0
$\begingroup$

The first question is:

If a group has a subgroup of each possible index, is it cyclic?

(1) For a finite group, if, for every divisor of the order of the group, there exists a unique subgroup of that order, then the group is cyclic.

(2) For an infinite cyclic group, there exists a unique subgroup of each finite index.

Is the converse of (2) true? So here is my second question:

If a group has a unique subgroup of each possible index, is it cyclic?

It is not given that the group is countable (although if the conjecture above is true then it has to be).

This is not a homework problem.

$\endgroup$
  • $\begingroup$ Your remark "although if it is true then group has to be countable" is not true - or are you just saying that the integers are countable? There exist uncountable groups with subgroups of every index. $\endgroup$ – user1729 Jan 15 '15 at 9:06
  • $\begingroup$ Also, what have you tried? Your question is likely to be closed unless you give context: what you have done, is it a homework question, or from your research? Were you just wondering it?... $\endgroup$ – user1729 Jan 15 '15 at 9:09
  • $\begingroup$ @user1729. You said 'There exist uncountable groups with subgroups of every index.' I want example of this. And the example you thinking has unique subgroup of each finite index or not $\endgroup$ – Sushil Jan 15 '15 at 11:32
  • $\begingroup$ You might be interested in Prufer quasi-cyclic groups. These are non-cyclic $p$-groups with a unique subgroup of index $p^n$ for every $n\in\mathbb{N}$. $\endgroup$ – user1729 Jan 22 '15 at 9:14
3
$\begingroup$

As stated, of course not. Take for instance the direct product of two copies of the integers.

$\endgroup$
1
$\begingroup$

For $n\in\Bbb Z^+$ let $C_n$ be the cyclic group of order $n$. Let

$$G=\left\{x\in\prod_{n\ge 2}C_n:\{n\ge 2:x_n\ne 0\}\text{ is finite}\right\}\;.$$

Then $\{x\in G:x_n=0\}$ is a subgroup of $G$ of index $n$, but every element of $G$ has finite order, so $G$ is not cyclic.

$\endgroup$
0
$\begingroup$

As a counter-example to (2), take any infinite simple group, for example Thompson's group $T$, then form the product $T\times \mathbb{Z}$. The resulting group has a unique subgroup of each finite index but is not cyclic. There also exist uncountable simple groups, so this construction can be extended to give you an uncountable group with unique subgroups of each finite index (and subgroups of each infinite index if your simple group has cardinality $\aleph_1$).

However, this general idea results in groups with lots of non-trivial subgroups of infinite index. I am pretty sure that this always happens: if an infinite group is non-cyclic then it must contain more than one non-trivial subgroup of infinite index. But, at the moment, a proof eludes me.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.