Group Example: x + y / 1 + x y

Theorem

The set of real numbers $\left\{{x \in \R: -1 < x < 1}\right\}$, under the operation:

$x \circ y = \dfrac {x + y} {1 + x y}$

is a group.

Proof

Let $-1 < x, y, z < 1$.

We check the group axioms in turn:

G1: Associativity

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({x \circ y}\right) \circ z\)	\(=\)	\(\displaystyle \frac {\frac {x + y} {1 + x y} + z} {1 + \frac {x + y} {1 + xy} z}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {x + y + z + x y z} {1 + x y + x z + y z}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x \circ \left({y \circ z}\right)\)	\(=\)	\(\displaystyle \frac {x + \frac {y + z} {1 + y z} } {1 + x \frac {y + z} {1 + y z} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {x + y + z + x y z} {1 + x y + x z + y z}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({x \circ y}\right) \circ z\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

G2: Identity

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x\)	\(=\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \frac {x + y} {1 + x y}\)	\(=\)	\(\displaystyle \frac {0 + y} {1 + 0 y}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac y 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Similarly, putting $y = 0$ we find $x \circ y = x$.

So $0$ is the identity.

G3: Inverses

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x \circ -x\)	\(=\)	\(\displaystyle \frac {x + \left({-x}\right)} {1 + x \left({-x}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Similarly, putting $x = -y$ gives us $\left({-y}\right) \circ y = 0$.

So each $x$ has an inverse $-x$.

G0: Closure

First note that: $-1 < x, y < 1 \implies x y > -1 \implies 1 + x y > 0$.

Next:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle -1\)	\(<\)	\(\displaystyle x, y < 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 0\)	\(<\)	\(\displaystyle \left({1 - x}\right) \left({1 - y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 0\)	\(<\)	\(\displaystyle 1 + x y - \left({x + y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 0\)	\(<\)	\(\displaystyle \frac {1 + x y - \left({x + y}\right)} {1 + x y}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 0\)	\(<\)	\(\displaystyle \frac {1 + x y} {1 + x y} - \frac {x + y} {1 + x y}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 0\)	\(<\)	\(\displaystyle 1 - \frac {x + y} {1 + x y}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \frac {x + y} {1 + x y}\)	\(<\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Finally:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle -1\)	\(<\)	\(\displaystyle x, y < 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 0\)	\(<\)	\(\displaystyle \left({1 + x}\right) \left({1 + y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 0\)	\(<\)	\(\displaystyle 1 + x y + \left({x + y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 0\)	\(<\)	\(\displaystyle \frac {1 + x y + \left({x + y}\right)} {1 + x y}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 0\)	\(<\)	\(\displaystyle \frac {1 + x y} {1 + x y} + \frac {x + y} {1 + x y}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle 0\)	\(<\)	\(\displaystyle 1 + \frac {x + y} {1 + x y}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle -1\)	\(<\)	\(\displaystyle \frac {x + y} {1 + x y}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Thus $-1 < x, y < 1 \implies -1 < x \circ y < 1$, and we see that in this range, $\circ$ is closed.

Thus the given set and operation form a group.

$\blacksquare$

Sources

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