If collection of subgroups of $G$ form a chain , then $G$ is cyclic?

Question

If $G$ is a group such that for any two subgroups $H,K$ either $H\le K$ or $K\le H$ holds ,then $G$ is cyclic.

I am going to prove this is true for finite groups and false for infinite groups.

Proof (For finite group)

Let $G$ be a finite group satisfying the condition.

Step 1: To show $G$ must be a $p$ group.

Let $|G|$ has at least two prime factors $p$ and $q$ (say, with some powers) then a Sylow- $p$ subgroup and a Sylow- $q$ subgroup is neither a subgroup of one another and so the condition does not hold , a contradiction.

So if $G$ is finite and satisfies the condition then it is $p$ group.

Step 2: To show $G$ is cyclic

Now , let $|G|=p^n$ where $n\ge 2$

Then let $H$ be a subgroup of order $p^{n-1}$ which exists by Sylow's Theorem.

Then let $a\in G\setminus H$ and $K=\langle a \rangle$ .

Now $K\le H$ is not possible since $a\notin H$.

So, $H\le K$ must hold. Then either $|K|=p^{n-1}$ or $p^n$.

If $|K|=p^{n-1}$ then $H=K$ which again contradicts $a\notin H$

So we must have $|K|=p^{n} \Rightarrow G=K$ which proves $G$ is cyclic.

So the result follows .

An example for falsity in infinite groups

Let $G=\{z \in \mathbb{C} : z^{2^n}=1 , n\in \mathbb{N}\}$ i.e $G$ consists of all square roots, fourth roots of unity. etc

$G$ forms a group with respect to multiplication.

Proof : Lef $z_1,z_2 \in G$

Then $z_i^{2^{n_i}}=1,i=1,2$ for some $n_1,n_2 \in \mathbb{N}$

Then $(z_1z_2)^{2^{n_1+n_2}}=(z_1^{2^{n_1}}) ^{2^{n_2}}(z_2^{2^{n_2}}) ^{2^{n_1}} =1$

Of course, there is the identity $1$.

Let us define $A_n =\{z\in \mathbb{C} : z^{2^n}=1\}$ where $n=1,2,3.,$

These $A_n$'s form a chain of subgroups given by

$A_1 \subset A_2 \subset A_3...$

I just need to show there are no other proper subgroups.

Let $A\lt G$ and be finite.

Let $n=\displaystyle\max_i \{i: z^{2^i}=1 $ for the least $i, z \in A\}$

Then $A_n \subseteq A$ since $A$ is a group.

The reverse inclusion $A\subseteq A_n$ is also true by the above chain and nature of $n$

So $A=A_n$

Now , if $A$ be infinite then the set $\{i:z^{2^i}=1 $ for the least $i ,z \in A\}$ is unbounded above and hence must it must contain $A_n$ for every $n$ which is nothing but $G$

But evidently $G$ is not cyclic, not even finitely generated.

Is there any logical fallacy in my work? Do you have a better proof/solution ? Is ther any generalisation of this result?

Sorry if my work is clumsy as I am not used to writing sophisticated proofs. Thanks a lot for your time.

Answer 1

Your proof for finite groups is correct. However, it might be easier to point out that if $G$ is generated by $x_1,\dots,x_n$ then the subgroups $\langle x_i\rangle$ form a chain. Thus all but one of the $x_i$ is unnecessary, and $G=\langle x\rangle$.

Your example is also correct. My proof above shows that, in such a group, every finitely generated subgroup is cyclic. Such groups are called locally cyclic. The easiest example of a locally cyclic group is the rationals under addition, but as you can see from that, not every locally cyclic group has your property. I'm not sure that I would accept your proof that $A=A_n$ though, as you don't seem to have demonstrated it beyond saying that it's clear. $A_n\subseteq A$ as $A$ is a group doesn't seem to make sense.

You need to note that $A_n$ contains all elements of $G$ whose order divides $2^n$. Then $A\leq A_n$ by choice of $n$, and since $A_n$ is cyclic (and generated by any element of $A_n\setminus A_{n-1}$) $A$ contains all of $A_n$, since it contains a generator of $A$, again by choice of $n$.

Notice that you may replace $2$ by $p$ in your example. These groups are called Prüfer groups.

Edit: Notice that you can describe your group as a quotient of the dyadic rationals. Indeed, back in the 1930s, Baer proved that every locally cyclic group (which is necessarily abelian) is a quotient of a subgroup of the rational numbers.

Answer 2

You can shorten the finite case (since you don't actually need the prime-power order conclusion for now) by choosing $x\in G$ with $o(x)$ maximal. Then if $\langle x\rangle\neq G$ there exists $y\in G-\langle x\rangle$ and so $\langle y\rangle$ is not contained in $\langle x\rangle$, hence $\langle x\rangle\subsetneq\langle y\rangle$, which contradicts the maximality of $o(x)$.

And in the infinite case, note that the subgroup lattice of the infinite cyclic group is not a chain. Hence the group must be torsion, and you can now show there is some prime $p$ such that every element must have $p$-power order (if $g$ has order $n=n_1n_2$, $n_1,n_2>1$ coprime, then $\langle g^{n_1}\rangle$ and $\langle g^{n_2}\rangle$ are nontrivial and have coprime orders, so violate the chain condition). This gives the Prüfer groups as the only example.