Jordan implies powers up to the fifth are well-defined

This article gives the statement and possibly, proof, of an implication relation between two magma properties. That is, it states that every magma satisfying the first magma property (i.e., Jordan magma) must also satisfy the second magma property (i.e., magma in which powers up to the fifth are well-defined)
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Statement

Suppose $(S,*)$ is a Jordan magma, i.e., it is a commutative magma satisfying the following identity:

$\!(x*y)*(x*x)=x*(y*(x*x))\ \forall \ x,y\in S$

Then cubes, fourth powers, and fifth powers are well-defined in $S$. In other words, we have the following for all $x\in S$:

$\!(x*x)*x=x*(x*x)$

This quantity is denoted by $x^{3}$. Note that we will henceforth abbreviate $x*x$ as $x^{2}$.

$\!x*x^{3}=x^{2}*x^{2}=x^{3}*x$

This quantity is denoted by $x^{4}$.

$\!x*x^{4}=x^{2}*x^{3}=x^{3}*x^{2}=x^{4}*x$

This quantity is denoted by $x^{5}$.

Related facts

Related facts for non-associative rings

• Jordan implies power-associative: When we are dealing with non-associative rings instead of magmas, the condition of being a Jordan ring is strong enough to guarantee being a power-associative ring, which means that all powers (including sixth and higher powers) are well-defined.

Proof

Since $S$ is commutative, some of the equalities are direct. We do the proof in three stages.

The cube is well-defined

To prove: $\!x^{2}*x=x*x^{2}$.

Proof: This follows from the fact that since the magma is commutative, $x^{2}$ and $x$ commute.

The fourth power is well-defined

To prove: <math! x * x^3 = x^2 * x^2 = x^3 * x</math>.

Proof: Since the magma is commutative, we have $x*x^{3}=x^{3}*x$. It thus suffices to show the equality $x*x^{3}=x^{2}*x^{2}$.

To do this, we write:

$\!x^{2}*x^{2}=(x*x)*(x*x)=x*(x*(x*x))=x*x^{3}$

In an intermediate step, we used Jordan's identity, setting $y=x$.

The fifth power is well-defined

To prove: $x*x^{4}=x^{2}*x^{3}=x^{3}*x^{2}=x^{4}*x$.

Proof: Since the magma is commutative, we obtain $x*x^{4}=x^{4}*x$ and $x^{2}*x^{3}=x^{3}*x^{2}$. It thus suffices to show that $x^{3}*x^{2}=x*x^{4}$.

To do this, we write:

$\!x^{3}*x^{2}=(x*x^{2})*(x*x)=x*(x^{2}*(x*x))=x*x^{4}$

In an intermediate step, we used Jordan's identity, setting $y=x^{2}$.