To prove that every element of $G$ has finite order. [duplicate]

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

Can i get some hints to get started?

Thanks.

EDIT

As per suggestion of user Slade, i have proceeded according to this

If $g$ has infinite order, then the cyclic group that $g$ generates will have elements of the form $\{g,g^{2},g^{3},......)$.

The claim is that no two elements are same, i.e. there are no repetitions. To show this let us assume that there are repetitions, say $a^{i}=a^{j}$, for some $i,j$. Then $a^{i-j}=e$ and this contradiction of $a$ having infinite order.

What should i do next?

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Suppose that $G$ has an element $g$ of infinite order. What are the subgroups of the cyclic group that $g$ generates?

• Coudnot quite follow you – Taylor Ted Mar 24 '15 at 19:40
• @K.Dutta Which part? – Slade Mar 25 '15 at 1:04
• If g has infinite order ,then cyclic group that g generates will have elements of form {g,$g^{2}$,$g^{3}$,......) .Claim is that no two elements are same .ie there are no repetitions .To show this let us assume that there are repetitions say $a^{i}=a^{j}$ for some i,j .Then $a^{i-j}=e$ this contradicts of us having infinite order of a .What should i do next – Taylor Ted Mar 25 '15 at 3:14
• @K.Dutta Don't forget the identity element, and don't forget inverses. Next, you should find all nontrivial subgroups of an infinite cyclic group, and calculate their intersection. – Slade Mar 25 '15 at 4:41
• @K.Dutta Incorrect; $1$ is not a multiple of $2$, for example. There is only one integer that is a multiple of all other integers. – Slade Mar 25 '15 at 6:06

Observe that in infinite cyclic group that intersection is just the identity. Conclude that your group has not a cyclic infinite subgroup. This evidently equivalent to being every element of finite order

Hint by absurd, take an element $a$ without a finite order and consider $I$ is the intersection of subgroups of $<a>$ different from $\{e\}$. prove that : $$I=\{e\}$$ (if $x=a^m \in I$ then $x\in <a^{m+1}>\cap <a^{m+1}>\cdots$)

A direct approach would be to take some $g\neq e$ which is contained in all nontrivial subgroups (which exists by hypothesis), and note that for any $x\neq e$, the set $\{e,x,x^2,x^3,\ldots\}$ forms a nontrivial subgroup and hence contains $g$. Thus, for any $x\neq e$ we have $x^n=g$ for some $n$. We can first use this to show that, letting $x=g^2$ we have either that $g^{2}=e$ or $g^{2n}=g$ for some $n$ - either of which means $g$ has finite order - that is, it satisfies $g^m=e$ for some $m>0$. However, since any $x$ other than the identity has a finite $n$ such that $x^n=g$ we have $x^{nm}=g^m=e$, hence every element has finite order.