Find all finite groups $G$ s.t for any $a,b\in G$ either $a$ is a power of $b$ or $b$ is a power of $a$

I think i showed that all such groups are $Z_{p^n}$ for $p$ prime, is this correct? I first showed that the group must be cyclic by considering the element of the largest order $\langle a\rangle$ and achiveing contradiction if $\langle a\rangle\not= G$., and then that if $Z_n$ with $n$ composite then it does not have this property. as there are two disjoint cyclic subgroups of coprime orders.

Is this correct? Are all groups such groups $Z_{p^n}$?

  • $\begingroup$ Yes, this is correct. P.S. use \langle and \rangle, not < and >. $\endgroup$ Commented Aug 18, 2020 at 20:59
  • $\begingroup$ @ArturoMagidin you just told me I very possibly passed my qual! Thank you $\endgroup$
    – 2132123
    Commented Aug 18, 2020 at 21:00
  • 1
    $\begingroup$ That may depend on how you justified the “achieving a contradiction if $\langle a\rangle\neq G$”, of course... ;-) $\endgroup$ Commented Aug 18, 2020 at 21:01
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    $\begingroup$ @ArturoMagidin That is true :P you let $b \not \in \langle a \rangle$ then the property implies that $<b>$ contains $a$ and so has larger order. $\endgroup$
    – 2132123
    Commented Aug 18, 2020 at 21:03

2 Answers 2


Yes, if you take $a$ with maximal order and, by contradiction, there is $b\notin\langle a\rangle$, then $a=b^n$ for some $n>1$, so $b$ has larger order than $a$.

Therefore $G$ is cyclic.

Now we can prove that the order of $G$ must be a prime power: you cannot exclude “composite” (a minor slip, but relevant).

If $|G|$ is divisible by two distinct primes $p$ and $q$, then $G$ has subgroups of order $p$ and $q$, but these have trivial intersection, so the group cannot have the stated property.

A cyclic group of order $p^n$ ($p$ a prime) has the stated property.


This is correct. Well, apart from the "disjoint subgroups" thing. The subgroups are "almost disjoint", i.e., their intersection is reduced to the identity element, but they cannot be literally disjoint.


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