proof of example of medial quasigroup

We shall proceed by first showing that the algebraic systems defined in the parent entry (http://planetmath.org/MedialQuasigroup) are quasigroups and then showing that the medial property is satisfied.

To show that the system is a quasigroup, we need to check the solubility of equations. Let $x$ and $y$ be two elements of $G$. Then, by definition of $\cdot$, the equation $x\cdot z=y$ is equivalent to

$$f(x)+g(z)+c=y.$$

This is equivalent to

$$g(z)=y-c-f(x).$$

Since $g$ is an automorphism, there will exist a unique solution $z$ to this equation.

Likewise, the equation $z\cdot x=y$ is equivalent to

$$f(z)+g(x)+c=y$$

which, in turn is equivalent to

$$f(z)=y-c-g(x),$$

so we may also find a unique $z$ such that $z\cdot x=y$. Hence, $(G,\cdot )$ is a quasigroup.

To check the medial property, we use the definition of $\cdot$ to conclude that

$(x\cdot y)\cdot (z\cdot w)$$=$$(f(x)+g(y)+c)\cdot (f(z)+g(w)+c)$

$=$$f(f(x)+g(y)+c)+g(f(z)+g(w)+c)+c$

Since $f$ and $g$ are automorphisms and the group is commutative, this equals

$$f(f(x))+f(g(y))+g(f(z))+g(g(w))+f(c)+g(c)+c.$$

Since $f$ and $g$ commute this, in turn, equals

$$f(f(x))+g(f(y))+f(g(z))+g(g(w))+f(c)+g(c)+c.$$

Using the commutative and associative laws, we may regroup this expression as follows:

$$(f(f(x))+f(g(z))+f(c))+(g(f(y))+g(g(w))+g(c))+c$$

Because $f$ and $g$ are automorphisms, this equals

$$f(f(x)+g(z)+c)+g(f(y)+g(w)+c)+c$$

By defintion of $\cdot$, this equals

$$f(x\cdot z)+g(y\cdot z)+c,$$

which equals $(x\cdot z)\cdot (y\cdot z)$, so we have

$$(x\cdot y)\cdot (z\cdot w)=(x\cdot z)\cdot (y\cdot z).$$

Thus, the medial property is satisfied, so we have a medial quasigroup.