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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 $Z_{p^n}$?

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  • $\begingroup$ 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
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    $\begingroup$ 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

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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.

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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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