# Correspondence Theorem

• April 5th 2010, 07:41 AM
slevvio
Correspondence Theorem
Use the correspondence theorem to show that

the only subgroups of $\mathbb{Z}$ are $\mathbb{Z}$ and $ \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 $ \le \not\le \mathbb{Z}$ [since <m> is the only proper subgroup of <r>, so <r> cannot be a subgroup of $\mathbb{Z}$] , and $| / | = | / | = s$[I understand this bit] .

f 1 < s then $ \not= $, 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
• April 5th 2010, 07:53 AM
slevvio
also surely <r> is a subgroup of the integers?
• April 5th 2010, 11:17 AM
tonio
Quote:

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: $G$ is a group whose only sbgps. are $G,\,\{1\}\Longleftrightarrow |G|=p=$ a prime.

Tonio
• April 5th 2010, 01:23 PM
Drexel28
Quote:

Originally Posted by slevvio
Use the correspondence theorem to show that

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

Quote:

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

Tonio

Could this possibly be saying that the only non-trivial ideals $I$ of $\left(\mathbb{Z},\cdot,+\right)$ are $I=\langle p\rangle$ with $p$ prime?
• April 5th 2010, 03:11 PM
FancyMouse
Quote:

Originally Posted by Drexel28
Could this possibly be saying that the only non-trivial ideals $I$ of $\left(\mathbb{Z},\cdot,+\right)$ are $I=\langle p\rangle$ with $p$ 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.
• April 5th 2010, 03:21 PM
Drexel28
Quote:

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.
• April 5th 2010, 06:37 PM
tonio
Quote:

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