I have this problem about subgroups of the tensor product of an abelian group $A$ with itself which arises from a complete different setting. I fell into this question studying quandles and quandle coverings, but the question can be stated just in terms of abelian groups.

Let $A$ be an Abelian group and $\alpha$ an automorphism of $A$ such that $1-\alpha$ is invertible.

I need to compute the following subgroups in $A\otimes A$:

$B=\langle x\otimes y-\alpha(y)\otimes x,\quad x,y\in A\rangle$

The goal is to find some condition on $A$ and $\alpha$ in order to have that $B=A\otimes A$. For instance, if $A$ is a cyclic group then $B=A\otimes A$. Moreover, I know that $B$ is the whole group when $A$ is elementary abelian different from $\mathbb{Z}_2\times \mathbb{Z}_2$ and $\alpha$ has order $|A|-1$ (it follows by an equivalent property in quandles setting).

A counterexample is given when $\alpha=-1$. Since $B=\langle x\otimes y+y\otimes x,\quad x,y\in A\rangle$ is a subgroup of the element fixed by the flip $\tau$ ($\tau(x\otimes y)=y\otimes x$) and then if $A$ is not cyclic, this is a proper subgroup.

Thank you. M.

I have this problem about subgroups of the tensor product of an abelian group $A$ with itself which arises from a complete different setting. I fell into this question studying quandles and quandle coverings, but the question can be stated just in terms of abelian groups.

Let $A$ be an Abelian group and $\alpha$ an automorphism of $A$ such that $1-\alpha$ is invertible.

I need to compute the following subgroups in $A\otimes A$:

$B=\langle x\otimes y-\alpha(y)\otimes x,\quad x,y\in A\rangle$

The goal is to find some condition on $A$ and $\alpha$ in order to have that $B=A\otimes A$. For instance, if $A$ is a cyclic group then $B=A\otimes A$. Moreover, I know that $B$ is the whole group when $A$ is elementary abelian different from $\mathbb{Z}_2\times \mathbb{Z}_2$ and $\alpha$ has order $|A|-1$ (it follows by an equivalent property in quandles setting).

A counterexample is given when $\alpha=-1$. Since $B=\langle x\otimes y+y\otimes x,\quad x,y\in A\rangle$ is a subgroup of the element fixed by the flip $\tau$ ($\tau(x\otimes y)=y\otimes x$) and then if $A$ is not cyclic, this is a proper subgroup.

Thank you. M.

P.S. I forgot to say that $A$ is finite.

I have this problem about subgroups of the tensor product of an abelian group A with itself which arises from a complete different setting. I fell into this question studying quandles and quandle coverings, but the question can be stated just in terms of abelian groups.

Let A be an Abelian group and $\alpha$ an automorphism of A such that $1-\alpha$ is invertible.

I need to compute the following subgroups in $A\otimes A$:

$B=\langle x\otimes y-\alpha(y)\otimes x,\quad x,y\in A\rangle$

The goal is to find some condition on A and $\alpha$ in order to have that $B=A\otimes A$. For instance if A is a cyclic group then $B=A\otimes A$. Moreover I know that $B$ is the whole group when A is elementary abelian different from $\mathbb{Z}_2\times \mathbb{Z}_2$ and $\alpha$ has order $|A|-1$ (it follows by a equivalent property in quandles setting).

A counterexample is given when $\alpha=-1$. Since $B=\langle x\otimes y+y\otimes x,\quad x,y\in A\rangle$ is a subgroup of the element fixed by the flip $\tau$ ($\tau(x\otimes y)=y\otimes x$) and then if A is not cyclic, this is a proper subgroup.

Thank you. M.

I have this problem about subgroups of the tensor product of an abelian group $A$ with itself which arises from a complete different setting. I fell into this question studying quandles and quandle coverings, but the question can be stated just in terms of abelian groups.

Let $A$ be an Abelian group and $\alpha$ an automorphism of $A$ such that $1-\alpha$ is invertible.

I need to compute the following subgroups in $A\otimes A$:

$B=\langle x\otimes y-\alpha(y)\otimes x,\quad x,y\in A\rangle$

The goal is to find some condition on $A$ and $\alpha$ in order to have that $B=A\otimes A$. For instance, if $A$ is a cyclic group then $B=A\otimes A$. Moreover, I know that $B$ is the whole group when $A$ is elementary abelian different from $\mathbb{Z}_2\times \mathbb{Z}_2$ and $\alpha$ has order $|A|-1$ (it follows by an equivalent property in quandles setting).

A counterexample is given when $\alpha=-1$. Since $B=\langle x\otimes y+y\otimes x,\quad x,y\in A\rangle$ is a subgroup of the element fixed by the flip $\tau$ ($\tau(x\otimes y)=y\otimes x$) and then if $A$ is not cyclic, this is a proper subgroup.

Thank you. M.