The set is the interval $(0,1)$ of real numbers. I want to show that $(0,1)$ forms a group in such a way that the inverse of $a \in (0,1)$ is $1-a$.
I’ve tried many variations of the form $a\cdot b = {f(a,b)\over{g(a,b)}}$ to no avail. A nice identity would be $1/2$ however I’ve not been able to find such an operation. Any help would be greatly appreciated.
The way to understand questions of this form generally is via transport of structure. Namely, $(0, 1)$ is in bijection with $\mathbb{R}$, and even in bijection with $\mathbb{R}$ in such a way that $a \mapsto 1 - a$ is sent to $r \mapsto -r$, for example via
$$(0, 1) \ni a \mapsto \tan \pi \left( a - \frac{1}{2} \right) \in \mathbb{R}.$$
Now we can just use the ordinary addition operation on $\mathbb{R}$, which has inverse $r \mapsto -r$, and transport it back to $(0, 1)$ via the inverse of the above map. This gives the group operation
$$a \cdot b = \boxed{ \frac{\arctan \left( \tan \pi \left( a - \frac{1}{2} \right) + \tan \pi \left( b - \frac{1}{2} \right) \right) }{\pi} + \frac{1}{2} }.$$
I don't know if this simplifies further or if a different bijection would give something nicer, though.
Send $(0,1)$ to $\mathbb{R}$ with the bijection
$$x\mapsto \log\left(\frac{x}{1-x}\right). $$
then use the additive group on $\mathbb{R}$.
Then indeed your nice identity and inverse properties are satisfied.
This is the classic log-odds transformation used in Probability / Statistics.
=== a shorter cleaner tl;dr answer ====
All three answers more or less do the following.
If you have a set $S$ and a group $<G, \cdot>$ and you can have a bijective mapping from $f:S \to G$, then you can create a group operation in the follow way.
For any $a,b \in S$, you will have $f(a), f(b) \in G$. You can do the group operation on $f(a)$ and $f(b)$ to get $f(a)\cdot f(b) = \gamma \in G$. And then you can map from $c\in G$ back to $S$ but $f^{-1} (\gamma) = c\in S$.
Thus we can define a group operation on $S$ as $a\odot b = f^{-1}(f(a)\cdot f(b))$
This will satisfy conditions of a group:
$\cdot$ is clearly binary and as $\cdot$ is associative:
$(a\odot b)\odot c=$
$f^{-1}(f(a\odot b)\cdot f(c))=$
$f^{-1}(f(f^{-1}(f(a)\cdot f(b)))\cdot f(c))=$
$f^{-1}((f(a)\cdot f(b))\cdot f(c))=$
$f^{-1}(f(a)\cdot (f(b)\cdot f(c)))=$
$f^{-1}(f(a)\cdot f(f^{-1}(f(b)\cdot f(c))))=$
$f^{-1}(f(a) \cdot f(b\odot c))=$
$a\odot (b\odot c)$
so $\odot$ is a associative.
If $e$ is the identity in $G$ the $f^{-1}(e)$ is clear the $\odot$ identity:
$f^{-1}(e) \odot a = f^{-1}(f(f^{-1}e)\cdot f(a)=f^{-1}(e\cdot f(a))=f^{-1}(f(a)) = a$.
And for any $a \in S$ if $f(a) =\alpha\in G$ and $\alpha^{-1}$ is the inverse of $\alpha$ in $G$ then $f^{-1}(\alpha^-1)$ will be the $\odot$ inverse of $a$:
$a\cdot f^{-1}(\alpha^{-1}) = f^{-1}(f(a)\cdot f(f^{-1}(\alpha^{-1})))=f^{-1}(\alpha \cdot \alpha^{-1})=f^{-1}(e)=$ the $\odot$ identity.
....
So as $(0,1) \sim \mathbb R \sim (0,\infty)$ any bijection for $(0,1)$ to $<\mathbb R, +>$ or to $<\mathbb R_{>0}, \times>$ will work.
In my original answer I showed (in excrutiating detail) have $f(x) = \frac {1-x}x=\frac 1x -1$, a bijection from $(0,1)\to \mathbb R_{>0}$ will yield a group operation:
$a\odot b =f^{-1}(f(a)\times f(b))= \frac {ab} {1-a-b+2b}$ will yield a group operation with group identity $f^{-1}(1) = \frac 12$ and for any $a\in (0,1)$ then $f^{-1}(\frac 1{f(a)}) = 1-a$ is the inverse of $a$.
Read below if you want more.
You can of course do something equivalent with the $\arctan$ function or then $\log (\frac x{1-x})$ to map $(0,1)$ to $\mathbb R$ which is a group with addition. (I don't know if they also yield $\frac 12$ and $1-a$ as identity and $1-a$ as inverse but I bet they do. In fact I'm pretty certain $\log\frac x{1-x}$ yields the exact same group element by element as mine does.)
=== original answer below ====
Step 1: Think of any other group that has the same cardinality of $(0,1)$. Two obvious examples are $< \mathbb R, +>$ and $<(0,\infty), \times>$. There are of course others but these are the only two that I believe everyone familiar with with.
Step 2: Think of a bijective mapping from $(0,1)$ two the other group. It doesn't have to be sophisticated or clean. it can be anything so long as it is bijective.
A simple way could noting we stretch $(0,1)$ to $(1,\infty)$ via $h(x) = \frac 1x$ and we go then shift $(1,\infty)$ down to $(0,\infty)$ via $g(c) = x -1$ so we can map $(0,1)\to (0, \infty)$ via $f(x)= \frac 1x - 1=\frac {1-x}x$. (I'll leave it as an exercise that that is one to one and onto.)
We have that $<(0,\infty), \times >$ is a group with identity equal to $1$ and $f$ maps $(0,1)$ to it.
Step three: We can then define the $\cdot$ operation by saying the product of two elements is the pre-image of the product of the mapped elements.
If other words... as $f(x) = \frac {1-x}x$ will map $(0,1)$ to $(0, \infty)$ then we can define $a \dot b$ as $f^{-1}( f(a) \times f(b))$.
If we want to spell it out specifically: $f(x) = \frac {x-1}x$ so $f^{-1}(x) = \frac 1{x+1}$ and so $a\cdot b = \frac 1{\frac {1-a}a\frac {1-b}b +1}=\frac {ab}{(1-a)(1-b) +ab}=\frac {ab}{1-a-b+2ab}$
The identity is $f^{-1}(1) = \frac 1{1+1} = \frac 12$.
And the inverse of $a$ would be: Well $f(a) = \frac {1-a}a$ and the multiplicative inverse of that is $\frac a{1-a}$ so $a^{-1} = f^{-1}(\frac a{1-a})=\frac 1{\frac a{1-a} + 1}=\frac {1-a}{a+(1-a)}=1-a$
If as an excercise we want to prove this is a group:
If $a,b \in (0,1)$ then $\frac 1a, \frac 1b \in (1,\infty)$ and $\frac 1a -1, \frac 1b - 1 \in (0, \infty)$ and so $(\frac 1a -1)(\frac 1b-1)\in (0, \infty)$ and $\frac 1{(\frac 1a -1)(\frac 1b-1)}\in (0, \infty)$.
$(a\cdot b)\cdot c = \frac {ab}{1-a-b+2ab}\cdot c=$
$\frac {\frac {ab}{1-a-b+2ab}c}{1-\frac {ab}{1-a-b+2ab}-c + 2c\frac {ab}{1-a-b+2ab}}=$
$\frac {abc}{(1-a-b+2ab)-ab -c(1-a-b-2ab) +2abc}=$
$\frac {abc}{1-a-b-c+ab +ac +bc + 4abc}$.
and similarly $a\cdot (b\cdot c)$ can be evaluated to equal the same.
$\frac 12 \cdot a = \frac {\frac 12 a}{1-\frac 12 -a +2\cdot \frac 12a}=$
$\frac {\frac 12 a}{\frac 12 -a+a}=a$.
$(1-a)\cdot a = \frac {a(1-a)}{1 - (1-a) -a + 2a(1-a)}= \frac{a(1-a)}{2a(1-a)}=\frac 12$
