Use the correspondence theorem to show that

the only subgroups of $\displaystyle \mathbb{Z}$ are$\displaystyle \mathbb{Z}$ and $\displaystyle <m> \iff $m is prime

I understand how to show this in the <= direction. I have a proof here for the => direction which i don't understand.

Conversely, suppose that the only subgroups of Z which contain <m> are <m> itself and Z.

If m = rs for natural numbers r,s with 1 < r [I am guessing here we suppse m is composite] then $\displaystyle <m> \le <r> \not\le \mathbb{Z} $ [since <m> is the only proper subgroup of <r>, so <r> cannot be a subgroup of $\displaystyle \mathbb{Z}$] , and $\displaystyle |<r> / <m> | = | <r> / <rs> | = s $[I understand this bit] .

f 1 < s then $\displaystyle <r> \not= <m>$, hence s = 1. I understand that s = 1 would prove that m must be prime, but where does this part come from? Where in the hypothesis does it say that <r>= <m> ? Isn't the whole point that they are not equal?

Any help with this would be appreciated, thank you