Almost have a group law: $(x,y)*(a,b) = (xa + yb, xb + ya)$ with rational components.

Question

Let $\Bbb{Q}$ be the set of rationals. Let $C \subset \Bbb{Q}^2$ be defined as the set of all pairs $(x,y)$ such that $x + y = 1$. Define $(x,y)*(a,b) = (xa + yb, xb + ya)$. Notice that $C$ is closed under such an operation by observing that $(x + y)(a + b) = xa + yb + xb + ya = 1$.

Also $(1, 0)$ is a two-sided identity for the operation since:

$$ (1,0)*(a,b) = (1a + 0b, 1b + 0a) = (a,b) \\ (x,y)*(1,0) = (x1 + y0, x0 + y1) = (x,y) $$

Now does there exist a general inverse $(x,y)^{-1} = (a,b)$ with respect to $*$?

For that we would need $(xa + yb, xb + ya) = (1,0)$ or in augmented matrix form:

$$ \begin{pmatrix} x & y & | & 1 \\ y & x & | & 0 \\ \end{pmatrix} $$

which is similar via row reduction to the matrix:

$$ \begin{pmatrix} 1 & 0 & | & \frac{1}{x} + \frac{y}{x(x^2 - y^2)} \\ 0 & 1 & | & \frac{-y}{x^2 - y^2} \end{pmatrix} $$ as long as $x \neq 0$ and $(x-y)(x+y) = x^2 - y^2 \neq 0$ or $x \neq y$ since $x + y = 1$ by definition. Thus add the two constraints $0 \neq x \neq y$ to the definition of $C$.

We then need to show that $*$ respects these constraints. If $0 \neq x \neq y$ and $0 \neq a \neq b$, then we have: $(x,y)*(a,b) = (xa + yb, xb + ya)$ and if $xa + yb = 0$ then $xb + ya = 1$.

I can show easily that $xa + yb \neq xb + ya$ since if so, then $x(a - b) - y(a - b) = (x - y)(a-b) = 0$ and so at least one of $a = b$ or $x = y$, a contradiction.

What I am having trouble showing is that $xa + yb \neq 0$. If this is impossible to show - there exist solutions such that $x a + yb = 0$, then what constraint can I add in order to turn $C$ into a group?

Answer 1

A slightly different viewpoint.

Your structure is captured by the usual multiplication on the set of $2\times 2$ matrices of the form $\pmatrix {x & y\\ y & x}$ which have $1$-eigenvector $(1 1)$. This gives the closure easily, the associativity, and the fact that the identity is given by $x=1, y=0$. It's clear, too, that inverses exist provided the matrix is non-singular, $x^2\not= y^2$, that is $x\not=\frac12$.

But the structure is even easier. All these matrices can be simultaneously diagonalised as they all have $(1 1)$ and $(1 -1)$ as eigenvectors, the eigenvalues being $1$ and $x-y$. So with respect to that basis we're looking at the set of matrices $\pmatrix{1 &0\\0 & x-y}$ under matrix multiplication. Removing the $x=y$ case we are left with the set of all $\pmatrix{1 &0\\0 & t}, t\not=0$, clearly isomorphic with the group $\mathbb{Q}^{*}$.

Answer 2

In your set, $x=1-y$, so we might as well drop the first coordinate and re-define $$y*b=(1-y)b+y(1-b)=y+b-2yb.$$ As you have observed, $0$ is an identity for this commutative operation. Moreover $$y*(b*t)=y+b+t-2yb-2yt-2bt+4ybt=(y*b)*t.$$ Does $y$ have an inverse? Solving $y+b-2yb=0$ for $b$ gives $b=y/(2y-1)$, that is yes for $y\ne1/2$. So this isn't a group, but we can hope to create one by discarding $1/2$.

The only possible catch now is if $b*y=1/2$ for $b$, $y\ne1/2$. But $b*y=1/2$ implies $$2y+2b-4yb=1$$ implies $$(1-2y)(1-2b)=0$$ implies $y=1/2$ or $b=1/2$.

Answer 3

Writing $\phi(h)$ for $(\frac{1+h}2,\frac{1-h}2)$, we have $$ \phi(a)*\phi(b)=\left(\frac{1+a}2\frac{1+b}2+\frac{1-a}2\frac{1-b}2,\,\ldots\right)=\left(\frac{1+ab}2,\,\ldots\right)=\phi(ab).$$ Thus, if we drop $(\frac12,\frac12)$ from your original definition of $C$, we obtain a group structure that is - via $\phi$ - isomorphic to $(Q^\times,\cdot)$.