Group/Examples/x+y over 1+xy
Theorem
Let $G = \set {x \in \R: -1 < x < 1}$ be the set of all real numbers whose absolute value is less than $1$.
Let $\circ: G \times G \to \R$ be the binary operation defined as:
- $\forall x, y \in G: x \circ y = \dfrac {x + y} {1 + x y}$
The algebraic structure $\struct {G, \circ}$ is a group.
Isomorphic to Real Numbers
$\struct {G, \circ}$ is isomorphic to the additive group of real numbers $\struct {\R, +}$.
Proof
Let $-1 < x, y, z < 1$.
We check the group axioms in turn:
Group Axiom $\text G 1$: Associativity
| \(\ds \paren {x \circ y} \circ z\) | \(=\) | \(\ds \frac {\frac {x + y} {1 + x y} + z} {1 + \frac {x + y} {1 + xy} z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {x + y + z + x y z} {1 + x y + x z + y z}\) |
| |||||||||||
| \(\ds x \circ \paren {y \circ z}\) | \(=\) | \(\ds \frac {x + \frac {y + z} {1 + y z} } {1 + x \frac {y + z} {1 + y z} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {x + y + z + x y z} {1 + x y + x z + y z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {x \circ y} \circ z\) |
Thus $\circ$ has been shown to be associative.
$\Box$
Group Axiom $\text G 2$: Existence of Identity Element
| \(\ds x\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {x + y} {1 + x y}\) | \(=\) | \(\ds \frac {0 + y} {1 + 0 y}\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac y 1\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds y\) |
Similarly, putting $y = 0$ we find $x \circ y = x$.
So $0$ is the identity.
$\Box$
Group Axiom $\text G 3$: Existence of Inverse Element
| \(\ds x \circ -x\) | \(=\) | \(\ds \frac {x + \paren {-x} } {1 + x \paren {-x} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) |
Similarly, putting $x = -y$ gives us $\paren {-y} \circ y = 0$.
So each $x$ has an inverse $-x$.
$\Box$
Group Axiom $\text G 0$: Closure
First note that:
- $-1 < x, y < 1 \implies x y > -1 \implies 1 + x y > 0$
Next:
| \(\ds -1\) | \(<\) | \(\ds x, y < 1\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 0\) | \(<\) | \(\ds \paren {1 - x} \paren {1 - y}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 0\) | \(<\) | \(\ds 1 + x y - \paren {x + y}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 0\) | \(<\) | \(\ds \frac {1 + x y - \paren {x + y} } {1 + x y}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 0\) | \(<\) | \(\ds \frac {1 + x y} {1 + x y} - \frac {x + y} {1 + x y}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 0\) | \(<\) | \(\ds 1 - \frac {x + y} {1 + x y}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {x + y} {1 + x y}\) | \(<\) | \(\ds 1\) |
Finally:
| \(\ds -1\) | \(<\) | \(\ds x, y < 1\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 0\) | \(<\) | \(\ds \paren {1 + x} \paren {1 + y}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 0\) | \(<\) | \(\ds 1 + x y + \paren {x + y}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 0\) | \(<\) | \(\ds \frac {1 + x y + \paren {x + y} } {1 + x y}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 0\) | \(<\) | \(\ds \frac {1 + x y} {1 + x y} + \frac {x + y} {1 + x y}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 0\) | \(<\) | \(\ds 1 + \frac {x + y} {1 + x y}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds -1\) | \(<\) | \(\ds \frac {x + y} {1 + x y}\) |
Thus:
- $-1 < x, y < 1 \implies -1 < x \circ y < 1$
and we see that in this range, $\circ$ is closed.
$\Box$
Thus the given set and operation form a group.
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Definition of Group Structure: $\S 26 \lambda$