# Finite number of subgroups $\Rightarrow$ finite group

I'm trying to prove that any group $G$ of infinite order has an infinite number of subgroups.

I think that if the group has an element of infinite order, then it's easy because I can take the groups generated by the powers of this element.

What if it doesn't? Every element generates a cyclic subgroup. Every element belongs to at least one cyclic subgroup (that generated by itself). So the group is the union of its cyclic subgroups. If all these are finite, we would have to have an infinite collection of subgroups anyway.

Is that correct?

• Yup, that's correct :) – Zev Chonoles Feb 21 '11 at 0:09
• +1 for showing your work. – Arturo Magidin Feb 21 '11 at 0:13

• Generally, many different elements will give the same subgroup; e.g., $g$ and $g^{-1}$ give the same cyclic subgroup. So you must argue somehow that you still have infinitely many, even after taking into account the repeats. – Arturo Magidin Feb 21 '11 at 1:51