I need help to prove this result:
"Let $G$ be a group such that the intersection of all its subgroups other than $\{1\}$ is a subgroup different from $\{1\}$. Then all its elements have a finite order".
I know I must think of an element $g$ of infinite order. That will imply that every subgroup has infinite order (because it will contain an element like $g^k$). After this, I don't know which step I can take. Can someone help?