# The Additive Group of Rational Numbers and The Multiplicative Group of Positive Rational Numbers are Not Isomorphic

## Problem 510

Let $(\Q, +)$ be the additive group of rational numbers and let $(\Q_{ > 0}, \times)$ be the multiplicative group of positive rational numbers.

Prove that $(\Q, +)$ and $(\Q_{ > 0}, \times)$ are not isomorphic as groups.

## Proof.

Suppose, towards a contradiction, that there is a group isomorphism
$\phi:(\Q, +) \to (\Q_{ > 0}, \times).$

Then since $\phi$ is in particular surjective, there exists $r\in \Q$ such that $\phi(r)=2$.
As $r$ is a rational number, so is $r/2$.

It follows that we have
\begin{align*}
2&=\phi(r)=\phi\left(\, \frac{r}{2}+\frac{r}{2} \,\right)\\
&=\phi\left(\, \frac{r}{2} \,\right)\cdot\phi\left(\, \frac{r}{2} \,\right) &&\text{ because $\phi$ is a homomorphism}\\
&=\phi\left(\, \frac{r}{2} \,\right)^2.
\end{align*}

It yields that
$\phi\left(\, \frac{r}{2} \,\right)=\pm \sqrt{2}.$

However, this is a contradiction since $\phi\left(\, \frac{r}{2} \,\right)$ must be a positive rational number, yet $\sqrt{2}$ is not a rational number.

We conclude that there is no such group isomorphism, and hence the groups $(\Q, +)$ and $(\Q_{ > 0}\times)$ are not isomorphic as groups.

### More from my site

#### You may also like...

##### The Existence of an Element in an Abelian Group of Order the Least Common Multiple of Two Elements

Let $G$ be an abelian group. Let $a$ and $b$ be elements in $G$ of order $m$ and $n$, respectively....

Close