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?

  • 1
    $\begingroup$ Consider an element of infinite order $g$ and then $<g>$. Since the intersection of all subgroups is nontrivial, every subgroup of $G$ must contain at least one element of $<g>$, which is a power of $g$, thus having infinite order. $\endgroup$ – Marra Mar 30 '12 at 2:03

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

  • $\begingroup$ Well, it must be $\{1\}$ because all subgroups of $<g>$ are cyclic. Since the intersection is a group, I can take an arbitrary power of $g$ and show that it doesn't belong to the intersection. $\endgroup$ – Marra Mar 30 '12 at 2:16
  • $\begingroup$ It is $\{1\}$ but «because all subgroups of $(g)$ are cyclic» is not a proof of that. I cannot see in what way the second sentence in your comment is related to neither your question not my answer :) $\endgroup$ – Mariano Suárez-Álvarez Mar 30 '12 at 2:18
  • $\begingroup$ Well, you can suppose that there's a $k$ integer such that $g^k$ is in the intersection of all subgroups of G. If it is in the intersection, then it must belong to <g^l> where $k$ and $l$ are relatively primes. But that's not possible; and then, since the intersection is a group, it must be the trivial group $\{ 1\}$, right? With this I can conclude the exercise. $\endgroup$ – Marra Mar 30 '12 at 2:24
  • $\begingroup$ I'm trying to proof that this intersection of all subgroups of $<g>$ is $\{1\}$. $\endgroup$ – Marra Mar 30 '12 at 2:25
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    $\begingroup$ @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?...... $\endgroup$ – user1729 Mar 30 '12 at 8:50

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 !


Suppose there exist an element $g \in G$ of infinite order.

The infinite cyclic groups $(g^n)$ generated by $g^n$ for all $n \in \mathbb N$ are subgroups of $G$.

Any common element of the subgroups $(g^n)$ must have order a multiple of every $n \in \mathbb N$.

$0 \in \mathbb N$ is the only multiple of every $n\in \mathbb N$. (To see this, $0 = n \cdot 0$ for all $n \in \mathbb N$ and any $m \ne 0 \in \mathbb N$ is not a multiple of $m+1 \in \mathbb N$.)

Therefore $g^0 = 1$ is the only element in the intersection of the subgroups $(g^n)$ hence the only element in the intersection of all subgroups of $G$.


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