Intersection of two subgroups

Question

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

Answer 1

No we can't: consider $\langle 1\rangle=\langle 2\rangle$ in $\Bbb Z/3\Bbb Z$.

Answer 2

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

Answer 3

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.