Exponential on Real Numbers is Group Isomorphism

Theorem

Let $\left({\R, +}\right)$ be the Additive Group of Real Numbers.

Let $\left({\R_{> 0}, \times}\right)$ be the Multiplicative Group of Positive Real Numbers.

Let $\exp: \left({\R, +}\right) \to \left({\R_{> 0}, \times}\right)$ be the mapping:

$x \mapsto \exp \left({x}\right)$

where $\exp$ is the exponential function.

Then $\exp$ is a group isomorphism.

Proof

Proof 1

From Exponent of Sum we have:

$\forall x, y \in \R: \exp \left({x + y}\right) = \exp x \cdot \exp y$

That is, $\exp$ is a group homomorphism.

Then we have that Exponential is Strictly Increasing and Convex.

From Strictly Monotone Function is Bijective, it follows that $\exp$ is a bijection.

So $\exp$ is a bijective group homomorphism, and so a group isomorphism.

$\blacksquare$

Proof 2

First we confirm that:

• The real numbers form a group under addition

• The positive real numbers form a group under multiplication.

We have that $\exp$ is surjective since for all $y \in R_{> 0}$ there exists $x = \ln \left({y}\right) \in R$ which satisfies $\exp \left({x}\right) = y$.

Also, $\exp$ is injective since it is strictly increasing.

Therefore, $\exp$ is a bijection.

Let $x, y \in R$.

Then:

$\exp \left({x + y}\right) = \exp \left({x}\right) \exp \left({y}\right)$

from Exponent of Sum.

So $\exp$ is a homomorphism and a bijection and is therefore an isomorphism.

$\blacksquare$

Sources

• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 62 \alpha$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 47$: Illustration