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

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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 Order of Isomorphic Image 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 Order of Isomorphic Image 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

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