# Intersections of all subgroups is a nontrivial subgroup, so every element has finite order.

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?

• 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. – 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?

• 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. – Marra Mar 30 '12 at 2:16
• 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 :) – Mariano Suárez-Álvarez Mar 30 '12 at 2:18
• 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. – Marra Mar 30 '12 at 2:24
• I'm trying to proof that this intersection of all subgroups of $<g>$ is $\{1\}$. – Marra Mar 30 '12 at 2:25
• @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?...... – 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 $$G$$.