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Let $G$ be a finite soluble group, let $x$ be a non-identity element of $G$, let $v$ be an $n$-tuple in $G$, and let $w$ be a word in $n+2$ variables. Does there exist $(G,x,v,w)$ satisfying the following equations?

$w(1,1,v)=w(x,1,v)=w(1,x,v)=1; w(x,x,v)=x.$

As for why the soluble condition is there: let $w(a,b,c) = [a,c^{-1}bc]$, and suppose we have chosen $x,v \in G$ such that $[x,v^{-1}xv]=x$. This seems quite a plausible equation in a simple group (at first glance, anyway) but obviously has no non-trivial solutions in a soluble group, because the left-hand side is further down the derived series than the right-hand side. But perhaps a more complicated word does not have this limitation.

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up vote 3 down vote accepted

After thinking about it some more, I realise it can't be done. Suppose $x$ is contained in a normal subgroup $M$ of $G$, but not in $M'=[M,M]$. Then modulo $M'$ we have

$w(x,1,v) = w(1,1,v)m_1; w(1,x,v) = w(1,1,v)m_2; w(x,x,v) = w(1,1,v)m_1m_2$

where $m_1,m_2 \in M$ (just 'pull through' all occurrences of the first two variables up to conjugacy, eg $xd$ can be rewritten $dx^d$). The first set of equations force $m_1 = m_2 = 1$ (all modulo $M'$), so $w(x,x,v)$ is in $M'$, in particular $w(x,x,v)$ does not equal $x$.

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Which is a good thing, because if it could be done, it would follow from the usual proof of Barrington’s theorem (and the Barrington–Thérien classification) that $\mathrm{NC}^1=\mathrm{TC}^0=\mathrm{AC}^0[m]$ for some $m$. – Emil Jeřábek Jun 14 '14 at 18:17

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