This question recently popped up on the front page, so I entered $S_3$ in UACalc, and calculated the $2$-generated relatively free group $F_{S_3}(x,y)$. UACalc informed me that the order of this group is $972=2^2\cdot 3^5$. (It gave me the multiplication table too, but it was too big to be of much use to me.) I then worked out the internal structure of the group, using UACalc to check my intermediate calculations.

Next, I followed the zentralblatt link given in Keith Dennis's answer, to see if what I'd done could be found in Kovacs' paper. That paper predicts the order of $F_{S_3}(x,y)$ to be $6377292 = 2^2\cdot 3^{13}$, not $972$.

I started snooping around with google, and found a copy of Kovacs' paper on his website with a red marginal comment saying that his original formula is wrong. The marginal comment explains how the formula should be modified, but unfortunately the margin was too small to contain the proof of the correction. Luckily for me, the corrected formula gives $972$ for the order of this group.

So, if you are interested in this problem, don't use the formula in Kovacs' paper, but get the copy from his website.

Let me make a conjecture about the structure of $F_{S_3}(n)$. I have verified this conjecture by hand + computer for $n =1, 2$, and it is consistent with Kovacs' corrected formula for all $n$. Moreover, the conjecture is so simple, that if someone pokes at it enough they will likely find a proof or disproof.

Let $\mathbb F_3$ be the $3$-element field and let $R = \mathbb F_3[\oplus^n \mathbb Z_2]$ be the group ring with coefficients in $\mathbb F_3$ of the rank $n$ free group, $\oplus^n \mathbb Z_2$, in the variety ${\mathcal V}(\mathbb Z_2)$ of elementary $2$-abelian groups.

Let $V=\left(\oplus^{n-1} R\right)\oplus \mathbb F_3$ be the $R$-module that is a direct sum of $n-1$ copies of the regular representation of $\oplus^n \mathbb Z_2$ over $\mathbb F_3$ along with one extra copy of the trivial representation. Additively $V$ is an elementary abelian $3$-group, but as an $R$-module it comes equipped with an action of $\oplus^n \mathbb Z_2$ by automorphisms.

The conjecture is that the resulting semidirect product $V\rtimes (\oplus^n \mathbb Z_2)$ is $F_{S_3}(n)$.

You can see from the description that the $\mathbb F_3$-dimension of $V$ is $(n-1)2^n+1$, and so the group has order $2^n\cdot 3^{(n-1)2^n+1}$. This is the modified Kovacs formula for the variety ${\mathcal V}(S_3)$. If the group I describe is $n$-generated, and if the modified Kovacs formula is correct, then the conjecture is true. I also note that the group I construct has the correct size for its center and its commutator subgroup.

This question recently popped up on the front page, so I entered $S_3$ in UACalc, and calculated the $2$-generated relatively free group $F_{S_3}(x,y)$. UACalc informed me that the order of this group is $972=2^2\cdot 3^5$. (It gave me the multiplication table too, but it was too big to be of much use to me.) I then worked out the internal structure of the group, using UACalc to check my intermediate calculations.

Next, I followed the zentralblatt link given in Keith Dennis's answer, to see if what I'd done could be found in Kovacs' paper. That paper predicts the order of $F_{S_3}(x,y)$ to be $6377292 = 2^2\cdot 3^{13}$, not $972$.

I started snooping around with google, and found a copy of Kovacs' paper on his website with a red marginal comment saying that his original formula is wrong. The marginal comment explains how the formula should be modified, but unfortunately the margin was too small to contain the proof of the correction. Luckily for me, the corrected formula gives $972$ for the order of this group.

So, if you are interested in this problem, don't use the formula in Kovacs' paper, but get the copy from his website.

Let me make a conjecture about the structure of $F_{S_3}(n)$. I have verified this conjecture by hand + computer for $n =1, 2$, and it is consistent with Kovacs' corrected formula for all $n$. Moreover, the conjecture is so simple, that if someone pokes at it enough they will likely find a proof or disproof.

Let $\mathbb F_3$ be the $3$-element field and let $R = \mathbb F_3[\oplus^n \mathbb Z_2]$ be the group ring with coefficients in $\mathbb F_3$ of the rank $n$ free group, $\oplus^n \mathbb Z_2$, in the variety ${\mathcal V}(\mathbb Z_2)$ of elementary $2$-abelian groups.

Let $V=\left(\oplus^{n-1} R\right)\oplus \mathbb F_3$ be the $R$-module that is a direct sum of $n-1$ copies of the regular representation of $\oplus^n \mathbb Z_2$ over $\mathbb F_3$ along with one extra copy of the trivial representation. Additively $V$ is an elementary abelian $3$-group, but as an $R$-module it comes equipped with an action of $\oplus^n \mathbb Z_2$ by automorphisms.

The conjecture is that the resulting semidirect product $V\rtimes (\oplus^n \mathbb Z_2)$ is $F_{S_3}(n)$.

You can see from the description that the $\mathbb F_3$-dimension of $V$ is $(n-1)2^n+1$, and so the group has order $2^n\cdot 3^{(n-1)2^n+1}$. This is the modified Kovacs formula for the variety ${\mathcal V}(S_3)$. If the group I describe is $n$-generated, and if the modified Kovacs formula is correct, then the conjecture is true. I also note that the group I construct has the correct isomorphism type of its center and its quotient modulo its commutator subgroup.

This question recently popped up on the front page, so I entered $S_3$ in UACalc, and calculated the $2$-generated relatively free group $F_{S_3}(x,y)$. UACalc informed me that the order of this group is $972=2^2\cdot 3^5$. (It gave me the multiplication table too, but it was too big to be of much use to me.) I then worked out the internal structure of the group, using UACalc to check my intermediate calculations.

Next, I followed the zentralblatt link given in Keith Dennis's answer, to see if what I'd done could be found in Kovacs' paper. That paper predicts the order of $F_{S_3}(x,y)$ to be $6377292 = 2^2\cdot 3^{13}$, not $972$.

I started snooping around with google, and found a copy of Kovacs' paper on his website with a red marginal comment saying that his original formula is wrong. The marginal comment explains how the formula should be modified, but unfortunately the margin was too small to contain the proof of the correction. Lucky for me, the corrected formula gives $972$ for the order of this group.

So, if you are interested in this problem, don't use the formula in Kovacs' paper, but get the copy from his website.

Let me make a conjecture about the structure of $F_{S_3}(n)$. I have verified this conjecture by hand + computer for $n =1, 2$, and it is consistent with Kovacs' corrected formula for all $n$. Moreover, the conjecture is so simple, that if someone pokes at it enough they will likely find a proof or disproof.

Let $\mathbb F_3$ be the $3$-element field and let $R = \mathbb F_3[\oplus^n \mathbb Z_2]$ be the group ring with coefficients in $\mathbb F_3$ of the rank $n$ free group, $\oplus^n \mathbb Z_2$, in the variety ${\mathcal V}(\mathbb Z_2)$ of elementary $2$-abelian groups.

Let $V=\left(\oplus^{n-1} R\right)\oplus \mathbb F_3$ be the $R$-module that is a direct sum of $n-1$ copies of the regular representation of $\oplus^n \mathbb Z_2$ and one extra copy of the trivial representation. Additively $V$ is an elementary abelian $3$-group, but as an $R$-module it comes equipped with an action of $\oplus^n \mathbb Z_2$ by automorphisms.

The conjecture is that the resulting semidirect product $V\rtimes (\oplus^n \mathbb Z_2)$ is $F_{S_3}(n)$.

You can see from the description that the $\mathbb F_3$-dimension of $V$ is $(n-1)2^n+1$, and so the group has order $2^n\cdot 3^{(n-1)2^n+1}$. This is the modified Kovacs formula for the variety ${\mathcal V}(S_3)$. If the group I describe is $n$-generated, and if the modified Kovacs formula is correct, then the conjecture is true. I also note that the group I construct has the correct size for its center and its commutator subgroup.

This question recently popped up on the front page, so I entered $S_3$ in UACalc, and calculated the $2$-generated relatively free group $F_{S_3}(x,y)$. UACalc informed me that the order of this group is $972=2^2\cdot 3^5$. (It gave me the multiplication table too, but it was too big to be of much use to me.) I then worked out the internal structure of the group, using UACalc to check my intermediate calculations.

Next, I followed the zentralblatt link given in Keith Dennis's answer, to see if what I'd done could be found in Kovacs' paper. That paper predicts the order of $F_{S_3}(x,y)$ to be $6377292 = 2^2\cdot 3^{13}$, not $972$.

I started snooping around with google, and found a copy of Kovacs' paper on his website with a red marginal comment saying that his original formula is wrong. The marginal comment explains how the formula should be modified, but unfortunately the margin was too small to contain the proof of the correction. Luckily for me, the corrected formula gives $972$ for the order of this group.

So, if you are interested in this problem, don't use the formula in Kovacs' paper, but get the copy from his website.

Let me make a conjecture about the structure of $F_{S_3}(n)$. I have verified this conjecture by hand + computer for $n =1, 2$, and it is consistent with Kovacs' corrected formula for all $n$. Moreover, the conjecture is so simple, that if someone pokes at it enough they will likely find a proof or disproof.

Let $\mathbb F_3$ be the $3$-element field and let $R = \mathbb F_3[\oplus^n \mathbb Z_2]$ be the group ring with coefficients in $\mathbb F_3$ of the rank $n$ free group, $\oplus^n \mathbb Z_2$, in the variety ${\mathcal V}(\mathbb Z_2)$ of elementary $2$-abelian groups.

Let $V=\left(\oplus^{n-1} R\right)\oplus \mathbb F_3$ be the $R$-module that is a direct sum of $n-1$ copies of the regular representation of $\oplus^n \mathbb Z_2$ over $\mathbb F_3$ along with one extra copy of the trivial representation. Additively $V$ is an elementary abelian $3$-group, but as an $R$-module it comes equipped with an action of $\oplus^n \mathbb Z_2$ by automorphisms.

The conjecture is that the resulting semidirect product $V\rtimes (\oplus^n \mathbb Z_2)$ is $F_{S_3}(n)$.

You can see from the description that the $\mathbb F_3$-dimension of $V$ is $(n-1)2^n+1$, and so the group has order $2^n\cdot 3^{(n-1)2^n+1}$. This is the modified Kovacs formula for the variety ${\mathcal V}(S_3)$. If the group I describe is $n$-generated, and if the modified Kovacs formula is correct, then the conjecture is true. I also note that the group I construct has the correct size for its center and its commutator subgroup.