A ternary relation on a group

Question

Let $G$ be a group. Consider the ternary relation $R \subset G^3$ defined by

$$R(x,y,z) \Leftrightarrow x y^{-1} z x^{-1} y z^{-1} = 1 $$

Show that $R$ is a symmetric relation, that is if $R(x,y,z)$, then $R(x',y',z')$ for all permutations $(x',y',z')$ of $(x,y,z)$.

Notes

1. The expression above can be encountered in Pappus' theorem. There the group $G$ is the $2\times 2$ lower triangular matrices.

2. For the case when $R$ is a ternary equivalence relation, see the paper "Groups with a ternary equivalence relation". This is where it was stated that the symmetry holds for all groups $G$ ( a surprise! ).

$\bf{Added:}$

There were some very crisp answers. From what I understand, one idea seems to be this:

Consider the $n$-ary relation on a group $G$

$$E(x_1, \ldots, x_n) \Leftrightarrow x_1 \cdots x_n= 1$$

Then $E$ is invariant under cyclic permutations, and also

$$x_1 \cdots x_n = 1 \implies x_n^{-1} \cdots x_1^{-1} = 1$$

These two properties take care of the proof. We also get

$$R(x,y,z) \implies R(x^{-1}, y^{-1}, z^{-1})$$

and $$R(x,y,z) \implies R(a x, a y , a z), R(xa, y a, z a)$$

I've found another way to express $R$:

Note that if $x y^{-1} = w$, $z x^{-1} = v$, $y z^{-1} = u$, then $x = w y$, $z = v x$, $y = u z$. So $w$, $v$, $y$ are (left) displacements. Now it is easy to see that

$$u v w = y z^{-1} \cdot z x^{-1} \cdot x y^{-1} = 1$$

But the condition says that on top of this we have $w v u = 1$. This is equivalent to $u$, $v$, $w$ commute. So that is the meaning of $R$ : the (left) displacements corresponding to $x$, $y$, $z$ commute. It is now easy to see why this condition is symmetric. Moreover it is equivalent to the condition for the right displacements.

This gets us back to the problem where this expression originates. In the Pappus theorem, the projections from one line to another through one of the points $X$, $Y$, $Z$ are elements of $PGL(2, k)$. We have the condition $R$ if and only if the points $X$, $Y$, $Z$ are collinear. That means that the relation $R$ has the property

$$R(x,y,z) \& R(x,z, t) \implies R(x,y,t)$$

Now, the paper quoted above does state some theorems, but the fact is this: $R$ satisfies the above ( similar to collinearity) if and only if the commutant of any non-trivial element is abelian. This can be expressed as follows: the relation $x \simeq y$ on $G \backslash \{e\}$ if $x$, $y$ commute is an $equivalence$ relation ( a CA-group).

One should check that for the group $G = PGL(2, k)$ it is true that the centralizer of any element $\ne 1$ is abelian .

Answer 1

Start with your $x y^{-1} z x^{-1} y z^{-1} = 1$. Multiply both sides by $z$ from the right, then multiply both sides by $z^{-1}$ from the left. You get $z^{-1} x y^{-1} z x^{-1} y = 1$. In this way you can cycle through the expression, moving elements from the right end to the left end and vice versa. Among the cyclic variants you get

$$ y z^{-1} x y^{-1} z x^{-1} = 1 \\ z x^{-1} y z^{-1} x y^{-1} = 1 $$

This shows $R(x,y,z)\Leftrightarrow R(y,z,x)\Leftrightarrow R(z,x,y)$. So three of the six permutations done, we need one order-reversing then we're done.

Start again with $x y^{-1} z x^{-1} y z^{-1} = 1$. Multiply from the left to cancel one term after the other. I.e. multiply first with $x^{-1}$ to get $y^{-1} z x^{-1} y z^{-1} = x^{-1}$ then with $y$ to get $z x^{-1} y z^{-1} = y x^{-1}$ and so on. Eventually you get

$$1 = z y^{-1} x z^{-1} y x^{-1}$$

which shows $R(x,y,z)\Leftrightarrow R(z,y,x)$. Combined with the cyclic argument from above, that accounts for the remaining permutations. Q.e.d.

Answer 2

Since $1=1^{-1}$ we have $(x y^{-1} z x^{-1} y z^{-1})^{-1} = zy^{-1}xz^{-1}yx^{-1} = 1$ which gives us $R(x,y,z)=R(z,y,x)$. With $x y^{-1} z x^{-1} y z^{-1} = 1$ we multiply on the left by $yz^{-1}$ and right by $zy^{-1}$ which gives $yz^{-1}xy^{-1}zx^{-1}=1$ and so $R(x,y,z)=R(y,z,x)$.

Now since we have three letters to permute, with a 3-cycle and a transposition acting on them they generate the entire symmetry group $S_3$ and we're done.

Answer 3

Multiply both sides of $xy^{-1}z x^{-1}yz^{-1} = 1$ by $zy^{-1}x$ on the right to get

\begin{equation}\tag{1}R(x,y,z) \Leftrightarrow xy^{-1}z = z y^{-1}x\end{equation}

Multiply (1) with $yz^{-1}$ on the left and $z^{-1}y$ on the right to get

\begin{equation}\tag{2}R(x,y,z) \Leftrightarrow yz^{-1}x = x z^{-1}y\end{equation}

Thus $R(x,y,z)$ is invariant under swapping $x$ and $z$ by $(1)$, and invariant under swapping $y$ and $x$ by $(2)$. This gives you invariance under all permutations of $x,y,z$.