# Unable to find example / contradiction of this question involving infinite order groups , non -abelian groups

This question was asked in my abstract algebra quiz and I was completely clueless in proving / contradicting any of the options and I am looking for help.

Question : Which of the following statements are right:

A There exists a group of infinite order whose every subgroup is of finite order except itself.

B There exists a non cyclic group of infinite order whose every subgroup is cyclic except itself.

C There exists a non abelian group such that that for all n $$\in \mathbb{N}$$ there exists an element of oirder n.

D there exists a non abelian group such that for all n $$\in \mathbb{N}$$- {0}, there are infinite number of elements of order n.

For abelian groups I tried $$\mathbb{Z} ,+$$ and $$\mathbb{Q} ,+$$ and there subgroups but no progress could be made . I have no stratergy or examples to prove / contradict any of the options although I have solved many problems in group theory.

For A and B, take $$\Bbb Z[\frac12]/\Bbb Z$$.

For the answers to C and D, here are the suggested example groups - however, it depends on tiny details of the problem statement whether they actually work:

For C, take $$(\Bbb Q/\Bbb Z)\times S_3$$.

For D take a product of infinitely many copies of the C example.

A,B: Let $$H$$ be a proper group of $$\Bbb Z[\frac12]/\Bbb Z$$. If $$\frac k{2^n}+\Bbb Z\in H$$ where wlog $$k$$ is odd, then by Bezout, there exist $$u,v\in\Bbb Z$$ with $$uk+2^nv=1$$, so $$\frac1{2^n}+\Bbb Z=\frac {u\cdot k}{2^n}+\Bbb Z\in H$$ and then clearly all $$\frac \ell{2^m}+\Bbb Z$$ with $$m\le n$$ are $$\in H$$. Hence for any proper subgroup, the denominators are bounded and if $$2^N$$ is the maximal denominator of an element of $$H$$, we see by the preceding argument that $$H=\langle\frac1{2^N}+\Bbb Z \rangle$$. In other words, every proper subgroup of the (non-cyclic) infinite group $$\Bbb Z[\frac12]/\Bbb Z$$ is a finite cyclic group.

C: The group is non-abelian because it has a subgroup isomorphic to $$S_3$$. And for $$n\in\Bbb N$$, the element $$(\frac1n+\Bbb Z,e)$$ has order $$n$$. Note: This assume that $$0\notin \Bbb N$$, the truth of which may depend on the author. If in your context, $$0$$ is an element of $$\Bbb N$$, then of course no such group can exist - because the order of a group element is always a positive integer (or $$\infty$$) and is never $$0$$.

D: It is clear from C that we can obtain an element of order $$n$$ by picking one in an arbitrary factor. Note: For $$n=1$$ this will always produce the neutral element - which is of course the only element of order $$1$$ in any group. So if we correct the problem statement to read "... for all $$n\in\Bbb N-\{0,1\}$$ ...", then the group specified above is an example. But if the problem statement is exactly as posted, there cannot exist such a group because there is always only one element of order $$1$$.

• For C and D can't we just take the group of all permutations of an infinite set?
– bof
Commented Nov 14, 2020 at 13:56
• @Hagen von Eitzen what is description of the group $\mathbb{Z}[1/2] /\mathbb{Z}$ ie what an arbitrary element of that group looks like?
– user775699
Commented Nov 16, 2020 at 11:16
• @HagenvonEitzen do yo have some time for answering my above comment?
– user775699
Commented Nov 16, 2020 at 19:38