0
$\begingroup$

(Q*,•) and (R*,•) is isomorphic.

This is false.

Why this problem false??

Would you ask to me counter-example?

$\endgroup$
1
  • $\begingroup$ Forget the operation, they do not even have the same number of elements. $\endgroup$
    – almagest
    May 21 '16 at 8:24
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.