Why are the integers and rationals not isomorphic under addition?

[icantadd]

Homework Statement

Prove that the integers (under addition) are not isomorphic to the rationals (under addition).

Homework Equations

Two groups are isomorphic if there is an isomorphism between them.

If there is an isomorphism from G to H, f : G --> H, then G is cyclic iff H is cyclic.

A group G is cyclic if $$\{ x^n | n \in \mathbb{Z} \} = G, for some x \in G$$ .

The Attempt at a Solution

The integers are generated by $$<1>$$. We can show that Z and Q are not isomorphic if we show that the rationals cannot be generated. Thus assume they are. Then there is an a such that

[tex]<a> = Q [\tex].<br /> [tex]0a = 0, 1a = \frac{l}{m} , 2a = \fract{2l}{m}[\tex].<br /> <br /> Because the rationals are dense there is a [tex]b \in Q s.t. \frac{l}{m} < b < \frac{2l}{m} [\tex]<br /> <br /> We must show that $$b = ka = \frac{kl}{m}, thus \frac{l}{m} < \frac{kl}{m} < \frac{2l}{m} [\tex]. <br /> <br /> Now I don't know what to do. The above is not a contradiction. Any ideas?$$[/tex][/tex][/tex]

 

[samkolb]

Suppose Q is cyclic.

Let p/q be a generator.

Can you find a rational number which is not an integer multiple of p/q?
Think of a rational involving p and q somehow.

 

[icantadd]

Suppose Q is cyclic.

Let p/q be a generator.

Can you find a rational number which is not an integer multiple of p/q?
Think of a rational involving p and q somehow.

sure, continuing

Observe p/2q

Then,

p/2q = p/q where p,q are not equal to zero. (if p is zero then the set is finite).

From the above conclude that
pq=2pq, therefore 1=2.
done.

 

[Dick]

No, no. <p/q> is the set of number k*p/q where k is an integer. Set kp/q=p/(2q) and derive a different contradiction.

 

[icantadd]

No, no. <p/q> is the set of number k*p/q where k is an integer. Set kp/q=p/(2q) and derive a different contradiction.

Thanks! Yeah, I don't know what I was thinking there. I went and got some food, came back and hit myself in the head on that one.