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Suppose that the order of $G$ is divisible by at least two distinct primes. Also, let $g\in G$ that order of $g$ is divisible by every prime divisor of $o(G)$ and $\forall x\in G$, $o(x)\mid o(g)$ or $o(g)\mid o(x)$.

With above conditions:

1 - What is the group $G$?

2 - Is it a cyclic group and $G=<g>$?

walkar
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asma
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1 Answers1

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The group need not be cyclic, let us take $G=\mathbb{Z}_p \times \mathbb{Z}_p \times \mathbb{Z}_q$ where $p,q $ are distinct primes.

Let us take the element $g=(1,1,1)$ note that the order of this element is $pq$, all possible orders of elements in this group are $p,q,pq$, so for every element in $x \in G$ we will have $o(x) | o(g)$, but clearly this group is not cyclic since there is no element of order $p^2q$.

baharampuri
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  • Thanks for your answer. But $g\neq 1$ in above question. So in this case, is not group G, cyclic? – asma Aug 06 '15 at 05:59
  • No it is not cyclic since you cannot generate the element $(1,0,0)$ from the element $(1,1,1)$ as $k(1,1,1)= (k \mbox{ mod } p , k \mbox{ mod } p , k \mbox{ mod } q )=(1,0,0)$ implies $k \mbox{ mod } p = 1 = 0$ which is a contradiction. And for $g \neq 1$ in the above question here you mean $g $ is not the identity of the group, and in the answer that I posted we have $g =(1,1,1)$ which is not the identity, it is $(0,0,0)$. – baharampuri Aug 06 '15 at 06:06
  • Thanks. I understand your answer. Only these groups that have above conditions, are special groups, that is they have special name? – asma Aug 06 '15 at 06:35
  • If the group is nilpotent then surely you have such a $g$, But a dihedral group of order $4\times 17$ has the rotation of order $2\times 17$ that satisfies all the properties you have mentioned but it is not nilpotent. – baharampuri Aug 06 '15 at 06:52