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I need help to prove this result:

"Let $G$ be a group such that the intersection of all its subgroups other than $\{1\}$ is a subgroup different from $\{1\}$. Then all its elements have a finite order".

I know I must think of an element $g$ of infinite order. That will imply that every subgroup has infinite order (because it will contain an element like $g^k$). After this, I don't know which step I can take. Can someone help?

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    Consider an element of infinite order $g$ and then $$. Since the intersection of all subgroups is nontrivial, every subgroup of $G$ must contain at least one element of $$, which is a power of $g$, thus having infinite order.2012-03-30

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If there is an element $g$ of infinite order, consider the intersection of all subgroups of $(g)$. What is it?

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    @GustavoMarra: To prove this, for an arbitrary $n\in\mathbb{Z}$ you need to find a subgroup which doesn't contain $n$. So...pick a number large than $n$, $|n|+1$ say...and then...can you think of a subgroup which contains $|n|+1$ but not $n$? If you are struggling, you could turn this on its head - instead of trying to find a subgroup which does not contain $n$, think about what the homomorphic images of $\mathbb{Z}$ are. They are modulo arithmetic, right? So, can you pick an integer $m$ such that $n\not\equiv 0\text{ mod }m$? So, what does this *mean*?......2012-03-30
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For otherwise there is an infinite cyclic subgroup $(a)$ of $G.$

Due to the non-triviality of the said subgroups, $a^k$ is in the said intersection (for some nonzero integer $k$).

However $(a^{2k})$ is a subgroup of $G$ without having $a^k$ in it !