Group/Examples/inv x = 1 - x

Theorem
Let $S = \left\{{x \in \R: 0 < x < 1}\right\}$.

Then an operation $\circ$ can be found such that $\left({S, \circ}\right)$ is a group such that the inverse of $x \in S$ is $1 - x$.

Proof
Define $f: \left({0 \,.\,.\, 1}\right) \to \R$ by:


 * $f \left({x}\right) := \displaystyle \log \left({ \frac {1 - x}{x} }\right)$

Let us show that $f$ is a bijection by constructing an inverse mapping $g: \R \to \left({0 \,.\,.\, 1}\right)$:


 * $g \left({z}\right) := \dfrac 1 {1 + \exp z}$

Lemma 2
Thus $f$ is a bijection.

Let $\left({\R, +}\right)$ be the additive group on $\R$.

Now define $\circ := +_f$ to be the operation induced on $\left({0 \,.\,.\, 1}\right)$ by $f$ and $+$, i.e.:


 * $x \circ y := f^{-1} \left({f \left({x}\right) + f \left({y}\right)}\right)$

Let us determine the behaviour of $\circ$ more explicitly:

We see that $\circ$ is commutative.

Let $x = \dfrac 1 2$, $\dfrac {1 - x} x = 1$ so that:

so that $\dfrac 1 2$ is the identity element for $\circ$.

Furthermore, putting $y = 1 - x$, the following obtains:

establishing $1 - x$ to be the inverse of $x$, as desired.

That $\circ$ in fact determines a group on $\left({0 \,.\,.\, 1}\right)$ follows from Pullback of Group is Group.