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This is Exercise 4.3.14 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE.

The Details:

Let $p$ be prime. A $p$-group is a group all of whose elements have a $p$ power order.

On page 12 of Robinson's book,

A torsion group [. . .] is a group all of whose elements have finite order.

On page 94, ibid.,

An element $g$ of an abelian group $G$ is said to be divisible in $G$ by a positive integer $m$ if $g=mg_1$ for some $g_1$ in $G$. [. . .]

An abelian group $G$ is said to be divisible of each element is divisible by every positive integer.

On page 106, ibid.,

A subgroup $H$ of an abelian group $G$ is called pure if

$$nG\cap H=nH$$

for all integers $n\ge 0$.

On page 107, ibid.,

Let $G$ be an abelian torsion group. A subgroup $B$ is called a basic subgroup if it is pure in $G$, it is the direct sum of cyclic groups, and $G/B$ is divisible.

On page 108, ibid.,

An additively written group is called bounded if its elements have boundedly finite orders.

The Question:

(Kulikov) Prove that an abelian torsion group has a unique basic subgroup if and only if it is divisible or bounded. [Hint: Let $B$ be the unique basic subgroup of the $p$-group $G$. Write $B=\langle x\rangle \oplus B_1$ and show that $G=\langle x\rangle \oplus G_1$ for some $G_1$. If $a\in G_1$ and $\lvert a\rvert\le\lvert x\rvert$, prove that the assignments $x\mapsto xa$ and $g_1\mapsto g_1$, $(g_1\in G_1)$, determine an automorphism of $G$. Deduce that $a\in B$.]

Thoughts:

The hint, in essence, I think, suggests that I use Exercise 4.3.8. I asked about that exercise here earlier:

Let $G$ be an abelian torsion group.

$\Rightarrow$

Suppose $B$ is a unique basic subgroup of $G$.

What allows us to assume $G$ is a $p$-group, following the hint?

$\Leftarrow$

It seems natural to use the following logical equivalence:

$$((D\lor B)\to U)\leftrightarrow ((D\to U)\land (B\to U)).$$

  • $(D\implies U)$

Suppose $G$ is divisible. Then for all $g\in G$ we have for all $m\in\Bbb N$ there exists an $h\in G$ with

$$g=mh.$$

What do I do next? I have nothing nontrivial to add.

  • $(B\implies U)$

Let $G$ be bounded. Then each $g\in G$ has boundedly finite order.

This is a stronger assertion than saying $G$ is a torsion group.

Again, I have nothing nontrivial to add.

I have skipped a few exercises from Section 4.3 of the book. This is primarily because I have spent too much time on the set; I forgot too much of what came before and I began to lose patience in myself. I chose this exercise because it seems interesting and I have experience with a similar problem (see above).

If I had more time, I think I could solve this myself. There's a lot going on in my life at the moment, though, so I cannot devote too much to one exercise, especially considering that the exercise is not marked as being referred to later on in the text.

Please help :)

Shaun
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    For bounded groups $G$, if $B\leq G$ and $G/B$ is divisible (and bounded), then $G/B=1$. Thus $B=G$ and hence is unique. – David A. Craven Oct 11 '21 at 19:45
  • Thank you, @DavidA.Craven :) – Shaun Oct 11 '21 at 19:50
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    Every quotient of a divisible group is divisible (if $x=n\cdot y$ then $x+H=n\cdot (y+H)$), so if $G$ is torsion divisible, we need that it has no non-trivial basic subgroups $B$. But torsion divisible groups are direct sums of quasicyclic groups (4.1.5), and pure subgroups are just direct summands by Exercise 4.3.3. Quasicyclic subgroups are not sums of cyclic subgroups, so $B=1$. – David A. Craven Oct 11 '21 at 19:59
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    Finally, I know Sylow's theorem is still not allowed, but for torsion abelian groups the set of all elements of order a power of $p$ is a subgroup, and a torsion group is the direct sum of these over all $p$. I'm sure Robinson uses this all the time. If $G=\bigoplus_p G_p$ has a unique basic subgroup $B$, then $B=\bigoplus_p B_p$, and so the question is whether $B_p$ is basic in $G_p$. Clearly it is a sum of cyclic groups, $G_p/B_p$ is divisible, and so pure is all that is needed. This is also clear from the fact that $B$ is pure in $G$. – David A. Craven Oct 11 '21 at 20:03
  • Please consider upgrading those comments to an answer, @DavidA.Craven. – Shaun Nov 20 '21 at 01:01

1 Answers1

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This is a CW answer adapting the comments by @DavidA.Craven above.

For bounded groups $G$, if $B\le G$ and $G/B$ is divisible (and bounded), then $G/B=1$. Thus $B=G$ and hence is unique.

Every quotient of a divisible group is divisible (if $x=n\cdot y$ then $x+H=n\cdot(y+H)$), so if $G$ is torsion divisible, we need that it has no non-trivial basic subgroups $B$. But torsion divisible groups are direct sums of quasicyclic groups (4.1.5), and pure subgroups are just direct summands by Exercise 4.3.3. Quasicyclic subgroups are not sums of cyclic subgroups, so $B=1.$

Finally, I know Sylow's theorem is still not allowed, but for torsion abelian groups the set of all elements of order a power of $p$ is a subgroup, and a torsion group is the direct sum of these over all $p$. I'm sure Robinson uses this all the time. If $G=\oplus_pG_p$ has a unique basic subgroup $B$, then $B=\oplus_pB_p$, and so the question is whether $B_p$ is basic in $G_p$. Clearly it is a sum of cyclic groups, $G_p/B_p$ is divisible, and so pure is all that is needed. This is also clear from the fact that $B$ is pure in $G$.

Shaun
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