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This is part of Exercise 4.3.5 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0 and this search, it is new to MSE.

(NB: I have left out the tag for a reason: the tools available here are entirely group theoretic.)

The Details:

A group $G$ satisfies ${\rm min}$ if every set $S$ of subgroups of $G$ has at least one minimal element (with respect to $H\le K$ for $H,K\in S$).

A group $G$ is a $p$-group, for prime $p$, if each element of $G$ has order a power of $p$.

Let $n\in\Bbb N$. Then $G[n]$ is the subgroup of a group $G$ of all elements $g\in G$ such that $ng=0$.

The Question:

An abelian $p$-group has finitely many elements of each order if and only if it satisfies ${\rm min}$.

Thoughts:

Let $p$ be prime. Suppose $G$ is an abelian $p$-group.

$(\Leftarrow)$

Suppose $G$ satisfies ${\rm min}$. Suppose, further, that $G$ has infinitely many elements of order $p^k$ for some $k\in\Bbb N$. I think it would help to consider $G[p^k]$; I'm not sure why.

I don't know where to go from here.

$(\Rightarrow)$

Suppose $G$ has finitely many elements of each order.

What do I do next?

I'm aware that it is possible to have an infinite group with finitely many elements of each order; for example, consider

$$\bigoplus_{p\text{ prime}}\Bbb Z_p.$$

The question is trivial if $G$ is finite.

This appears to be a question I could answer myself with more time. Thus hints are preferred over full solutions.

Please help :)

Shaun
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    Suppose that $G$ has infinitely many elements of order $p$, so they form an infinite-dimensional vector space. Can you find a collection of subspaces without min? – David A. Craven Aug 17 '21 at 22:56
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    An abelian group of bounded exponent is a direct sum of cyclic groups. If there are infinitely many elements of order $p^k$ for some $k\geq 0$, then the subgroup they generate is a direct sum of infinitely many cyclic groups, and then it is easy to construct a strictly descending chain of subgroups that does not stabilize. Conversely, if $G$ has the minimal property, then the subgroup of all elements of order dividing $p^k$ must be a finite direct sum of cyclic groups, so there are only finitely many elements of order $p^k$. – Arturo Magidin Aug 18 '21 at 01:56
  • @ArturoMagidin I don't think $G$ has bounded exponent. – David A. Craven Aug 18 '21 at 15:19
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    @DavidA.Craven: $G$ itself no, but the subgroup generated by elements of order at most $p^k$ surely does. – Arturo Magidin Aug 18 '21 at 15:21
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    @ArturoMagidin Since I don't think you can prove this inside a bounded exponent subgroup, there is an issue with your proof. And unless I'm going bonkers, isn't your converse the contrapositive, not the converse? – David A. Craven Aug 18 '21 at 15:38
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    @DavidA.Craven. I don't think I understand your first comment. What is "this"? $G$ has the min condition if and only if it has DCC on subgroups, and of course if $G$ has DCC on subgroups, so does any subgroup. So if a subgroup does not satisfy DCC, then neither does $G$. So if $G$ has infinitely many elements of order $p^k$ for some $k$, then $G$ does not have DCC. I think your second objection is valid, though... wrote it a bit in haste. – Arturo Magidin Aug 18 '21 at 15:42
  • Yeah, I proved the same thing twice: if $G$ has min, then it has finitely many elements of each order. – Arturo Magidin Aug 18 '21 at 15:44
  • @ArturoMagidin I meant the other direction. "This" is the whole theorem. That is what my first comment at the top does. You can do that in bounded exponent, indeed in exponent $p$. The converse is trivial for bounded exponent, so you cannot prove the general case by restricting to bounded exponent. – David A. Craven Aug 18 '21 at 15:44
  • @DavidA.Craven I still don't understand your first comment. The result is valid (in fact, easy) if $G$ is of bounded exponent (if only finitely many of each order, then $G$ is in fact finite, hence has min; if it has min, then the argument shows that the decomposition into direct sums is finite). Do you mean, "the result does not follow if you only prove it inside a bounded exponent subgroup"? – Arturo Magidin Aug 18 '21 at 15:47
  • @ArturoMagidin Yes. I don't think any proof of the converse that works entirely inside a bounded-exponent subgroup can work for all groups. – David A. Craven Aug 18 '21 at 15:48
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    @DavidACraven: Since I messed up the converse anyway, don't think it matters ;-) – Arturo Magidin Aug 18 '21 at 15:49
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    For the converse (finitely many elements of any given order implies min), use Kulikov's Theorem 4.3.14: if $G$ is reduced, let $G_n$ be the subgroup of elements of order dividing $p^n$. This is finite, so there is a bound $k(n)$ to the height of elements of $G_n$ (since $G$ has no nontrivial divisible subgroup). Since $G_1\leq G_2\leq\cdots$ and $G$ is the union of the $G_n$. Then $G$ is a direct sum of cyclic groups, and finiteness gives min. For arbitrary $G$, write as a direct sum of divisible and reduced; the divisible is itself a direct sum of Prufer groups, which must be finite. – Arturo Magidin Aug 18 '21 at 16:19
  • For completeness: "4.3.14 (Kulikov) An abelian $p$-group $G$ is a direct sum of cyclic groups if and only if there is an ascending chain of subgroups $G_1\le G_2\le\dots\le G_n\le\dots$ whose union is $G$ such that the height of a nonzero element of $G_n$ cannot exceed some positive integer $k(n)$" – Shaun Aug 18 '21 at 16:20
  • @ArturoMagidin You don't need any difficult stuff to do this. Hint: write $\Omega_i(G)$ for the subgroup generated by elements of order $p^i$. Then $\Omega_1(G)$ is elementary abelian of rank $n$, say. Notice that the rank of the e.a. group $\Omega_{i+1}(G)/\Omega_i(G)$ is at most $n$. This is enough to prove that $\Omega_1(G)$ finite implies min. I can post an answer if needed. – David A. Craven Aug 18 '21 at 17:05
  • Please do, @DavidA.Craven; I'm not familiar with elementary abelian groups. Robinson defines them in Exercise 1.4.8 but that's about it. – Shaun Aug 18 '21 at 17:11

1 Answers1

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Let $\Omega_i(G)$ denote the subgroup generated by all elements of order $p^i$. Recall that an elementary abelian group is simply a direct product of (possibly infinitely many) copies of the cyclic group $C_p$*. Its rank is the number of copies of $C_p$. Notice that raising to the power $p$ is a map from $\Omega_i(G)$ to $\Omega_{i-1}(G)$ for each $i$. In particular, it induces a homomorphism from the quotient groups $\Omega_{i+1}/\Omega_i(G)\to \Omega_i(G)/\Omega_{i-1}(G)$, and this homomorphism must be injective. Since all elements of this quotient have order $p$, it is an elementary abelian $p$-group, so we are interested in its rank. In particular, the rank of $\Omega_i/\Omega_{i-1}(G)$ is at most the rank of $\Omega_1(G)/\Omega_0(G)=\Omega_1(G)$. For $i\in \mathbb N$, write $r(G,i)$ for the rank of $\Omega_i(G)/\Omega_{i-1}(G)$ (which could be $\infty$), a weakly descending sequence for any fixed $G$, i.e., $r(G,i)\geq r(G,j)$ for all $i\leq j$. This applies for all subgroups of $G$ as well, and of course $r(G,i)\geq r(H,i)$ for any subgroup $H$ of $G$.

Thus if there are infinitely many elements of order $p^i$ for some $i$ then there are infinitely many of order $p$. These look like an infinite-dimensional $\mathbb{F}_p$-vector space. Take the set of all infinite subgroups of $\Omega_1(G)$. Certainly there are infinitely many of them, and they cannot have a minimal element, just by removing one basis element at a time. Thus $G$ cannot have min.

Conversely, let $$H_1>H_2>H_3>\cdots$$ be an infinite descending chain of subgroups of $G$, but that $G$ has only finitely many elements of order $p$ (and hence of order $p^i$ for all $i$). For each $i$, let $f(i)$ denote the smallest order of an element of $H_i\setminus H_{i+1}$, which exists since all elements of $G$ have finite order. Notice that $f:\mathbb N\to\mathbb N$ cannot take the same value $m$ infinitely often, as then there would be infinitely many elements of order $p^m$ in $G$.

Note that $r(H_i,j)\geq r(H_{i+i},j)$ for all $i$ and $j$. If $f(1)=m_1$, then we see that $r(H_1,m_1)>r(H_2,m_1)$, which means that the latter is strictly less than $r$, and indeed therefore $r(H_i,m)<r$ for all $i\geq 2$ and $m\geq m_1$. We will now repeat the process, but we have to be a little careful.

Since the sequence $f(i)$ diverges, we can produce a weakly increasing subsequence, with indices $i_1,i_2,\dots$, and values $m_j=f(i_j)$. We have $$ r\geq r(H_{i_1},m_1)>r(H_{i_1+1},m_1)\geq r(H_{i_2},m_2)>r(H_{i_2+1},m_2)\geq r(H_{i_3},m_3)>\cdots.$$ Such a sequence cannot have more than $r$ strict inequalities, and so we obtain a contradiction.

*Here direct product means direct sum in the categorical sense, sometimes a restricted direct product. How about: this is an abelian group all of whose elements have order $p$.

David A. Craven
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    What is $n$? Do you mean $p$? – Shaun Aug 18 '21 at 19:11
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    @Shaun thanks, edited. I'm still going through changing things. I find it very difficult to proof-read these answers before posting, for some reason. – David A. Craven Aug 18 '21 at 19:12
  • @Shaun Damn, yes. OK, all of those are probably swapped round. – David A. Craven Aug 18 '21 at 19:12
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    Direct sum, not direct product. Otherwise, there wouldn't be a countable elementary abelian group... – Arturo Magidin Aug 18 '21 at 19:15
  • Basically, this $r(G,i)$ counts the number of elements of order $p^i$ mod the number of order $p^{i-1}$. Because the multiplication map is injective, this gets smaller as $i$ increases. It also gets smaller as you do down the chain of subgroups. Careful choice of where you look obtains the contradiction. – David A. Craven Aug 18 '21 at 19:16
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    @ArturoMagidin solved that problem by switching to multiplicative notation, which eliminates the issue. I'm a group theorist, not an abelian group theorist! – David A. Craven Aug 18 '21 at 19:17
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    No, it's still wrong to say a direct product. A direct product of countably many copies of a nontrivial group is uncountable. There is no way to express $\oplus_{n\in\mathbb{N}}C_p$ as a direct product. You could say "restricted direct product", of course. Some use "product" for restricted and "cartesian" for arbitrary, but I don't think anyone uses "direct product" for restricted product, do they? – Arturo Magidin Aug 18 '21 at 19:18
  • Usage is conflicting E.g.,, this question uses "direct product" for cartesian product. Wikipedia gives "infinite direct product" as the cartesian. Robinson uses "product" (no "direct") for the restricted direct product, Hungerford uses "restricted product" and "product", Lang uses product for the cartesian product. – Arturo Magidin Aug 18 '21 at 19:23
  • @ArturoMagidin Now I remember. It depends on who you talk to. – David A. Craven Aug 18 '21 at 19:24
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    I have attempted to solve the problem with a possibly more confusing footnote. – David A. Craven Aug 18 '21 at 19:26
  • Hang on, shouldn't $$\cdot^p:\Omega_i(G)\to \Omega_{i+1}(G)?$$ – Shaun Aug 19 '21 at 12:32
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    @Shaun $\Omega_i(G)$ consists of elements of order dividing $p^i$. So raising to the power $p$ maps to the set of elements of order dividing $p^{i-1}$, namely $\Omega_{i-1}(G)$. There's also the notation $\mho^i(G)$, which consists of all $p^i$th powers of $G$, also written $G^{p^i}$. These are in some sense dual to one another. – David A. Craven Aug 19 '21 at 13:07