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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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  • $\begingroup$ 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? $\endgroup$ – M. Shahryari Jan 8 '14 at 13:44
  • $\begingroup$ 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 $F_V(x,y)\cong S_3\times C_6$. $\endgroup$ – Arturo Magidin Jan 9 '14 at 3:30
  • $\begingroup$ @: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 $S_3\times C_6$. $\endgroup$ – M. Shahryari Jan 9 '14 at 4:52
  • $\begingroup$ 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. $\endgroup$ – Arturo Magidin Jan 9 '14 at 5:04
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    $\begingroup$ For those who, like me, had no idea what "variety" means in this context or what a "relatively free" group is, see encyclopediaofmath.org/index.php/Variety_of_groups and encyclopediaofmath.org/index.php/Free_group — I think people asking a question should make the minimal effort of providing such links or explanations. $\endgroup$ – Gro-Tsen Sep 16 '18 at 11:50
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Let me provide by hand free groups in the variety $V_p$ generated by the dihedral group $D_{2p}$ of order $2p$, $p$ odd prime.

First observe that $D_{2p}$ satisfies the group identities $x^{2p}=1$, $[x^2,y^2]=1$.

In any group, let $G^2$ be the set of squares and $G_p$ the set of elements of order dividing $p$. For any group, $G_p\subset G^2$. For $G$ satisfying the identity $x^{2p}=1$, we have the reverse inclusion, and hence $G_p=G_2$.

In addition, the identity $[x^2,y^2]=1$ implies that for $G\in G_p$, any product of squares belongs to $G_p$, so the subgroup generated by $G^2(=G_p)$ is contained in $G_p$. This means that $G_p$ is a subgroup, obviously a normal elementary abelian $p$-subgroup. Since $G_p=G^2$, $G/G^2$ is a 2-group. If $G$ is finite, we deduce that $G=G_p\rtimes Q$, where $Q$ is any 2-Sylow subgroup, and $Q$ is elementary abelian (say of order $2^k$); for convenience write $Q=Q_k$.

The irreducible $\mathbf{F}_p[Q_k]$-modules are all 1-dimensional, given by the $2^k$ homomorphisms $\chi:Q\to C_2$; denote it by $V_\chi$. So we can write $G_p=\bigoplus_{\chi\in\mathrm{Hom}(Q,C_2)}V_\chi^{n_\chi}$.

Now let $G=G(k)$ be the free group of rank $k$ in this variety. Clearly $G/G^2$ is isomorphic to $Q_k$, so it remains to determine the multiplicities $n_\chi=n_\chi(k)$. The abelianization of $G(k)$ being isomorphic to $C_{2p}^k$, we have $n_0(k)=k$. We have a canonical homomorphism $\mathrm{Aut}(G)\to\mathrm{Aut}(Q_k)\simeq\mathrm{GL}_k(\mathbf{F}_2)$. It is surjective: indeed if $(x_1,\dots,x_k)$ is a basis of $G$, then mapping $x_i$ to $x_ix_j$ ($i\neq j$ fixed) and fixing all other basis elements, induces the corresponding elementary matrix, and these generate $\mathrm{GL}_k(\mathbf{F}_2)=\mathrm{SL}_k(\mathbf{F}_2)$. It follows that all $n_\chi(k)$, for $\chi\neq 0$, are equal, say to some number $n(k)$, to determine.

First, we have $n(k)<k$ for all $k\ge 1$. To show this, fix $\chi\neq 0$, and we have to check that $V_\chi^k\rtimes Q_k$ is not generated by $k$ elements. Modding out by the kernel of $\chi$, this amounts to showing it when $k=1$, i.e., $\mathbf{F}_p^k\rtimes_{\pm}C_2$ is not $k$-generated. Indeed, consider $k$ generators $u_1,\dots,u_k$. We can suppose that $u_1\notin \mathbf{F}_p^k$. Then, replacing $u_i$ with $u_iu_1$, we can suppose that $u_1\in \mathbf{F}_p^k$ for all $k\ge 2$. Then, for if $k\ge 2$, modding out by the subgroup generated by $u_k$ (which is normal) shows that $\mathbf{F}_p^{k-1}\rtimes_{\pm}C_2$ is $(k-1)$-generated. We can continue until $k=1$ and deduce that $\mathbf{F}_p\rtimes_{\pm}C_2$ is 1-generated, which is a contradiction since it is not abelian.

Now I claim that $n(k)=k-1$. That is,

the free group $G(k)$ on $k$-generators in the variety $V_k$ is isomorphic to $((\bigoplus_{\chi\neq 0}V_\chi^{k-1})\oplus V_0^k)\rtimes Q_k$, where $Q_k\simeq C_2^k$, and where $\chi$ is meant to range over $\mathrm{Hom}(Q_k,C_2)$, and $V_\chi$ is $\mathbf{F}_p$ endowed with the action $q\cdot v=\chi(q)v$ of $Q_k$.

Given the above, it remains (a) to prove that this group is $k$-generated, which implies that $G(k)$ is indeed free in the variety generated by the identities $x^{2p}$ and $[x^2,y^2]$, and that $G(k)$ indeed belongs to the variety generated by $D_{2p}$.

(b) is easy: indeed, given any nontrivial $g$ element of $G(k)$, we have to find a homomorphism $G(k)\to D_{2p}$ such that $g$ is not in the kernel. If $g$ does not belong to $G(k)_p$, this is clear (map onto $Q_k$, and then to an element of order 2). Otherwise there exists a quotient of $G(k)$ of the form $V_\chi\rtimes Q_k$, and killing the kernel of $\chi$, we get, if $\chi\neq 0$ $V_\chi\rtimes C_2\simeq D_{2p}$, and if $\chi=0$ we get $V_0\simeq C_p$.

(a) (1) $H_\chi:=V_\chi^{k-1}\rtimes Q_k$ is $k$-generated, by some $k$-tuple mapping onto the canonical basis of $Q_k=C_2^k$. First suppose that $\chi$ is the $k$-projection. Then this writes as $C_2^{k-1}\times(\mathbf{F}_p^{k-1}\rtimes_{\pm} C_2)$, which is generated by $(s_1q_1,s_2q_1,\dots,s_{k-1}q_{k-1},q_k)$, where $(s_i)$ is the canonical basis of $\mathbf{F}_p^{k-1}$ and $(q_i)$ is the canonical basis of $C_2^k$. In general, this yields a generating $k$-tuple mapping onto some basis of $C_2^k$ (usually not the original basis). Using elementary operation, we deduce a generating $k$-tuple of $H_\chi$ mapping exactly onto the canonical basis of $C_2^k$.

(2) $V_0^k\rtimes Q_k$ has the same property: this is obvious since it is isomorphic to $C_{2p}^k$.

We can conclude. For every $\chi$, we have a generating subset of $V_\chi^{n_\chi}$ of the form $((v_{\chi 1},q_1),\dots,(v_{\chi,k},q_k))$. Write $v_i=(v_{\chi,i})_\chi\in V=\bigoplus_\chi V_\chi^{n_\chi}$. I claim that $(v_1q_1,\dots,v_kq_k)$ generates $V\rtimes Q_k$. Since $(v_iq_i)^p=q_i$, the subgroup it generates contains $Q_k$, hence has the form $V'\rtimes Q_k$ with $V'$ some $Q_k$-submodule. Since $V'$ maps onto each isotypic component $V_\chi^{n_\chi}$, we have $V'=V$. So $G(k)\simeq (\bigoplus_{\chi\neq 0}V_\chi^{k-1}\oplus V_0^k)\rtimes Q_k$.

Note: Since $\bigoplus_\chi V_\chi$ is isomorphic to the regular representation $R_k$ (of $Q_k$ over $\mathbf{F}_p$), this can be rewritten as $(R_k^{k-1}\oplus V_0)\rtimes Q_k$ (which has order $p^{(k-1)2^k+1}2^k$). This confirms Keith Kearnes's conjecture, which motivated this additional answer.

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    $\begingroup$ @M.Shahryari thanks for the edits. However, the abelianization of the $G(k)$ is really $C_6^k$, not $C_2^k$. Indeed, $C_3^k$ belongs to the variety and is $k$-generated, so has to be a quotient of $G(k)$. $\endgroup$ – YCor Sep 23 '18 at 7:42
  • $\begingroup$ thank you, you are right. I am also not sure if the map $x_i\to x_ix_j$ (others are fixed) can be really extend to an automorphism. It seems true but why? $\endgroup$ – M. Shahryari Sep 23 '18 at 7:48
  • $\begingroup$ It can be extended to an endomorphism (by freeness). And so does the map $x_i\mapsto x_ix_j^{-1}$ (others fixed). Composing these maps in both directions yields the identity map. $\endgroup$ – YCor Sep 23 '18 at 7:58
  • $\begingroup$ One more question: you mean $C_{2p}^k$ not $C_6^k$, am I right? $\endgroup$ – M. Shahryari Sep 23 '18 at 10:26
  • $\begingroup$ Sure, it's $C_{2p}$. I had $p=3$ in mind since the general case is exactly the same. $\endgroup$ – YCor Sep 23 '18 at 10:32
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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).


Edit.

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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  • $\begingroup$ It seems very nice, do you mean the group of all term functions with coefficients from $G$? can I have a reference? $\endgroup$ – M. Shahryari Jan 9 '14 at 2:16
  • $\begingroup$ 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. $\endgroup$ – M. Shahryari Jan 9 '14 at 2:42
  • $\begingroup$ @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 $x(12)x^2(123)$ $\endgroup$ – Anton Klyachko Jan 9 '14 at 12:29
  • $\begingroup$ 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 $S_3$. $\endgroup$ – M. Shahryari Jan 9 '14 at 12:51
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    $\begingroup$ 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). $\endgroup$ – Benjamin Steinberg Jan 9 '14 at 16:11
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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.

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See

Kovács, L. G., Free groups in a dihedral variety, Proc. R. Ir. Acad., Sect. A 89, No. 1, 115-117 (1989). ZBL0697.20012.

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    $\begingroup$ Would you summarize the contents of the link for those who do not have access? $\endgroup$ – András Bátkai Sep 11 '18 at 17:53
  • $\begingroup$ @AndrásBátkai: It does not strike me as particularly useful; it states that if you have a dihedral group of order $2^{d+1}e$ with $e$ odd, then you can view it as a subdirect product of a dihedral group of order $2e$ and one of order $2^{d+1}$; if $\mathfrak{U}$ is the variety generated by the first, and $\mathfrak{B}$ the variety generated by the second, and $\mathfrak{V}$ is the variety generated by your original dihedral, then $F_r(\mathfrak{V})$ is "the subdirect product of $F_r(\mathfrak{U})$ and $F_r(\mathfrak{B})$ amalgamating precisely $F_r(\mathfrak{A}_2)$ (cont) $\endgroup$ – Arturo Magidin Sep 11 '18 at 22:18
  • $\begingroup$ (cont) where $\mathfrak{A}_2$ is the variety of abelian groups of exponent $2$. But in this case, it just says that the relatively free group of rank $r$ in $\mathrm{Var}(S_3)$ is a subdirect product of the free group of rank $r$ in $\mathrm{Var}(S_3)$ and the $C_2^r$, amalgamating precisely the latter, i.e., it says nothing at all. As far as structure, it further notes that $\mathrm{Var}(S_3) = \mathrm{A}_3\mathrm{A}_2$. This suffices to calculate the size (the objective of the paper), but not to describe the group, I think.... $\endgroup$ – Arturo Magidin Sep 11 '18 at 22:22

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