# Prove every element of $G$ has 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 $\{e\}$. Prove that every element of $G$ has finite order.

Assume that $a$ has infinite order in $G$. Then let $y\neq e$ belong to the intersection of all subgroups of $G$ different from $\{e\}$. Then $y=a^n$ and $y$ will also belong to subgroup generted by $a^2$ hence $y=a^{2m}$ for some integers $m$. This contradicts that $a$ has infinte order.

Is this correct?

• How do you find that $y=a^n=a^{2m}$? – YTS Jan 8 '16 at 15:56
• $y$ will also belong to subgroup generted by $a^2$ – Eklavya Jan 8 '16 at 16:01

## 2 Answers

To express your idea differently:

If $G$ has an element $a$ of infinite order, then the intersection of all nontrivial subgroups is trivial.

Indeed, this is true for $\mathbb Z$, which is isomorphic to $\langle a \rangle$.

Therefore, if the intersection of all nontrivial subgroups is nontrivial, then $G$ cannot have an element of infinite order and so all elements must have finite order.

Not completely true, but it is the idea precise that $y$ is in the group generated by $a$ so $y=a^n$. $y$ is also in the group generated by $a^{2n}$ so $y={a^{2n}}^m=a^{2nm}$, so $a^{2nm-n}=1$ contradiction

• because it is in the intersection of all subgroup of $G$ and the subgroup generated by $a^{2n}$ is a subgroup of $G$. – Tsemo Aristide Jan 8 '16 at 16:20