First to recap the various types of nonassociativity, expressed with associators:
Associative: (x,y,z)=id
Jacobi: ((x,y,z),(y,z,x),(z,x,y))=id
Jordan: (x,y,xx)=id
Lebiniz: ?
Power associative: (x^n,x^m,x^p)=id for any n,m,p in integers
Alternative: (x,x,y)=id,(y,x,x)=id
Flexible: (x,y,x)=id
Elastic: (x,y,x)=id and (x,y,xx)=id and x(xx)y = (xx)(xy)
For a field $K$ and $1<n\in \mathbb{N}$ let $A\in K^{(n-1)\times n}$ aa matrix with rank $n-1$. For a row vector $z\in K^{1\times n}$ let $\left (\frac{A}{z}\right )\in K^{n\times n}$ be the matrix that we get if we add as the $n$-th row of the matrix $A$ the vector $z$.
To show that there is a...
Here is the theorem giving me trouble:
Here is theorem 6, which is invoked in the proof of theorem 6:
Here is the proof of theorem 8:
Here is the sentence giving me trouble so far:
"Since $\varphi$ is an $R$-module homomorphism, the generators of the subgroup $H$ in equation (3) all ...
