TITLE: Intersections of all subgroups is a nontrivial subgroup, so every element has finite order.
QUESTION [8 upvotes]: 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?

REPLY [4 votes]: 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 $G$.