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## 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 [tex] \{ x^n | n \in \mathbb{Z} \} = G, for some x \in G [/tex] .

## The Attempt at a Solution

The integers are generated by [tex] <1> [/tex]. 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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