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](https://mathoverflow.net/questions/153743/relatively-free-algebras-in-a-variety-generated-by-a-single-algebra)



relatively free groups in $Var(S_3)$
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.


