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This is part of Exercise 4.3.7 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE.

[NB: I did not include the tag for a reason.]

The Details:

Presentations are defined in a previous chapter of Robinson's book; however, many of the tools available in combinatorial group theory are not discussed; I cannot, for instance, use Tietze transformations in this exercise.

The tools available are:

  • Basic facts about free groups.

  • Basic facts about reduced words.

  • The normal form of an element.

  • The projective property of free groups.

  • von Dyck's Theorem.

  • Basic facts about finitely presentated groups; viz., B.H. Neumann's theorem and P. Hall's theorem.

This list is not exhaustive. There's a whole section of the book on varieties but I doubt it's relevant.

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 $G$. If $p^h$ is the largest power of the prime $p$ dividing $g$, then $h$ is called the $p$-height of $g$ in $G$: should $g$ be divisible by every power of $p$, we say that $g$ has infinite $p$-height in $G$. [. . .]

An abelian group $G$ is said to be divisible if each element is divisible by every positive integer. This is equivalent to saying that each element of $G$ has infinite $p$-height for all primes $p$.

On page 96, ibid.,

An abelian group is said to be reduced if it has no nontrivial divisible subgroups.

On page 106, ibid.,

In speaking of the height of an element in an abelian $p$-group $G$ we always mean the $p$-height.

The Question:

Let $G$ be generated by $x_1, x_2,\dots$ subject to the defining relations $px_1=0$, $p^ix_{i+1}=x_1$ and $x_i+x_j=x_j+x_i$. Prove that $G$ is a countable reduced abelian $p$-group containing a nonzero element of infinite height.

Thoughts:

Since $x_i+x_j=x_j+x_i$, clearly $G$ is abelian. We can conclude that $G$ is a $p$-group because each of its generators is a $p$-th power multiple of $x_1$ by $p^ix_{i+1}=x_1$, and $x_1$ has order $p$ by $px_1=0$. That much is easy.

A naïve guess would be that $G$ is countable because it can be generated by countably many generators. I do not hold this view, simply out of caution; I don't know how to go about proving it yet. My suspicion is that this guess, if true, would allow a proof by contradiction.

Showing $G$ is reduced might also be done by contradiction. To this end, suppose $H\le G$ is divisible. Then each nontrivial $h\in H$ would permit, for lack of a better way of putting it, "too many prime divisors $q\neq p$".

As for the element $h$ of infinite height, I'm not sure. It has to follow from $px_1=0$ and $p^ix_{i+1}=x_1$, I suppose, and, again, for lack of a better way of putting it, perhaps $h$ is a "limit of the $x_i$ as $i$ tends to infinity" - but this is nonsense.

A word on the multifaceted nature of this question: I think demonstrating the properties of $G$ here might each be fairly simple and that they're eluding me because I am overthinking them; and as such, I think asking a separate question for each property would lead to tedious repetition and diminishing returns.

I think I could answer this myself if I gave it some more time, especially because my current degree is in combinatorial group theory. It's just that I've given it over a week now and I'm losing momentum in the book because of it.

The type of answer I'm hoping for is a set of strong hints or a full solution.

Please help :)

Shaun
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    (Rewritten, because some of the statements in the earlier version were wrong!) Just a couple of quick comments. Yes, every group with a countable set of generators is countable. That is basically because every element of the group can be expressed a word of finite length in the generators. Yes, the fact that $p^{i}x_{i+1}=x_1$ proves that $x_1$ is divisible by $p^i$ for all $i$, so it has infinite $p$-height. Reducibility is a bit harder. – Derek Holt Sep 02 '21 at 19:59
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    That a countably generated group is countable follows because every element of the group $\langle X\rangle$ is a finite product of elements of $X$ and their inverses. that means that the cardinality of such a group is at most $\aleph_0|X|$, so for $X$ countable, the generated group is countable. – Arturo Magidin Sep 02 '21 at 20:40
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    To show reducibility: none of elements $x_k$, $k\gt 1$, is a $p$th power. You can verify this by mapping $G$ to the cyclic group of order $p^k$, sending $x_k$ to the generator, and all other $x_i$ to the trivial element. If $x_k$ were a $p$th power, then its image would be a $p$th power in that group, which it plainly isn't. Thus, no subgroup containing any $x_k$ with $k\gt 1$ can be divisible. That only leaves $\langle x_1\rangle$ and its subgroups, but this is cyclic of order $p$ and hence not divisible. – Arturo Magidin Sep 02 '21 at 20:47
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    @ArturoMagidin To complete the proof of reducibility, you would need to show that every subgroup contains some $x_k$ for $k>1$. The subgroup $\langle x_2+px_3\rangle$ does not obviously do so. (A minor modification of your proof does work though.) – David A. Craven Sep 02 '21 at 21:27
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    @DavidA.Craven: Good point. Thanks. – Arturo Magidin Sep 02 '21 at 21:32
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    For any fixed $k\gt 1$, you can map the subgroup generated by $x_1,\ldots,x_k$ to the cyclic group of order $p^{k+1}$ by sending $x_k$ to the generator, $x_{k-1}$ to the $p$th power of the generator, etc. Any subgroup not contained in $\langle x_1\rangle$ contains a nontrivial element in such a subgroup, but under the map it lands in an element of finite height, so the subgroup is not divisible. – Arturo Magidin Sep 02 '21 at 21:39
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    @ArturoMagidin That should work. A much smaller modification is to let $\sum a_ix_i$ be some element in the subgroup, and just map to the quotient you described, where $k$ is such that $a_k\neq 0$. OK, $p$th powers might exist, but since this is a cyclic $p$-group, some $p$-power won't work. This might involve fiddling with the generators though to avoid this all elements of order $p$ are the same problem, so perhaps your method is better. – David A. Craven Sep 02 '21 at 21:41
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    I can fix that by noting that the intersection of the kernels of all of your quotients is $\langle x_1\rangle$, so any element outside of that has non-trivial image under one of these homomorphisms. – David A. Craven Sep 02 '21 at 21:45
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    Since $G/\langle x_1\rangle\cong\langle \bar{x_2}\rangle\oplus\langle \bar{x_3}\rangle\oplus\ldots$ is reduced then $G$ is also reduced. Here $\bar{x_i}=x_i+\langle x_1\rangle$. – kabenyuk Sep 03 '21 at 04:44
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    @DavidA.Craven: I think this gets rid of the complications (similar to @kabenyuk's): if $H$ is a divisible subgroup of $G$, then $pH$ is divisible. But $pG\cong \oplus_{i=1}^{\infty}C_{p^i}$ (which is straightforward to to establish, I think) is reduced, so $pH\leq pG$ is trivial, and since $H$ is divisible, this means $H$ is trivial. – Arturo Magidin Sep 03 '21 at 17:51

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