\begin{document}

\maketitle
\begin{abstract}
Let $G$ be a finite nilpotent group of class at most $2$, and let $w=w(x_1,\ldots,x_n)$
be a group word in $n$ variables. Then we prove that the number of solutions in $G\times\overset{n}{\cdots}\times G$ to the equation $w=1$ is at least $|G|^{n-1}$. This result, also independently obtained bt Matthew Levy [L], solves a special case of a conjecture of Alon Amit.
\end{abstract}


\section{Introduction}
If $w=w(x_1,\ldots,x_n)$ is a group word, then for every group $G$ we have a
verbal mapping
\[
\begin{matrix}
w & : & G^{(n)} & \longrightarrow & G
\\
& & (g_1,\ldots,g_n) & \longmapsto & w(g_1,\ldots,g_n)
\end{matrix}
\]
by substitutions of the $n$-tuples.\\

We use the notation $N(G,w)$ for the number of solutions to the equation $w=1$ in
the group $G$, that is, for the cardinality of $w^{-1}(1)$.
More generally, if $S$ is a subset of $G^{(n)}$, we write $N(S,w)$ for the number of
solutions to $w=1$ lying in $S$.
Thus $N(G,w)$ can be seen as a shortcut for $N(G^{(n)},w)$. \\

If $G$ is abelian, then the verbal mapping $w:G^{(n)}\rightarrow G$ is a
homomorphism and so each non-empty fibre $w^{-1}(g)$ is a coset of the kernel
$w^{-1}(1)$. Besides, since $G$ is also finite (a) $\left\vert w^{-1}
(1)\right\vert \geq\left\vert G^{(n)}\right\vert /\left\vert G\right\vert $
and (b) all non-empty fibres have the same size. If $G$ is finite and nilpotent, but not abelian, then (b) is no longer true in
general. Whether (a) still holds is an open problem (that it does is
conjectured by Amit [A]). The aim of this paper is to show that the conjecture of Amit is true for finite groups of nilpotency class at most 2.\\

 \begin{thm}
 Let $G$ be a finite group of nilpotency class at most $2$, and let $w=w(x_1,\ldots,x_n)$
be a group word. Then $N(G,w)\ge |G|^{n-1}$.
 \end{thm} 
 If the theorem holds for two groups $A$ and $B$of nilpotency class at most 2, then it holds for their direct product $G = A\times B$ since the word values can be solved componentwise.
 
  Let $\g = \a\cdot \b=(a_1, ..., a_n)\cdot (b_1, ..., b_n)\in A^{(n)}\times B^{(n)}=G^{(n)}$, then  
$w(\g) = w(\a).w(\b)$ which implies $N(G,w)\geq N(A,w).N(B,w)$
and the result follows from $|G| = |A|.|B|$. 
 Any finite nilpotent group is a direct product of its Sylow subgroups. Hence it will be enough to show the theorem for $p$-groups:
 
 \begin{thm}
 Let $G$ be a finite $p$-group of nilpotency class at most $2$, and let $w=w(x_1,\ldots,x_n)$
be a group word. Then $N(G,w)\ge |G|^{n-1}$.
 \end{thm} 
 
This result was also independently obtained bt Matthew Levy [L].
\section{Useful lemmas}

If $w_1,\ldots,w_k$ is a finite family of words in the same
variables $x_1,\ldots,x_n$, then we can consider the mapping
\[
\begin{matrix}
f_{w_1,\ldots,w_k} & : & G^{(n)} & \longrightarrow & G^{(k)}
\\
& & (g_1,\ldots,g_n) & \longmapsto & (w_1(g_1,\ldots,g_n),\ldots,w_k(g_1,\ldots,g_n)).
\end{matrix}
\]
On the other hand, if $w=w(x_1,\ldots,x_n)$ and $u_1,\ldots,u_n$ are words in the
same variables $y_1,\ldots,y_k$, then we can define the \emph{composition\/}
$w(u_1,\ldots,u_n)$, which results from substituting each word $u_i$ for the indeterminate
$x_i$ in the expression for $w$.
In other words, if $F$ is the free group on $x_1,\ldots,x_n$ and $F^*$ is the free group
on $y_1,\ldots,y_k$, then $w(u_1,\ldots,u_n)$ is the image of $w$ under the homomorphism
$F\rightarrow F^*$ sending each $x_i$ to $u_i$.

\begin{lem}
\label{reduction}
Let $w=w(x_1,\ldots,x_n)$ and $u_1,\ldots,u_n$ be group words, and put
$w'=w(u_1,\ldots,u_n)$.
If $G$ is a group for which the map $f_{u_1,\ldots,u_n}$ is bijective,
then we have $N(S,w')=N(S,w)$ for every subset $S$ of $G^{(n)}$.
\end{lem}


\begin{proof}
This is clear, since $f_{w'}$ is the composition of $f_{u_1,\ldots,u_n}$ and
$f_w$.
\end{proof}


The previous lemma will be used in conjunction with the following one.
Recall that, if $G$ is a finite $p$-group and $m$ is an integer which is not
divisible by $p$, then the map $g\mapsto g^m$ is bijective.
This is an immediate consequence of B\'ezout's identity.


\begin{lem}
\label{equivalent}
Let $u_1,\ldots,u_n$ be words of the form $u_i=x_i^{m_i} v_i$, where $v_i$ is
a word which only depends on $x_{i+1},\ldots,x_n$.
If $G$ is a finite $p$-group and $(m_i,p)=1$ for every $i=1,\ldots,n$, then
the mapping $f_{u_1,\ldots,u_n}$ is bijective on $G^{(n)}$.
\end{lem}


\begin{proof}
Since $f_{u_1,\ldots,u_n}$ goes from $G^{(n)}$ to $G^{(n)}$, it suffices to prove
that it is an injective map.
If $(g_1,\ldots,g_n)$ and $(h_1,\ldots,h_n)$ are two tuples with the same image
under $f_{u_1,\ldots,u_n}$, we prove that $g_i=h_i$ for all $i=1,\ldots,n$ by reverse
induction on $i$.
For $i=n$ this is an immediate consequence of the remark preceding this lemma,
since the word $u_n$ is of the form $x_n^{m_n}$ with $(m_n,p)=1$.
Assume now that $g_{i+1}=h_{i+1},\ldots,g_n=h_n$, and let us see that $g_i=h_i$.
Since $u_i(g_i,\ldots,g_n)=u_i(h_i,\ldots,h_n)$ and $u_i=x_i^{m_i} v_i$ for a word
$v_i$ involving only $x_{i+1},\ldots,x_n$, it follows that $g_i^{m_i}=h_i^{m_i}$,
and we conclude that $g_i=h_i$ as before.
\end{proof}


Of course, the previous lemma admits variations which follow by reordering the
variables.
For example, if $n=3$ and $u_1=x_1$, $u_2=x_2x_1$, $u_3=x_3x_2$, then the map
$f_{u_1,u_2,u_3}$ is bijective on $G^{(3)}$.


\vspace{5pt}


The following result will also be used on several occasions.
The proof is obvious.


\begin{lem}
\label{simpler word}
Let $w$ be a group word in the free group $F$, and let $G$ be a finite $p$-group
of order $p^n$ and class $c$.
If $w'$ is a word such that $w\equiv w' \pmod{\gamma_{c+1}(F)F^{p^n}}$, then
$f_w=f_{w'}$ on $G^{(n)}$.
\end{lem}


As a consequence, if $G$ is a finite $p$-group of class $2$ and we want to study the
number $N(G,w)$ for a group word $w$, we may assume without loss of generality that
$w$ is of the form
\[
x_1^{k_1}\ldots x_n^{k_n} \prod_{1\le i<j\le n} \, [x_i,x_j]^{k_{ij}}.
\]
Also, if $w=u^k v$ for some words $u$ and $v$ and for some exponent $k$, then we
may replace $k$ with any other integer $k'$ such that $k\equiv k'\pmod{|G|}$.


\vspace{5pt}


In the next theorem we consider a particular kind of words, which will be crucial
for the proof of our main result.
We need the following very simple lemma.


\begin{lem}
\label{[g,G]}
Let $G$ be a nilpotent group of class at most $2$, and let $g$ be an element of $G$.
Then $[g,G]=\{[g,h]\mid h\in G\}$, and every element of $[g,G]$ can be represented
in $|C_G(g)|$ different ways in the form $[g,h]$ with $h\in G$.
Hence $|[g,G]|=|G:C_G(g)|$.
\end{lem}


\begin{proof}
The equality $[g,G]=\{[g,h]\mid h\in G\}$ follows from the fact that the commutator
is multilinear in a nilpotent group of class at most $2$.
On the other hand, since
\[
[g,h_1]=[g,h_2] \Longleftrightarrow g^{h_1}=g^{h_2} \Longleftrightarrow
h_1h_2^{-1}\in C_G(g) \Longleftrightarrow C_G(g)h_1=C_G(g)h_2,
\]
the elements which give the same commutator with $g$ form a right coset of $C_G(g)$.
Thus the number of different representations of a commutator with $g$ coincides
with the order of $C_G(g)$.
\end{proof}

\section{Proof of the theorem}
Let's first show the proof for a particular word which is the main step in the main proof:
\begin{thm}
\label{particular case}
Let $G$ be a finite $p$-group of class at most $2$, and let $w$ be a group word
of the form
\begin{equation}
\label{particular word}
w = x_1^m \prod_{2\le i<j\le n} \, [x_i,x_j]^{k_{ij}}.
\end{equation}
Then $N(G'\times G^{(n-1)},w)\ge |G|^{n-1}$.
\end{thm}


\begin{proof}
We argue by induction on $n$, the case $n=1$ being trivial.
Due to the form of the word $w$, it is impossible to have $n=2$, so we suppose
now that $n\ge 3$.
In particular, we may assume that not all the exponents $k_{ij}$ are $0$.
Choose the pair $(r,s)$ such that the $p$-part of the exponent $k_{rs}$ is
as small as possible.
By renaming the variables, we may assume that $r=n-1$ and $s=n$.
For simplicity, let us write $k$ for $k_{n-1,n}$.
For every pair $(i,j)$ with $2\le i<j\le n$, let $k'_{ij}$ be a solution
to the congruence
\[
k'_{ij}k\equiv k_{ij} \pmod{|G|}.
\]
(Such a solution exists by the choice of $(r,s)$, and because $|G|$ is a power
of $p$.)
By the remark preceding Lemma 5, we may assume that
\[
w = x_1^m \prod_{2\le i<j\le n} \, [x_i,x_j]^{k'_{ij}k}.
\]
In the previous expression, we collect all commutators which contain either
$x_{n-1}$ or $x_n$, and get
\[
w = x_1^m v [x_{n-1},x_n]^k [x_{n-1},v_{n-1}]^k [v_n,x_n]^k,
\]
for some words $v$, $v_{n-1}$, and $v_n$ involving only the variables $x_i$
with $2\le i\le n-2$.
Thus
\[
w = x_1^m v' [x_{n-1}v_n,x_nv_{n-1}]^k,
\]
where $v'=[v_{n-1},v_n]^k v$ again only involves $x_2,\ldots,x_{n-2}$.
Now, by using Lemmas 3 and 4 with $u_{n-1}=x_{n-1}v_n$,
$u_n=x_nv_{n-1}$ and $u_i=x_i$ for $i=1,\ldots,n-2$ (see the remark after Lemma
4), we get
$N(G'\times G^{(n-1)},w)=N(G'\times G^{(n-1)},w')$ for the word
\[
w' = x_1^m v' [x_{n-1},x_n]^k = x_1^m v' [x_{n-1}^k,x_n].
\]

Since $w''=x_1^m v'$ is of the form in the statement of the theorem and involves $n-2$
variables, the result is true for $w''$.
On the other hand,
\begin{align*}
N(G'\times G^{(n-1)}
&,w)
\\
&=
\big| \{ (g_1,\ldots,g_n)\in G'\times G^{(n-1)}
\mid w''(g_1,\ldots,g_{n-2})=[g_{n-1}^k,g_n]^{-1} \} \big|.
\end{align*}
By Lemma 6, each element of the commutator subgroup $[g_{n-1}^k,G]$ is
of the form $[g_{n-1}^k,g_n]$, and there are $|C_G(g_{n-1}^k)|$ choices of $g_n$
giving rise to the same element.
Consequently,
\begin{align*}
&N(G'\times G^{(n-1)},w) =
\\
& \sum_{g_{n-1}\in G} \, |C_G(g_{n-1}^k)| \,
\big| \{ (g_1,\ldots,g_{n-2})\in G'\times G^{(n-3)}
\mid w''(g_1,\ldots,g_{n-2})\in [g_{n-1}^k,G] \} \big|.
\end{align*}
Now, if we put $\overline G=G/[g_{n-1}^k,G]$, we have
\begin{multline*}
\big| \{ (g_1,\ldots,g_{n-2})\in G'\times G^{(n-3)}
\mid w''(g_1,\ldots,g_{n-2})\in [g_{n-1}^k,G] \} \big|
\\
=
|[g_{n-1}^k,G]|^{n-2} N(\overline G'\times \overline G^{(n-3)},w'').
\end{multline*}
Since the result is true for $w''$, we get
\begin{align*}
N(G'\times G^{(n-1)},w)
&\ge
\sum_{g_{n-1}\in G} \, |C_G(g_{n-1}^k)| \, |[g_{n-1}^k,G]|^{n-2} \, |\overline G|^{n-3}
\\
&=
\sum_{g_{n-1}\in G} \, |C_G(g_{n-1}^k)| \, |[g_{n-1}^k,G]| \, |G|^{n-3}
\\
&=
\sum_{g_{n-1}\in G} \, |G|^{n-2}
=
|G|^{n-1},
\end{align*}
where we have used again Lemma 6.
\end{proof}


\begin{thm}
\label{general case}
Let $G$ be a finite $p$-group of class at most $2$, and let $w=w(x_1,\ldots,x_n)$
be a group word.
Then $N(G,w)\ge |G|^{n-1}$.
\end{thm}


\begin{proof}
By working in the abelian group $F/F'$, we can write $w=(x_1^{k_1}\ldots x_n^{k_n})^m v$,
where $v\in F'$, $m$ is a power of $p$, and at least one $k_i$ is not divisible by $p$.
After renaming the variables if necessary, we may assume that $i=1$.
If we put
\[
u_1=x_1^{k_1}\ldots x_n^{k_n},\ u_2=x_2,\ \ldots \ ,\ u_n=x_n,
\]
then we can apply Lemmas 3 and 4 to obtain $N(G,w)=N(G,w')$,
where now $w'$ is of the form $w' = x_1^m v'$, for some word $v'\in F'$.
Since $G$ has class at most $2$, we may assume that $v'$ is of the form
\[
\prod_{1\le i<j\le n} \, [x_i,x_j]^{k_{ij}}.
\]
We can write $v'=[x_1,v_1']v_2'$, where $v_1'$ and $v_2'$ depend only on $x_2,\ldots,x_n$,
and where $v_2'\in F'$.
If we put $w''=x_1^m v_2'$, then $N(G,w)\ge N(G'\times G^{(n-1)},w'')$.
Now the result follows by applying Theorem 7.
\end{proof}
\textbf{References}
\begin{flushleft}
[A] A. Amit, On equations in nilpotent groups, unpublished\medskip


[L] M. Levy, On the probability of satisfying a word in nilpotent groups of class 2, \emph{http://arxiv.org/abs/1101.4286}\medskip


\end{flushleft}


\end{document}
{"config": "arxiv", "file": "1406.5395.tex"}