Group Example: 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 $1 - x$.

Invalid Example

• The first thing we can do is find the identity.

From Identities all Self-Inverse, the identity must satisfy $x = \left({1-x}\right)$, which means the identity must be $1/2$.

• Now we investigate an equation $x \circ \left({1 - x}\right) = 1/2$.

Symmetry of the set about the element $x = 1/2$ suggests we might want to average. So try it out:

Let $\ast$ be defined as $x \ast y = \left({x + y}\right) / 2$.

• Next, note that $\left({x + \left({1 - x}\right)}\right) / 2 = 1/2$ as required.

• It is straightforward to check that $\left({S, \ast}\right)$ is closed.

• Note, however, that:

• $x \ast \left({y \ast z}\right) = \left({2 x + y + z}\right) / 4$
• $\left({x \ast y}\right) \ast z = \left({x + y + 2 z}\right) / 4$

Thus $x \ast \left({y \ast z}\right) \not \equiv \left({x \ast y}\right) \ast z$ and $\ast$ is not associative.

So $\ast$ can not be the operation we are after.

Sources

• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 26 \mu$