Let $A$ be an algebra, not necessarily associative multiplicatively. The nucleus of $A$ is: $$\mathcal{N}(A):=\lbrace a\in A\mid [a,A,A]=[A,a,A]=[A,A,a]=0 \rbrace,$$ where $[\ , , ]$ is the associator bracket. In other words, the nucleus is the set of elements that multiplicatively associate with all elements of $A$ An element $a\in A$ is nuclear if $a\in\mathcal{N}(A)$

$\mathcal{N}(A)$ is a Jordan subalgebra of $A$ To see this, let $a,b\in \mathcal{N}(A)$ Then for any $c,d\in A$ \begin{eqnarray} [ab,c,d] &=& ((ab)c)d-(ab)(cd) = (a(bc))d-(ab)(cd) \\ &=& a((bc)d)-(ab)(cd) = a(b(cd))-(ab)(cd) \\ &=& a(b(cd))-a(b(cd)) = 0 \end{eqnarray}Similarly, $[c,ab,d]=[c,d,ab]=0$ and so $ab\in\mathcal{N}(A)$

Accompanying the concept of a nucleus is that of the center of a nonassociative algebra $A$ (which is slightly different from the definition of the center of an associative algebra): $$\mathcal{Z}(A):=\lbrace a\in \mathcal{N}(A)\mid [a,A]=0 \rbrace,$$ where $[\ , ]$ is the commutator bracket.

Hence elements in $\mathcal{Z}(A)$ commute as well as associate with all elements of $A$ Like the nucleus, the center of $A$ is also a Jordan subalgebra of $A$