# Intersection of two subgroups

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

• In fact, it's false. – Rafael Gonzalez Lopez Aug 7 '18 at 17:44
• 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 $\langle(1,1)\rangle$. – Henning Makholm Aug 7 '18 at 17:54
• Of course you can say it if you want to, but it would be false. – Derek Holt Aug 7 '18 at 19:29
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No we can't: consider $\langle 1\rangle=\langle 2\rangle$ in $\Bbb Z/3\Bbb Z$.

• Thank you Arnaud for your answer. – Amine El Bouzidi Aug 9 '18 at 12:24

No, let $H=\langle 2 \rangle$ and $K=\langle 3 \rangle$ as subgroups of $\mathbb{Z}$. Then $H \cap K = \langle 6 \rangle$.

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

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.