$H$ and $K$ are subgroups of a group $G$ with $H$ doesn't equal to $K$ and $|H| = |K| = 11$. Prove that $H \cap K = \{e\}$
Have no idea about this question. Dont know if there is some theorem about it.
$H$ and $K$ are subgroups of a group $G$ with $H$ doesn't equal to $K$ and $|H| = |K| = 11$. Prove that $H \cap K = \{e\}$
Have no idea about this question. Dont know if there is some theorem about it.
Hint: $H\cap K$ is a subgroup of $H$. What can you say about the subgroups of a group of prime order?