If intersection of all the non-singleton subgroups of a group is not single ton , then is every element of the group is of finite order ? Let $G$ be a group such that the intersection of all its subgroups which are different from $\{e\}$ is a subgroup different from $\{e\}$ , then is it true  that every element in $G$ has finite order ? I was trying to argue by contradiction , say $x \in G $ has infinite order , but actually couldn't go anywhere , Please help . 
 A: I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
This problem is Problem 2 on p.46.
I solved this problem as follows:

Let $H$ be the intersection of all subgroups of $G$ which are different from $\{e\}$.
Since $H\neq \{e\}$ by the assumption, there exists $a\in H$ such that $a\neq e$.
$(a):=\{x\in G\mid x=a^k, k\in\mathbb{Z}\}$ is a subgroup of $G$ which is
different from $\{e\}$.
So, $H\subset (a)$ by the definition of $H$.
Since $a$ is an element of $H$, $(a)\subset H$.
Therefore, $H=(a)$.
Assume that $o(a)=+\infty$, where $o(a)$ is the order of $a$.
Then, $o(a^2)=+\infty$.
$(a^2):=\{x\in G\mid x=a^{2k}, k\in\mathbb{Z}\}$ is a subgroup of $G$ which is
different from $\{e\}$.
So, $H=(a)\subset (a^2)$ by the definition of $H$.
Since $a\in H=(a)$, $a\in (a^2)$.
So, we can write $a=a^{2k}$ for some $k\in\mathbb{Z}$.
This means $o(a)$ is finite.
This is a contradiction.
So, $o(a)$ must be finite.
Let $b$ be an arbitrary element of $G$.
If $b=e$, then $o(b)$ is finite.
Suppose that $b\neq e$.
Assume that $o(b)=+\infty$.
$(b):=\{x\in G\mid x=b^k, k\in\mathbb{Z}\}$ is a subgroup of $G$ which is
different from $\{e\}$.
So, $H=(a)\subset (b)$ by the definition of $H$.
Since $a\in H=(a)$, $a\in (b)$.
So, we can write $a=b^k$ for some $k\in\mathbb{Z}$.
Since $o(b)=+\infty$, $o(a)=+\infty$ must hold.
This is a contradiction.
So, $o(b)$ must be finite.

A: Suppose $x \in G$ has infinite order. Then $\langle x \rangle \cong \mathbb{Z}$. The intersection of all the nontrivial subgroups of $\mathbb{Z}$ is $\{0\}$ ($0$ is the only number divisible by all the other numbers), thus the intersection of all the nontrivial subgroups of $G$ is reduced to $\{e\}$.
