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Let $\Gamma$ be a sofic group. Gabor Elek and Endre Szabo showed here, that for any field $k$ and $a,b \in k[\Gamma]$ with $ab=1$ one has $ba=1$. Hence, coming up with a cleverly chosen group where this fails would provide a counterexample. Note that $k=\mathbb C$ is not a good start since Kaplansky showed long ago that the implication holds for fields of characteristic zero. However, for $k= GF(2)$ one might be lucky.

Let us consider $k=GF(2)$ for now. One strategy could be to start with $a = \sum_{g \in F} g$ and $b = \sum_{h \in K} h$ for some finite sets $F,K \subset G$. If $ab=1$, then a number of relations must hold: We quickly convince ourselves that $F$ and $K$ must have an odd number of elements and there exists a self-matching of the set $F \times K \setminus (f,k)$ such that matched pairs $(f',k') \sim (f'',k'')$ satisfy $f'k' = f''k''$ and $f=k^{-1}$ for the special unmatched pair. You can now turn everything around and start with an abstract group with generators $F \cup K$ and relations as above coming from an arbitrarily choosen self-matching. In the group ring of this abstract group, we will have $ab=1$, but why do we have $ba=1$? I was working on this for a while but could not come up with a counterexample. On the other hand, if $F$ and $K$ are large, I cannot believe that $ba=1$ will always hold.

Let $\Gamma$ be a sofic group. Gabor Elek and Endre Szabo showed here, that for any field $k$ and $a,b \in k[\Gamma]$ with $ab=1$ one has $ba=1$. Hence, coming up with a cleverly chosen group where this fails would provide a counterexample. Note that $k=\mathbb C$ is not a good start since Kaplansky showed long ago that the implication holds for fields of characteristic zero. However, for $k= GF(2)$ one might be lucky.

Let us consider $k=GF(2)$ for now. One strategy could be to start with $a = \sum_{g \in F} g$ and $b = \sum_{h \in K} h$ for some finite sets $F,K \subset G$. If $ab=1$, then a number of relations must hold: We quickly convince ourselves that $F$ and $K$ must have an odd number of elements and there exists a self-matching of the set $F \times K \setminus (f,k)$ such that matched pairs $(f',k') \sim (f'',k'')$ satisfy $f'k' = f''k''$ and $f=k^{-1}$ for the special unmatched pair. You can now turn everything around and start with an abstract group with generators $F \cup K$ and relations as above coming from an arbitrarily chosen self-matching. In the group ring of this abstract group, we will have $ab=1$, but why do we have $ba=1$? I was working on this for a while but could not come up with a counterexample. On the other hand, if $F$ and $K$ are large, I cannot believe that $ba=1$ will always hold.

Let $\Gamma$ be a sofic group. Gabor Elek and Endre Szabo showed here, that for any field $k$ and $a,b \in k[\Gamma]$ with $ab=1$ one has $ba=1$. Hence, coming up with a cleverly choosen group where this fails would provide a counterexample. Note that $k=\mathbb C$ is not a good start since Kaplansky showed long ago that the implication holds for fields of characteristic zero. However, for $k= GF(2)$ one might be lucky.

Let us consider $k=GF(2)$ for now. One strategy could be to start with $a = \sum_{g \in F} g$ and $b = \sum_{h \in K} h$ for some finite sets $F,K \subset G$. If $ab=1$, then a number of relations must hold: We quickly convince ourselves that $F$ and $K$ must have an odd number of elements and there exists a self-matching of the set $F \times K \setminus (f,k)$ such that matched pairs $(f',k') \sim (f'',k'')$ satisfy $f'k' = f''k''$ and $f=k^{-1}$ for the special unmatched pair. You can now turn everything around and start with an abstract group with generators $F \cup K$ and relations as above coming from an arbitrarily choosen self-matching. In the group ring of this abstract group, we will have $ab=1$, but why do we have $ba=1$? I was working on this for a while but could not come up with a counterexample. On the other hand, if $F$ and $K$ are large, I cannot believe that $ba=1$ will always hold.

Let $\Gamma$ be a sofic group. Gabor Elek and Endre Szabo showed here, that for any field $k$ and $a,b \in k[\Gamma]$ with $ab=1$ one has $ba=1$. Hence, coming up with a cleverly chosen group where this fails would provide a counterexample. Note that $k=\mathbb C$ is not a good start since Kaplansky showed long ago that the implication holds for fields of characteristic zero. However, for $k= GF(2)$ one might be lucky.

Let us consider $k=GF(2)$ for now. One strategy could be to start with $a = \sum_{g \in F} g$ and $b = \sum_{h \in K} h$ for some finite sets $F,K \subset G$. If $ab=1$, then a number of relations must hold: We quickly convince ourselves that $F$ and $K$ must have an odd number of elements and there exists a self-matching of the set $F \times K \setminus (f,k)$ such that matched pairs $(f',k') \sim (f'',k'')$ satisfy $f'k' = f''k''$ and $f=k^{-1}$ for the special unmatched pair. You can now turn everything around and start with an abstract group with generators $F \cup K$ and relations as above coming from an arbitrarily choosen self-matching. In the group ring of this abstract group, we will have $ab=1$, but why do we have $ba=1$? I was working on this for a while but could not come up with a counterexample. On the other hand, if $F$ and $K$ are large, I cannot believe that $ba=1$ will always hold.

Let $\Gamma$ be a sofic group. Gabor Elek and Endre Szabo showed here, that for any field $k$ and $a,b \in k[\Gamma]$ with $ab=1$ one has $ba=1$. Hence, coming up with a cleverly choosen group where this fails would provide a counterexample. Note that $k=\mathbb C$ is not a good start since Kaplansky showed long ago that the implication holds for fields of characteristic zero. However, for $k= GF(2)$ one might be lucky.

Lets consider $k=GF(2)$ for now. One strategy could be to start with $a = \sum_{g \in F} g$ and $b = \sum_{h \in K} h$ for some finite sets $F,K \subset G$. If $ab=1$, then a number of relations must hold: We quickly convince ourselves that $F$ and $K$ must have an odd number of elements and there exists a self-matching of the set $F \times K \setminus (f,k)$ such that matched pairs $(f',k') \sim (f'',k'')$ satisfy $f'k' = f''k''$ and $f=k^{-1}$ for the special unmatched pair. You can now turn everything around and start with an abstract group with generators $F \cup K$ and relations as above coming from an arbitrarily choosen self-matching. In the group ring of this abstract group, we will have $ab=1$, but why do we have $ba=1$? I was working on this for a while but could not come up with a counterexample. On the other hand, if $F$ and $K$ are large, I cannot believe that $ba=1$ will always hold.

Let $\Gamma$ be a sofic group. Gabor Elek and Endre Szabo showed here, that for any field $k$ and $a,b \in k[\Gamma]$ with $ab=1$ one has $ba=1$. Hence, coming up with a cleverly choosen group where this fails would provide a counterexample. Note that $k=\mathbb C$ is not a good start since Kaplansky showed long ago that the implication holds for fields of characteristic zero. However, for $k= GF(2)$ one might be lucky.

Let us consider $k=GF(2)$ for now. One strategy could be to start with $a = \sum_{g \in F} g$ and $b = \sum_{h \in K} h$ for some finite sets $F,K \subset G$. If $ab=1$, then a number of relations must hold: We quickly convince ourselves that $F$ and $K$ must have an odd number of elements and there exists a self-matching of the set $F \times K \setminus (f,k)$ such that matched pairs $(f',k') \sim (f'',k'')$ satisfy $f'k' = f''k''$ and $f=k^{-1}$ for the special unmatched pair. You can now turn everything around and start with an abstract group with generators $F \cup K$ and relations as above coming from an arbitrarily choosen self-matching. In the group ring of this abstract group, we will have $ab=1$, but why do we have $ba=1$? I was working on this for a while but could not come up with a counterexample. On the other hand, if $F$ and $K$ are large, I cannot believe that $ba=1$ will always hold.