Showing that if the intersection of all subgroups other than $\langle e \rangle$ is not $\langle e \rangle$, then every element is of finite order

Question

This is problem 2 on page 46 of I. N. Herstein's Topics in Algebra.

Let $G$ be a group such that the intersection of all its subgroups which are different from $\langle e\rangle$ is a subgroup different from $\langle e\rangle$. Prove that every element in $G$ has finite order.

Answer 1

Hint: Assume that not every element has finite order; i.e., that there is an element $a$ of infinite order. What can you say about the intersection of the non-trivial subgroups of the cyclic subgroup $\langle a\rangle$.

Answer 2

A simpler answer than my previous one:

Consider

$$H=\bigcap_{g\in G} \langle g\rangle$$

The intersection of cyclic groups is cyclic, so either

(1) $H=\langle h\rangle$ with $\text{ord}_G(h)=n<\infty$

(2) $H=\langle h\rangle\cong \Bbb Z$ whence $\langle h^2\rangle$ is a proper subgroup, contradicting that $H$ was the intersection of all cyclic subgroups of $G$.

So the intersection has finite order. Since $n\ne 1$ by assumption, every $g\in G$ has that $\langle h\rangle \le \langle g\rangle$, but every non-identity element of an infinite cyclic group has infinite order, and clearly $h\in\langle g\rangle$ does not, so $|\langle g\rangle|<\infty$

Answer 3

Let $a \neq e$ be an element from that intersection. If $a^2=e$, then $a$ has order $2$. Otherwise, $a \in \langle a^2 \rangle$, so $a=a^{2k}$, for some integer $k$, which implies $a^{2k-1}=e$, so $a$ has finite order. Either way $a$ has finite order.

Now let $g \in G$ be arbitrary. Since $a \in \langle g \rangle$, it follows that $g^{\ell}=a$ for some integer $\ell$, so $g^{\ell \cdot \mathrm{ord}(a)}=e$, so the order of $g$ is finite.