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Suppose $S_3$ is the symmetric group of order 6. Which elements of the variety $Var(S_3)$ are relatively free?

This question is related to my previous question Relatively free algebras in a variety generated by a single algebra

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In the first look it seems as a trivial question. If $V=Var(S_3)$, then clearly we have $F_V(x)=C_6$ the cyclic group of order 6. It maybe also true that $F_V(x, y)=S_3\times S_3\times C_6$. Now, what about $F_V(x_1, \ldots, x_n)$? If $G$ is and arbitrary group and $V=Var(G)$, then how we can express $F_V(x_1, \ldots, x_n)$ in terms of $G$ and known groups? – M. Shahryari Jan 8 '14 at 13:44
I think that your second question is in general difficult except for special cases, such as critical groups; for $G$ a finite nonabelian simple group, for example, it is not hard to show that the finitely generated groups in $\mathrm{Var}(G)$ is of the form $G^n\times K$, where $K\in\mathrm{Var}((\mathbf{HS}-1)(G))$ (see for example the proof of Lemma 3.2 in Sheila Oates's "Identical relations in groups", J. London Math. Soc. 38 (1963), 71-78). $S_3$ is critical (since the proper subfactors are all abelian) so perhaps something similar can be done. I expect $F_V(x,y)\cong S_3\times C_6$. – Arturo Magidin Jan 9 '14 at 3:30
@:Arturo Magidin: $F_V(x_1, \ldots, x_n)$ most be the largest $n$-generator element of $V$. So, $F_V(x,y)$ is not $S_3\times C_6$, since we have also $C_6\times C_6\in V$ which is not a quotient of $S_3\times C_6$. – M. Shahryari Jan 9 '14 at 4:52
Fair enough, but it must also be $2$-generated, so it cannot be $S_3\times S_3\times C_6$, which cannot be generated by $3$ elements. So it cannot be of the form $S_3^n\times A$ for some abelian $A$. It must involve some different nonabelian group. Of course, $V$ is finitely based by Powell-Oats, so in principle you would be able to write down a basic set of laws and figure it out explicitly. – Arturo Magidin Jan 9 '14 at 5:04
Likely the next candidate for the free 2-algebra is $S_3^2 \times C_3^2$, or some similar product. – The Masked Avenger Jan 9 '14 at 6:29
up vote 4 down vote accepted

The relatively free group $F_{var(G)}(x_1,\dots,x_n)$ is isomorphic to the group of all polynomial functions $G^n\to G$, where a function is called polynomial if it can be expressed via the multiplication and inverses of its arguments; the polynomial functions form a group with respect to the pointwise multiplication.

$F_{var(S_3)}(x,y)$ is not $S_3\times S_3\times C_6$ because the latter group is not two-generated (since it maps onto the elementary abelian group of order 8).


Theorem. The group $F$ of polynomial functions $G^n\to G$ is the relatively free group in $var(G)$ of rank $n$. A free basis of $F$ consists of the functions $f_1(x_1,\dots,x_n)=x_1,\dots,f_n(x_1,\dots,x_n)=x_n$.

Proof. Clearly, $F\in var(G)$. Suppose that we have a relation $w(f_1,\dots,f_n)=1$ in $F$. By definition, this means that the the function the fuction $G^n\to G$ sending $(g_1,\dots,g_n)$ to $w(g_1,\dots,g_n)$ is the constant function identitically equal to 1. Thus, $w(x_1,\dots,x_n)=1$ is an identity (law) in the group $G$. This completes the proof.

Example. The rank-one group $F_{var(S_3)}(x)$ consists of the following 6 functions from $S_3$ to $S_3$: $$ x\mapsto 1,\ x\mapsto x,\ x\mapsto x^2,\ x\mapsto x^3,\ x\mapsto x^4,\ x\mapsto x^5. $$ Note that, according to the definition above, we cannot use constants in formulas for polynomial functions; so, for instance, the function $x\mapsto (12)x$ is not polynomial.

Similarly, the $F_{var(S_3)}(x,y)=\{1,x,y,xy,x^2y,\dots\}$ but I do not know how many different polynomyal functions is there and so I do knot know the order of this group (though this is a question of direct calculation).

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It seems very nice, do you mean the group of all term functions with coefficients from $G$? can I have a reference? – M. Shahryari Jan 9 '14 at 2:16
By a search in web, I learned that the polynomial function groups have usually large orders. For example in the case of $G=S_3$, there are 324 polynomial functions of one variable and if you are right, then we should have $|F_V(x)|=324$. But as I said above, $F_V(x)=C_6$. So, I am not sure that your answer is correct. There may be a gap. – M. Shahryari Jan 9 '14 at 2:42
@M.Shahryari, see the edit. I guess that 324 is the number of one -variable polynomial functions over $S_3$ in a different sense of the word polynomial; propably they mean polynomials with coefficients from the group --- such as $x(12)x^2(123)$ – Anton Klyachko Jan 9 '14 at 12:29
Thank you for the completed answer. So, you mean the coefficient-free polynomial functions while I saw the number of one-variable polynomial functions with coefficients from $S_3$. – M. Shahryari Jan 9 '14 at 12:51
Related: if you are allowed to use constants, then Mauer-Rhodes showed that a finite group $G$ is simple non-abelian iff for all n, every mapping $G^n\to G$ is realizable via a polynomial (with constants). – Benjamin Steinberg Jan 9 '14 at 16:11

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