Hi

**pooma**.

I recently proved this in one of

**NonCommAlg**’s Fun algebra problems.

. Let $\displaystyle x$ be a nontrivial element in the intersection of all the nontrivial subgroups. If $\displaystyle x^2=e$ then $\displaystyle x$ has finite order. Otherwise $\displaystyle x \in \langle x^2 \rangle$ and so $\displaystyle x=x^{2n}$ for some $\displaystyle n\in\mathbb Z$ and so $\displaystyle x^{2n-1}=e$ and so again $\displaystyle x$ has finite order.

Suppose $\displaystyle x$ has order $\displaystyle p.$ Now let $\displaystyle g$ be any element in the group $\displaystyle G.$ If $\displaystyle g=e$ then $\displaystyle g$ has finite order. Otherwise $\displaystyle x$ is in the subgroup generated by $\displaystyle g$ and so $\displaystyle x=g^m$ for some integer $\displaystyle m.$ It follows that $\displaystyle g^{mp}=x^p=e$ so that $\displaystyle g$ has finite order.

NB: You can go further that the order of $\displaystyle x$ must be a prime and that every nonidentity element of $\displaystyle G$ has order a power $\displaystyle p.$