# 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 .

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 $\{e\}$.
• Wait - I'm confused again. We want to show that the implication "not all elements have finite order $\Rightarrow$ intersection is trivial" does not hold. But you have just given an example where it does hold. – user1729 Nov 3 '14 at 16:21
• No, you want to show that that implication does hold. The statement you want to prove is: "If the intersection of all nontrivial subgroups is nontrivial, then every element of $G$ has finite order." The contrapositive of this statement, which is logically equivalent to the original statement, is: "If not every element of $G$ has finite order, then the intersection of all nontrivial subgroups is trivial." Proving that statement is equivalent to proving the original one. – Greg Martin Nov 3 '14 at 17:13