Group Automorphism/Examples/Constant Product on Real Numbers

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Examples of Group Automorphisms

Let $\struct {\R, +}$ denote the real numbers under addition.

Let $\alpha \in \R$ be a real number.


Let $f: \R \to \R$ be the real function defined as:

$\forall x \in \R: \map f x = \alpha x$

Then $f$ is a (group) automorphism if and only if $\alpha \ne 0$.


Proof

Let $\alpha \in \R$.

We have that:

\(\ds \forall x, y \in \R: \, \) \(\ds \map f {x + y}\) \(=\) \(\ds \alpha \paren {x + y}\) Definition of $f$
\(\ds \) \(=\) \(\ds \alpha x + \alpha y\) Real Multiplication Distributes over Addition
\(\ds \) \(=\) \(\ds \map f x + \map f y\) Definition of $f$

Thus $f$ is a (group) homomorphism for all $\alpha \in \R$.


Sufficient Condition

Let $\alpha \in \R$ such that $\alpha \ne 0$.

Then $\dfrac 1 \alpha$ is the multiplicative inverse of $\alpha$ in the field of real numbers $\R$.

We have that:

\(\ds \forall x, y \in \R: \, \) \(\ds \map f x\) \(=\) \(\ds \map f y\)
\(\ds \leadsto \ \ \) \(\ds \alpha x\) \(=\) \(\ds \alpha y\) Definition of $f$
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds y\) multiplying both sides by $\dfrac 1 \alpha$

and so $f$ is injective by definition.

Then we have:

\(\ds \forall x \in \R: \, \) \(\ds x\) \(=\) \(\ds \alpha y\)
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 \alpha x\) \(=\) \(\ds y\) Definition of $\dfrac 1 \alpha$
\(\ds \leadsto \ \ \) \(\ds \forall x \in \R: \exists y \in \R: \, \) \(\ds x\) \(=\) \(\ds \map f y\) Definition of $f$

This demonstrates that $f$ is surjective.

So by definition $f$ is a bijection.

Thus $f$ is an (group) isomorphism from $\R$ to $\R$ and so a (group) automorphism.

$\Box$


Necessary Condition

Let $f$ be a (group) automorphism.

Aiming for a contradiction, suppose $\alpha = 0$.

Then we have:

$\forall x, y \in \R: \map f x = \map f y = 0$

and so $f$ is not injective.

Hence $f$ is not a bijection.

Therefore $f$ is not an automorphism.

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

$\blacksquare$


Sources

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