $G$ is a direct sum of cyclic $p$-groups, then every element $g\in G$ of order $p$ is contained in a cyclic and complementable subgroup.

Proof: Let $G$ be $\oplus_iC_i$ with $C_i$ cyclic. Out of all $i$ such that $g$ projects to a nontrivial element of the summand $C_i$, choose one such that $|C_i|$ is as small as possible, say $p^k$. The equation $g=x^{p^{k-1}}$ has a solution in $G$ because it has a solution in each summand to which $g$ projects nontrivially. Choose such an $x$. The group (of order $p^k$) generated by $x$ projects isomorphically to some $C_i$ and is therefore complementable.

  • 1
    $\begingroup$ Nice! So I guess for arbitrary $m,k$ the optimal value of $C(m,k)$ is $\prod_p p^{\nu_p(m) \nu_p(k)}$, where $\nu_p(m)$ denotes the number of times $p$ divides $m$. This answers my second question satisfactorily. Would still be interested in the first question though (specifically, an argument - possibly with non-optimal bounds - that avoids the classification of finite abelian groups, which was for instance used here to prove Claim 2). $\endgroup$
    – Terry Tao
    Dec 4, 2023 at 16:18
  • 1
    $\begingroup$ @Terry Tao Yes, I was hoping to avoid that. (Strictly speaking, I used it, not in proving Claim 2, but in deducing Claim 1 from Claim 2.) $\endgroup$ Dec 4, 2023 at 23:59
  • $\begingroup$ @Terry Tao I used the classification of finite abelian $p$-groups, together with Claim 2, to take care of the base case when $|H|=p$. Instead, I could have made do with the weaker statement that in a finite abelian group of exponent $p^a$ every element of order $p$ is in some cyclic and complementable subgroup. But one could also deduce the classification of finite abelian groups from that weaker statement, by an induction on the order of the group (without ever using the usual principle whose highbrow statement is that $\mathbb Z/n\mathbb Z$ is an injective module for itself). $\endgroup$ Dec 5, 2023 at 17:39
  • $\begingroup$ I'm provisionally accepting this answer on the grounds that it answers one of the two questions that I had, though it seems the other remains open. (From what I understand of Denis's argument it implicitly uses the fact that every finite abelian m-torsion group splits into the direct sum of injective ${\bf Z}/p^a {\bf Z}$-modules for various $p^a$ dividing $m$, which I view as a fact essentially equivalent to the classification of finite abelian groups.) $\endgroup$
    – Terry Tao
    Dec 5, 2023 at 22:29
  • $\begingroup$ Another proof of claim 2 : Choose $x \in G$ such that $x^{p^n} = g$ and $n$ maximal. Then $<x>$ is a pure subgroup of $G$ and therefore complementable. $\endgroup$
    – jjcale
    Dec 10, 2023 at 14:52

Relevant notion is that of essential submodule. $N \subset M$ is called essential, if $N$ intersects nontrivially with every non-zero submodule $M' \subset M$ (typical example is the socle of an indecomposable module over finite length local ring). Every submodule $N \subset M$ has essential envelope $E_M(N)$: $E_M(N)$ is the maximal submodule such that $N \subset E_M(N)$ is essential. From the definitions it's clear that direct summands are exactly those submodules which are their own essential envelopes.

To see how things can go wrong, look at the innocuously looking three-dimensional ring $Q = k[[x, y]]/(x, y)^2$ for any field $k$. There are arbitrarily large indecomposable $Q$-modules $M_l$. Underlying vector space of $M_l$ is $k[[A]]/(A^l) \oplus k[[\epsilon]]/(\epsilon^l)$; action is given by $x: (A^i, \epsilon^j) \mapsto (0, \epsilon^i)$, $y: (A^i, \epsilon^j) \mapsto (0, \epsilon^{i+1})$. There can be no bound of the form asked for modules over $Q$: minimal summand of $M_l$ containg the one-dimensional subspace generated by $\epsilon^{l-1}$ is the whole module, as it is indecomposable.

What's written below works for any finite length uniserial ring, not necessarily commutative.

Let $R = \Bbb Z/ p^{s}$, and $G$ an $R$-module — finite abelian $p$-group of exponent $p^s$. Further I'll look not at orders of groups, but at $p$-logarithms of orders, and use the word "size" for this logarithm: size of $K$ is $k$ if $p^k = |K|$.

I'll address the dual problem: for a given $K \subset G$, find the size of the smallest group $H$ such that $K \subset H \subset G$, and $H$ is a direct summand of $G$.

As the ring is uniserial, multiplication by $p$ — the generator of the maximal ideal — just "moves the module along itself"; think of modules, which have form $\Bbb Z/p^{l}$, as a bead chain, sliding along a rod of length $l$, and everything that does not fit on the rod falls off. I'll try to explain why this picture is actually true, unlike "bad" ring $Q$ discussed above.

Recall that radical $\operatorname{rad}(M)$ is the intersection of maximal submodules of $M$; socle $\operatorname{soc}(M)$ is the dual notion — sum of all simple submodules of $M$. Over $R$ those have easy descriptions: $\operatorname{rad}(M) = M \otimes_R (R/\operatorname{rad}(R)) = M \otimes_R \Bbb Z/p$, $\operatorname{soc}(M) = \operatorname{Ann_M}(p)$, i. e. kernel of multiplication by $p$. There's a natural isomorphism $M \otimes \Bbb Z/p = rad(M) \cong soc(M^*) = \operatorname{ker} p \subset M^*$, where $M^* := \mathrm{Hom}_{\Bbb Z}(M, \Bbb Q/\Bbb Z)$; duality is an equivalence between finite length right $R$-modules and finite length left $R$-modules. In particular, for a cyclic module both socle and radical are 1-dimensional vector spaces over $\Bbb Z/p$.

But! Because ideals $(p^i)$ of $R$ are linearly ordered, their annihilators in modules are also linearly ordered! From this we can derive (by Nakayama lemma applied to modules and their duals) that the only way to get short exact sequence $C \to M \to C'$ where $C, C'$ are cyclic and $M$ is indecomposable, is to "glue" socle of $C'$ to the radical of $C$ (i. e. multiplication by $p$ induces bijection between them in $M$), so the extension is also cyclic.

But if there are no non-cyclic indecomposable extensions between cyclic modules, all finite length indecomposable modules over uniserial ring are cyclic. One can choose as representatives of isomorphism classes either $(p)^k$, or $R/(p)^{s-k}$.

That's why uniserial property is important: direct decomposition of finite length modules over uniserial ring is completely governed by direct decomposition of its socle, and, equivalently-dually, by decomposition of its quotient by radical — because every module is an interated extension of its socle, and we know that the only form of such extension is "division by $p$". Beads cannot jump between rods; if something falls off a rod, it needed to come from the same rod.

But both socle and radical of $R$-modules are vector spaces over $\Bbb Z/p$, and their direct decomposition properties are trivial.

Let's denote the size of $K \otimes \Bbb Z/p$ as $d$ (its dimension over $\Bbb Z/p$).

We see that the size of $H$ is at most $ds$ as follows. Even if we are very generous, $K^*$ is always covered by a free module of rank $d$ (its size is $ds$), and conversely (because over $R$ projectives and injectives coincide), $K$ is always embedded in an injective module of size $ds$. $E_G(K)$ is always a submodule of injective envelope $E(K)$ — because injective envelope is the "absolute essential envelope" of $K$, i. e. you construct it looking not at intermediate modules $K \subset H \subset G$, but consider all abstract extensions $K \subset L$, in which $K$ is essential, and take colimit ("union") of all those $L$'s.

This bound is optimal, because it's attained on $(p^{s-1}\Bbb Z/ p^s)^d \subset (\Bbb Z/ p^s)^d$.

If you want to get rid of $d$ and get a bound exclusively in terms of $k$ and $s$, you get that size of $H/K$ is at most $k(s-1)$. It is attained on the same example in previous paragraph.

  • $\begingroup$ Can you explain more about how every group gets a unique radical? The non-uniqueness of roots was the biggest difficulty I had when trying to resolve this problem. For instance, in $({\bf Z}/p^2{\bf Z})^2$, it seems to me that the order $p$ subgroup $\langle (p,0) \rangle$ has multiple candidates for such a "radical"; there is $\langle (1,0) \rangle$, of course, but also $\langle (1,p) \rangle$ for instance. Also, do you mean $p^{dk}$ instead of $dp^k$? $\endgroup$
    – Terry Tao
    Dec 3, 2023 at 18:56
  • $\begingroup$ @TerryTao I hope that after complete rewriting my answer makes more sense. $\endgroup$
    – Denis T
    Dec 3, 2023 at 19:50
  • $\begingroup$ I'm sorry, I cannot follow the construction of the essential submodule and will probably need an explicit example. Let's take the one from my previous comment: $R = {\bf Z}/p^2$, $M = ({\bf Z}/p^2)^2$, and $N = \langle (p,0) \rangle$ is the size one submodule of $M$ generated by $(p,0)$. What is $E_M(N)$, viewed as a submodule of $M$? Is it only unique up to isomorphism? As I said my previous remark, both $\langle (1,0)\rangle$ and $\langle (1,p)\rangle$ seem to be candidates for this envelope. $\endgroup$
    – Terry Tao
    Dec 3, 2023 at 21:09
  • $\begingroup$ From what I can read up on about injective hulls, it seems that in order to embed the essential submodule of $N$ into $M$, it is necessary that $M$ is itself an injective $R$-module? This is basically the point in my proof where the classification of finite abelian groups came in. $\endgroup$
    – Terry Tao
    Dec 3, 2023 at 21:27
  • $\begingroup$ @TerryTao Obviously, the fact that every irreducible module is cyclic (aka classification of finite abelian groups) is NECESSARY to prove the bound you asked for; I've provided a counterexample that explains necessity of this property. I guess, you can try to hide using this fact in a proof, but it will still be there. // Essentiall hull is uniquely defined (as a submodule) in uniserial case; in general it's not even unique up to abstract isomorphism. I'm not claiming that my post contains all details — usually it takes ~40 pages in a book to state and prove properties of Nakayama algebras. $\endgroup$
    – Denis T
    Dec 6, 2023 at 1:23

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