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Can we classify the groups for which every maximal cyclic subgroup is of same order and intersection of any two maximal cyclic subgroups is identity?

For example in case of abelian groups $$G=\mathbb{Z}_p\times \mathbb{Z}_p \times \cdots \times \mathbb{Z}_p$$

Can we do same kind of classification for non abelian groups?

Is there any example or some contradictions such that no such non abelian group exists which satisfy both these properties. Any suggestion will be helpful.

Derek Holt
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PARVEEN
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    Welcome Parveen. It's better to make the title short. Instead, you can ask in the body. This could ease others to find and to view the question. Thanks – Mikasa Jan 25 '22 at 06:06
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    Direct product of finite number of groups is cylcic when greatest common divisor of their cradinalities is equal to 1. For maximal cyclic subgroup try to construct subgroups of maximum order such that greatest commom divisor of their cardinalities is 1 – prashant dattatrey Jan 25 '22 at 06:09
  • With non-abelian groups, thing get hairy pretty fast. Even if there are no cyclic subgroups with more than two elements (which automatically ensures your second condition), nothing good can be said. – Ivan Neretin Jan 25 '22 at 06:29
  • @IvanNeretin If there are no cyclic subgroups with more than two elements, then every nonidentity element has order 2, so the group is abelian. So even this is not possible for nonabelian groups. – ckefa Jan 25 '22 at 06:44
  • @ckefa Sorry, I was thinking of infinite groups. In the finite case, of course, what you said is right. – Ivan Neretin Jan 25 '22 at 06:58
  • Any group of prime exponent $p$ has this property, and there are nonabelian examples for $p > 2$. I wonder whether there are any more examples. – Derek Holt Jan 25 '22 at 09:14

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