# How to prove isomorphic? [duplicate]

(Q*,•) and (R*,•) is isomorphic.

This is false.

Why this problem false??

Would you ask to me counter-example?

• Forget the operation, they do not even have the same number of elements. May 21 '16 at 8:24

From your notation I guess that $\mathbb{Q}^*$ are the invertible elements in $\mathbb{Q}$ and you want to show that $\mathbb{Q}^*$ and $\mathbb{R}^*$ are not isomorphic as groups? Well, in that case, notice that $\mathbb{Q}^*=\mathbb{Q}_0$ and $\mathbb{R}^*=\mathbb{R}_0$, the first set is countable whereas the second is not. Therefore they cannot be isomorphic as that would yield that they are bijective.
There are several ways to prove this. An isomorphism is a function $\Phi$ that is a bijective homomorphism. By Cantor's diagonal argument, there is no bijection from $\mathbb{R}$ to $\mathbb{Q}$.
Also quoting @Louis in his answer here, that can also be applied to the multiplicative groups $\mathbb{R}^*$ and $\mathbb{Q}^*$
Let $\Phi: \mathbb{Q} \rightarrow \mathbb{R}$ homomorphism of additive groups. Then $\Phi$ is already determined by $\Phi(1)$ (as $\Phi(\frac{a}{b})= \Phi(\frac{1}{b})+ ... + \Phi(\frac{1}{b})$ ($a$ summands) and $\Phi(1)= \Phi(\frac{1}{b}) + ... + \Phi(\frac{1}{b})$, ($b$ summands))
Now, say $\sqrt{2} \cdot \Phi(1)$ doesn't have a preimage.