Originally Posted by

**Swlabr** I do not believe this is a proof. You assume $\displaystyle \phi(x) = x\phi(1)$, but this is not necessarily so. Essentially, you have produced a generic proof which says that a proper subgroup of a group cannot be isomorphic to the group itself. This is not so. An obvious counter-example is the infinite cartesian product of the integers. $\displaystyle \mathbb{Z} \times \mathbb{Z} \times \ldots \cong <1> \times \mathbb{Z} \times \ldots$.

However, the result you used is true, but needs to be proven.

Notice that for $\displaystyle x \in \mathbb{Z}$, $\displaystyle x\phi = x$. So the result now just needs to be shown for the denominators.

$\displaystyle x\phi = 1/p \Rightarrow (px)\phi = 1 \Rightarrow x = 1/p$ as the kernel is trivial.

Thus, $\displaystyle \frac{p}{q} \phi = \frac{p\phi}{q\phi} = \frac{p}{q}$ as required.

Or something along those lines...