Use the correspondence theorem to show that
the only subgroups of are and 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 [since <m> is the only proper subgroup of <r>, so <r> cannot be a subgroup of ] , and [I understand this bit] .
f 1 < s then , 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