Group Isomorphism/Examples/Real Power Function

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

Let $\struct {\R_{>0}, \times}$ be the multiplicative group of positive real numbers.

Let $\alpha \in \R_{>0}$ be a strictly positive real number.

Let $f: \struct {\R, +} \to \struct {\R_{> 0}, \times}$ be the mapping:
 * $\forall x \in \R: \map f x = \alpha^x$

where $\alpha^x$ denotes $\alpha$ to the power of $x$.

Then $f$ is a (group) automorphism $\alpha \ne 1$.

Proof
Let $\alpha \in \R_{>0}$.

We have by definition of power to a real number that:


 * $\alpha^x = e^{x \ln \alpha}$

From Natural Logarithm of 1 is 0:
 * $\ln \alpha = 0 \iff \alpha = 1$

First we note that:

Thus $f$ is a (group) homomorphism from $\struct {\R, +}$ to $\struct {\R_{> 0}, \times}$ for all $\alpha \in \R_{>0}$.

Sufficient Condition
Let $\alpha \in \R_{>0}$ such that $\alpha \ne 1$.

We have that:

and so $f$ is injective by definition.

Then we have:

This demonstrates that $f$ is surjective.

So by definition $f$ is a bijection.

Thus $f$ is an (group) isomorphism from $\struct {\R, +}$ to $\struct {\R_{> 0}, \times}$.

Necessary Condition
Let $f$ be a (group) automorphism.

$\alpha = 1$.

Then we have:


 * $\forall x, y \in \R: \map f x = \map f y = 1$

and so $f$ is not injective.

Hence $f$ is not a bijection.

Therefore $f$ is not an isomorphism.

Hence by Proof by Contradiction $f$ is not an automorphism.