1. ## Correspondence Theorem

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

2. also surely <r> is a subgroup of the integers?

3. Originally Posted by slevvio
also surely <r> is a subgroup of the integers?

The original question is obviously false. What it should be, imo, is: $\displaystyle G$ is a group whose only sbgps. are $\displaystyle G,\,\{1\}\Longleftrightarrow |G|=p=$ a prime.

Tonio

4. Originally Posted by slevvio
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
Originally Posted by tonio
The original question is obviously false. What it should be, imo, is: $\displaystyle G$ is a group whose only sbgps. are $\displaystyle G,\,\{1\}\Longleftrightarrow |G|=p=$ a prime.

Tonio
Could this possibly be saying that the only non-trivial ideals $\displaystyle I$ of $\displaystyle \left(\mathbb{Z},\cdot,+\right)$ are $\displaystyle I=\langle p\rangle$ with $\displaystyle p$ prime?

5. Originally Posted by Drexel28
Could this possibly be saying that the only non-trivial ideals $\displaystyle I$ of $\displaystyle \left(\mathbb{Z},\cdot,+\right)$ are $\displaystyle I=\langle p\rangle$ with $\displaystyle p$ prime?
No. nZ is always an ideal no matter n is prime or not. pZ just makes it maximal. But I believe that's unrelated to the question.

6. Originally Posted by FancyMouse
No. nZ is always an ideal no matter n is prime or not. pZ just makes it maximal. But I believe that's unrelated to the question.
I meant to say maximal.

7. Originally Posted by Drexel28
I meant to say maximal.

This, I think, is already assuming too much: the OP clearly wrote subgroups, nothing about rings, ideals or maximal.

Let us give the OP a chance to clear out his own question, shall we?

Tonio