Histo4432.

Subgroups of are of the form for some non-negative integer . Obviously, if and are subgroups of then is a subgroup of

Conversely, suppose is a subgroup of can be or Otherwise, there is some with Then Hence there is a positive integer such that for some integer and a positive integer such that for some integer

[Edit: The rest of the proof I originally posted is wrong. It led to the following conclusion (highlighted in red) which, asThePerfectHackerpoints out, is clearly incorrect. I am editing to remove my false proof.]

The subgroups of are therefore the direct products of subgroups of