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Let $G$ be a group and $H$, $K$ two cyclic subgroups of $G$ of the same order such that:$$ H=\langle x\rangle,~~K=\langle y\rangle~~ \text{and}~~x\neq y.$$ Can we said that $H\cap K$ is trivial? Thank you

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  • $\begingroup$ In fact, it's false. $\endgroup$ – Rafael Gonzalez Lopez Aug 7 '18 at 17:44
  • $\begingroup$ You can't even say that $H$ and $K$ are either equal or have trivial intersection: In $\mathbb Z_{p^2}\times\mathbb Z_p$ consider $\langle (1,0)\rangle$ versus $\langle(1,1)\rangle$. $\endgroup$ – Henning Makholm Aug 7 '18 at 17:54
  • $\begingroup$ Of course you can say it if you want to, but it would be false. $\endgroup$ – Derek Holt Aug 7 '18 at 19:29
  • $\begingroup$ Since you are new to here, just as a reminder for you: After you read the answers given, if there exists an answer that you accept it, click the tick beside the answer. If there does not exist, comment below the answer to state that which part of the answer you need more elaboration. $\endgroup$ – Alan Wang Aug 8 '18 at 13:58
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No we can't: consider $\langle 1\rangle=\langle 2\rangle$ in $\Bbb Z/3\Bbb Z$.

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  • $\begingroup$ Thank you Arnaud for your answer. $\endgroup$ – Amine El Bouzidi Aug 9 '18 at 12:24
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No, let $H=\langle 2 \rangle$ and $K=\langle 3 \rangle$ as subgroups of $\mathbb{Z}$. Then $H \cap K = \langle 6 \rangle $.

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  • $\begingroup$ Thank you Nicky for your answer $\endgroup$ – Amine El Bouzidi Aug 9 '18 at 12:24
  • $\begingroup$ No problem Amine, hope you learned something from all the answers. $\endgroup$ – Nicky Hekster Aug 9 '18 at 12:59
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Another example:

$\langle 1 \rangle=\langle -1 \rangle$ in $\Bbb{Z}$. So clearly $1\neq -1$ but $\langle 1 \rangle\cap\langle -1 \rangle\neq e$.

By the way, $H$ and $K$ have trivial intersection if $|H|$ and $|K|$ are coprime. This can be proven using Lagrange's Theorem.

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