# Additive Group of Real Numbers is Not Isomorphic to Multiplicative Group of Real Numbers

## Theorem

Let $\struct {\R, +}$ denote the additive group of real numbers.

Let $\struct {\R_{\ne 0}, \times}$ denote the multiplicative group of real numbers.

Then $\struct {\R, +}$ is not isomorphic to $\struct {\R_{\ne 0}, \times}$.

## Proof 1

Consider the element $-1 \in \struct {\R_{\ne 0}, \times}$.

We have that:

- $-1 \times -1 = 1$

From Real Multiplication Identity is One it follows that $-1$ is of order $2$ in $\struct {\R_{\ne 0}, \times}$.

From Group Isomorphism Preserves Order of Group Element, it is sufficient to demonstrate that there exists no element of order $2$ in $\struct {\R, +}$.

From Real Addition Identity is Zero, that means finding $x$ such that $x + x = 0$.

But:

- $x + x = 0 \implies x = 0$

and as $0$ is the identity of $\struct {\R, +}$, there is indeed no element of order $2$ in $\struct {\R, +}$.

The result follows from Group Isomorphism Preserves Order of Group Element.

$\blacksquare$

## Proof 2

From Real Numbers form Field, $\struct {\R, +, \times}$ forms a field.

The result then follows as an example of Additive Group and Multiplicative Group of Field are not Isomorphic.

$\blacksquare$

## Proof 3

There are two element of $\struct {\R_{\ne 0}, \times}$ which are self-inverse, that is, $-1$ and $1$.

However, there is only one element of $\struct {\R, +}$ which is self-inverse, that is, $0$.

Aiming for a contradiction, suppose there exists an isomorphism $f: \struct {\R_{\ne 0}, \times} \to \struct {\R, +}$.

Then:

\(\ds 0\) | \(=\) | \(\ds \map f 1\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \map f {-1 \times -1}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \map f {-1} + \map f {-1}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 2 \map f {-1}\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds \map f {-1}\) | \(=\) | \(\ds 0\) | |||||||||||

\(\ds \) | \(=\) | \(\ds \map f 1\) |

So we have demonstrated that $f$ is not an injection.

Hence $f$ is not a bijection and so not an isomorphism.

It follows from Proof by Contradiction that there can be no such isomorphism.

Hence the result.

$\blacksquare$

## Sources

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): Chapter $7$: Homomorphisms: Exercise $5$