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.

Can you please guide?


1 Answer 1


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 $1$.

  • $\begingroup$ For C and D can't we just take the group of all permutations of an infinite set? $\endgroup$
    – bof
    Commented Nov 14, 2020 at 13:56
  • $\begingroup$ @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? $\endgroup$
    – user775699
    Commented Nov 16, 2020 at 11:16
  • $\begingroup$ @HagenvonEitzen do yo have some time for answering my above comment? $\endgroup$
    – user775699
    Commented Nov 16, 2020 at 19:38

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