# Proving the order of elements of a group is finite

Let $G$ be a group such that 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.

Need hints to prove it or disprove it.

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## 2 Answers

Suppose $a\in G$ is an element of non-finite order. Then the intersection of $A_d=\{a^{nd}, n\in Z\}$ is $e$, contradicting the premise.

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Let $S$ be the smallest nontrivial subgroup of $G$, which exists by hypothesis, and let $a$ be any nonidentity element of $S$.

Show that $S=\langle a\rangle$, that $a$ has finite order, by contradiction, and that for any $x\in G$ we have $a\in\langle x\rangle$.

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"...the smallest non-trivial group..." +1 – DonAntonio May 3 '14 at 1:12