# Z and Q are not isomorphic

## 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].
[tex] 0a = 0, 1a = \frac{l}{m} , 2a = \fract{2l}{m}[\tex].

Because the rationals are dense there is a [tex] b \in Q s.t. \frac{l}{m} < b < \frac{2l}{m} [\tex]

We must show that [tex] b = ka = \frac{kl}{m}, thus \frac{l}{m} < \frac{kl}{m} < \frac{2l}{m} [\tex].

Now I don't know what to do. The above is not a contradiction. Any ideas?

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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.

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.

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.

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.