If a group is cyclic (and non-infinite), and there is a subgroup that contains more than just the identity, shouldn't the subgroup contain the same elements as the group? (I know the subgroup is also cyclic).
I.e. since a cyclic group is of the form $\{e,x,x^2,\ldots,x^n\}$, shouldn't any subgroup $H$ be of the form $\{e,x,x^2,\ldots,x^r\}$, where $r\leq n$?
(I was reading a proof where they treated the elements of the subgroup as distinct, and I don't see why that would be.)
Best Answer
Here's a simple example: take $\mathbb{Z}/4\mathbb{Z}$, and the subgroup $\{0,2\}$. This subgroup is nontrival and yet is not the whole of $\mathbb{Z}/4\mathbb{Z}$.
There are simple abelian groups, and those are of the form $\mathbb{Z}/p\mathbb{Z}$ when $p$ is prime. The only subgroups of these are $\{0\}$ and itself. But in general, an abelian group of composite order will never be simple.