# A note on words having same image on finite groups

Shrinit Singh  Department of Mathematics, Shiv Nadar Institution of Eminence,
Greater Noida, Dadri-201314, India<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

The question of whether two words in a free group that induce the same measure
on every finite group as word maps are automorphic remains open. In this work,
we study words whose images as word maps on every finite group are identical.
We establish that two words in $F_{2}$, where one of the words is $x^{n}$ or
$[x,y]^{n}$ for $n\in{\mathbb{Z}}$, having same image on every finite group,
are endomorphic to each other. Furthermore, we demonstrate that if the word
map corresponding to a word $w_{1}\in F_{n}$ has the same image as a test word
$w_{2}\in F_{n}$ on every finite group, then this is sufficient to ensure that
$w_{1}$ and $w_{2}$ induce the same measure on every finite group.

###### Key words and phrases:

free group, word map

###### 2020 Mathematics Subject Classification:

Primary 20F10

##  1\. Introduction

Let $F_{d}$ be a free group on $d-$generators. An element $w$ of the free
group is said to be a word and has an expression
$\prod_{j=1}^{s}x_{i_{j}}^{a_{j}}$, where
$i_{j}\in\\{1,2,\ldots,d\\},a_{j}\in{\mathbb{Z}}\setminus\\{0\\}$ and for
every $j\in\\{1,2,\ldots,s\\},$ $x_{i_{j}}\neq(x_{i_{j+1}})^{-1}.$ For any
group $G$, we use $G^{(n)}$ to denote direct product of $n$ copies of $G.$
Corresponding to the word $w\in F_{d}$ as above, we define a word map $w$ on a
group $G$ to be a map from $G^{d}$ to $G$ as follows
$w(g_{1},\ldots,g_{d})=\prod_{j=1}^{s}g_{i_{j}}^{a_{j}}.$ The image of the
word map $w$ on $G$ is denoted by $G_{w}.$

Let $w_{1},w_{2}\in F_{d}$ induce word maps on a finite group $G$. We say that
$w_{1}$ and $w_{2}$ induce same measure on $G$ if cardinality of
$w_{1}^{-1}(g)$ and $w_{2}^{-1}(g)$ for every $g\in G$ is same. Otherwise, we
say that they induce different measure. Two words $w_{1}$ and $w_{2}$ are
automorphic if there exists an automorphism $\phi$ of free group $F_{d}$ such
that $\phi(w_{1})=w_{2}.$ It is easy to note that one cannot distinguish
automorphic words just based on their measures on finite groups. A word $w\in
F_{d}$ is said to be profinitely rigid if a word $u\in F_{d}$ induces the same
measure as $w$ on every finite group, then $u$ must be automorphic to $w$.

Finite groups have been studied via word maps [3, 5, 7]. On the other hand,
one can talk about relation between words if corresponding word maps on every
finite group behave similarly. In this direction, the following conjecture is
well known:

######  Conjecture 1.1.

[2, 10] Every word in $F_{d}$ for $d\geq 2$, is profinitely rigid.

The conjecture, if true, suggests that for any finite set of words of a free
group in which no element from the set is automorphic to any other element in
the set, one can find a finite group where different words will induce
different measures. It is straightforward that every word in $F_{1}$ is
profinitely rigid. Although the conjecture is still open for $F_{d}$, $d\geq
2$, but recently, there has been developments in this direction. Puder and
Parzanchevski [8] proved that primitive words in any free group are
profinitely rigid. Recently, Hanany, Meiri and Puder [6] proved that
$x_{1}^{n}$ and $[x_{1},x_{2}]^{n}$, where $n\in{\mathbb{Z}}$, are profinitely
rigid in every free group $F_{d}$ for $d\geq 2$ .

In this article, we focus on the image of a word map on every finite group.
For example, let $w_{1}=x_{1}$ and $w_{2}=x_{1}^{2}x_{2}x_{1}^{-1}x_{2}^{-1}$
be two words in $F_{2}$, they have the same image in every group. But they
induce different measures on symmetric group $S_{3}$. By [8], we get that
$w_{1}$ and $w_{2}$ are not automorphic. The main goal of this article is to
understand relationship between words which have same image on every finite
group and under what condition these words will induce identical measure on
every finite group.

In this regard, we propose an intuitive question backed by some results from
this article. But before the question, let’s setup some definitions:

######  Definition 1.2.

Two elements $g$ and $h$ in a group $G$ are endomorphic equivalent if there
exists two endomorphisms $\phi_{1}$ and $\phi_{2}$ of $G$ such that
$\phi_{1}(g)=h$ and $\phi_{2}(h)=g.$

######  Lemma 1.3.

Two words $w_{1}$ and $w_{2}$ in a free group $F_{d}$ are endomorphic
equivalent if and only if they have same image as word maps on every group.

######  Definition 1.4.

We call a word $w\in F_{d}$ weakly profinitely rigid if any word $u\in F_{d}$
that have same image as the word $w$ on every finite group, then $w$ and $u$
are endomorphic equivalent.

In free group $F_{1}=\langle x_{1}\rangle$, any two words $x_{1}^{m}$ and
$x_{1}^{n}$ have same images on every finite group if and only if $m=\pm n$.
Then words $x_{1}^{m}$ and $x_{1}^{n}$ are automorphic. So every word in
$F_{1}$ is weakly profinitely rigid. Hence any two words in $F_{1}$ which have
same image on every finite group induce same measure on every finite group.
Although it is important to mention that it is not known that profinitely
rigidity of a word implies its weak profinite rigidity. We have the following
question:

######  Question 1.5.

Is every word in $F_{d}$ for $d\geq 2$ weakly profinitely rigid?

A positive answer to the above question implies that words can be
distinguished up to endomorphism equivalent if the image of word map on some
finite group is different. It would also affirmatively answer [11, Question
18], indicating that for a chiral word, there exists a finite group on which
the chiral word will exhibit chirality.

In the next section, we generalise weak profinite rigidity to elements of
arbitrary finitely generated group. We establish an equivalent notion of two
elements in a finitely generated group have same image on every finite group.

In the third section, we examine a word in a free (profinite) group that has
the property of being fixed by any (continuous) endomorphism, which then
necessarily becomes a (continuous) automorphism. Such words are known as test
words. We demonstrate that test words in a free group remain test words in its
profinite completion, where the free group is identified with its natural
image in its profinite completion.

In the final section, we explore weak profinite rigidity among words in free
groups. We proved that primitive words are weakly profinitely rigid. It is
worth noting that while the weak profinite rigidity of the word
$x_{1}^{2}x_{2}x_{1}^{-1}x_{2}^{-1}$ is established, its profinite rigidity is
unknown. For a given test word $w\in F_{n}$, we establish that any word $u\in
F_{n}$ having the same image as $w$ on every finite group implies that $w$ and
$u$ induce same probability measure on every finite group. We also show that
the notion of weak profinite rigidity coincides with that of profinite
rigidity for test words. As a result, the word $[x,y]^{d}$ where
$d\in{\mathbb{Z}}$ in $F_{2}$ is weakly profinitely rigid.

##  2\. weak Profinite Rigidity and its equivalent notions

We will prove lemma 1.3 which will be a quick corollary to the following
lemma.

######  Lemma 2.1.

Let $w$ and $w^{\prime}$ be two words in a free group $F_{d}$. There exists an
endomorphism $\phi$ of $F_{d}$ such that $\phi(w)=w^{\prime}$ if and only if
$G_{w^{\prime}}\subseteq G_{w}$ for any group $G.$

###### Proof.

Given $\phi(w)=w^{\prime}$ implies that
$w(\phi(x_{1}),\ldots,\phi(x_{d}))=w^{\prime}$. Now if
$g=w^{\prime}(g_{1},\ldots,g_{d})\in G_{w^{\prime}}$ where $g,g_{i}\in G$ for
$1\leq i\leq d$ then
$w(\phi(x_{1})(g_{1},\ldots,g_{d}),\ldots,\phi(x_{d})(g_{1},\ldots,g_{d}))=g$
and hence $G_{w^{\prime}}\subseteq G_{w}$.

To prove the converse part, let $G=F_{d},$ then $w^{\prime}\in(F_{d})_{w}$.
This implies that there exists $w_{1},\ldots,w_{d}\in F_{d}$ such that
$w(w_{1},\ldots w_{d})=w^{\prime}$. Sending $x_{i}$ to $w_{i}$ for $1\leq
i\leq d$ will give an endomorphism $\phi$ of $F_{d}$ with the property
$\phi(w)=w^{\prime}.$ ∎

If $w,w^{\prime}\in F_{d}$ are endomorphic equivalent then
$G_{w}=G_{w^{\prime}}$ for every group $G.$ It is clear that two words which
have same image on every group are endomorphic equivalent. We want to study
the problem whether two words which have same image on every finite group will
be endomorphic equivalent. To restrict the class of all finite group, we give
an easy lemma:

######  Lemma 2.2.

Let $w,u\in F_{d}.$ $w$ and $u$ have same image on every finite group if and
only if they have same image on every $d$-generated finite group.

###### Proof.

Suppose $w$ and $u$ have same image on every $d$-generated finite group. Let
$G$ be any finite group and $w(g_{1},\ldots,g_{d})=g\in G_{w}$. Then
$H=\langle g_{1},\ldots,g_{d}\rangle$ is $d$-generated finite group. And
$H_{w}=H_{u}$ implies $g\in H_{u}\subseteq G_{u}.$ ∎

It suffices to consider the weak profinite rigidity of a word $w\in F_{d}$
only within the class of $d$-generated finite group. Profinite completion is a
natural process that captures information of its finite quotient. Since all
$d$-generated finite groups are quotients of the $d$-generated free group
$F_{d}$, we need to transition from the free group to its profinite
completion. To do this, we generalise the notion of word map corresponding to
elements in arbitrary finitely generated group.

Given any $d$-tuple $g_{1},\ldots,g_{d}$ in $G$, there is a homomorphism
$\phi$ from $F_{d}$ to $G$ defined by $\phi(x_{i})=g_{i}$ for $1\leq i\leq d$
and conversely, any such homomorphism arises from a $d$-tuple in $G$ due to
the universal property of free groups. For $w\in F_{d},$ we have
$w(g_{1},\ldots,g_{d})=\phi(w).$

Let $P$ be a finitely generated group and $G$ any group. Any $\gamma\in P$
corresponds to a word map on $G,$ and the image of the word map $\gamma$ on
$G$ is denoted by $G_{\gamma}$. Let $\mathrm{Hom}(P,G)$ be the set of all
homomorphisms from $P$ to $G$, then we define
$G_{\gamma}=\\{\phi(\gamma)|\phi\in\mathrm{Hom}(P,G)\\}.$ We can also
interpret the measure induced by an element $\gamma\in P$ on a finite group
$G$ (see [6]).

######  Lemma 2.3.

Let $P$ be a finitely generated group. If $\gamma_{1},\gamma_{2}\in P$ is
endomorphically equivalent if and only if $G_{\gamma_{1}}=G_{\gamma_{2}}$ for
any group $G.$

###### Proof.

Suppose there exists an endomorphism $\chi$ of $P$ such that
$\chi(\gamma_{1})=\gamma_{2}.$ If $\phi(\gamma_{2}),$ for some
$\phi\in\mathrm{Hom}(P,G),$ is in the image of $\gamma_{2}$, then
$\phi\circ\chi(\gamma_{1})=\phi(\gamma_{2})$ and hence we get
$G_{\gamma_{2}}\subseteq G_{\gamma_{1}}.$ Similarly we can show
$G_{\gamma_{1}}\subseteq G_{\gamma_{2}}.$ For the other way, take $G=P$. Let
$\phi$ be the identity automorphism of $P.$ Then
$\gamma_{2}=\phi(\gamma_{2})\in P_{\gamma_{2}}=P_{\gamma_{1}}.$ Therefore
there exists $\chi\in\mathrm{Hom}(P,P)$ with $\chi(\gamma_{1})=\gamma_{2}$.
Similarly the remaining part can be shown. ∎

We can define weak profinite rigidity for elements in any finitely generated
group.

######  Definition 2.4.

An element $\gamma$ in a finitely generated group $P$ is said to be weakly
profinitely rigid if every element $\beta\in P$ with $G_{\gamma}=G_{\beta}$
for every finite group $G$ implies $\gamma$ and $\beta$ are endomorphically
equivalent.

######  Corollary 2.5.

Every element in finite group is weakly profinitely rigid.

###### Proof.

The proof follows similarly as the last part of proof of Lemma 2.3. ∎

Although, it is straightforward that every element in every finite group is
profinitely rigid. But two words which are endomorphic equivalent need not be
automorphic. For example, take two elements of order $2$ in group
$S_{3}\times{\mathbb{Z}}_{2}$, namely $((1\,2),0)$ and $((\,),1)$, they are
endomorphic to each other but they are not automorphic. Here is an example of
a group where no element is weakly profinitely rigid:

######  Example 2.6.

Let $S$ be a finitely generated infinite simple group. Then no element of $S$
is weakly profinitely rigid. As $\mathrm{Hom}(S,G)$, for any finite group $G,$
consists of only one element $\phi$ where $\phi(s)=e_{G}$ for all $s\in S$ and
$e_{G}$ is the identity element of $G.$ Then for every finite group $G,$
$G_{e_{S}}=G_{s}$ where $s\in S$ and $e_{S}$ is the identity element of $S.$
It is easy to see that $e_{S}$ and $s$, where $s$ is non trivial element, can
never be endomorphically equivalent.

######  Example 2.7.

Every non-residually finite group contains element which is not weakly
profinitely rigid. Elements from the intersection of all normal subgroup of
finite index will never be weakly profinitely rigid.

Whatever example of groups, which consist of non weakly profinitely rigid
elements, we have are non residually finite. So we are intrigued by the
following question:

######  Question 2.8.

Is every element in residually finite group weakly profinitely rigid?

In this context, we prove the following.

######  Lemma 2.9.

Every element of a finitely generated free abelian group is weakly profinitely
rigid.

###### Proof.

Let ${\mathbb{Z}}^{r}$ be any free abelian group of rank $r$. Any element
$(a_{1},\ldots,a_{r})\in{\mathbb{Z}}^{r}$ can be written as
$m(b_{1},\ldots,b_{r})$, where $m=gcd(a_{1},\ldots,a_{r})$ and $mb_{i}=a_{i}$
for all $i\in\\{1,2,\ldots,r\\}$. We notice that $(b_{1},\ldots,b_{r})$ is a
basis element of ${\mathbb{Z}}^{r}$. It is straighforward that image of
$(b_{1},\ldots,b_{r})$ as a word map on any finite group is the whole group.
Hence image of $(a_{1},\ldots,a_{r})$ on any group consists of all $m$th power
elements of $G$. Similarly any other element in ${\mathbb{Z}}^{r}$ which have
same images on every finite group as $(a_{1},\ldots,a_{r})$ must be of the
form $m(c_{1},\ldots,c_{r})$, where $(c_{1},\ldots,c_{r})$ is a primitive
element of ${\mathbb{Z}}^{r}$. It is not hard to see that
$(a_{1},\ldots,a_{r})$ and $m(c_{1},\ldots,c_{r})$ are automorphic as any two
basis elements in ${\mathbb{Z}}^{r}$ are automorphic. ∎

######  Remark 2.10.

Calvert, Dutta and Prasad [4] have shown that any two endomorphic equivalent
elements in countable abelian group are automorphic. Hence weak profinite
rigidity implies profinite rigidity for the elements of finitely generated
abelian group,.

Now we provide a concise overview of the profinite completion of a group. For
any group $P$, the basis for the pofinite topology on $P$ is the set of left
cosets of subgroups of finite index. The profinite completion of $P$, denoted
by $\hat{P}$, is the inverse limit of its finite quotients. A normal subgroup
$N$ of finite index in $P$ is denoted by $N\unlhd_{\textrm{f.i.}}P.$ Formally,
we write

$\hat{P}=\varprojlim_{N\unlhd_{\textrm{f.i.}}P}P/N.$

$\hat{P}$ possesses the topology of inverse limit, where each finite quotient
$P/N$ is a discrete topological group. With this topology, $\hat{P}$ becomes
compact, Hausdorff and totally disconnected topological group. Any compact,
Hausdorff and totally disconnected topological group is referred to as a
profinite group. There is a natural homomorphism $i:P\longrightarrow\hat{P}$
defined by mapping $\gamma$ to $(\gamma N)_{N\unlhd_{\textrm{f.i.}}P}.$

Let $G$ be a finite group endowed with discrete topology. By definition, we
assume that homomorphism from profinite group to a finite group is continuous.
For our purpose, we don’t need to assume continuity as all homomorphisms are
continuous when $P$ is finitely generated. Let ${\mathrm{Hom}}(\hat{P},G)$ be
the set of all continuous homomorphism from $\hat{P}$ to $G.$ By universal
property of profinite completion of group, there is one-one correspendence
between $\mathrm{Hom}(P,G)$ and $\mathrm{Hom}(\hat{P},G).$ To be more precise,
for every homomorphism $\phi\in\mathrm{Hom}(P,G)$ there exists unique
$\hat{\phi}\in\mathrm{Hom}(\hat{P},G)$ such that $\phi=\hat{\phi}\circ i$ [9,
Lemma 3.2.1].

Let $\gamma\in P$ and $g\in G$ where $P$ is finitely generated group and $G$
is finite group. We define

$\displaystyle\mathrm{Hom}_{\gamma,g}(P,G)$
$\displaystyle:=\\{\phi\in\mathrm{Hom}(P,G)\,|\,\phi(\gamma)=g\\},$
$\displaystyle K_{P}(G)$
$\displaystyle:=\bigcap_{N\unlhd_{\text{f.i.}}P,\frac{P}{N}\cong H\leq G}N,$
$\displaystyle J_{P}(G)$
$\displaystyle:=\bigcap_{\phi\in\mathrm{End}(P)}\phi^{-1}(K_{P}(G)).$

We claim that $J_{P}(G)$ is finite indexed fully invariant subgroup of $P$.
Since $P$ is finitely generated, there are finitely many subgroups of given
finite index. Hence it is easy to see that $K_{P}(G)$ is a finite indexed
normal subgroup of $P$. For any $\phi\in\mathrm{End}(P)$, define a map
$\chi:P/\phi^{-1}(K_{P}(G))\longrightarrow P/K_{P}(G)$ by mapping
$\chi(p\phi^{-1}(K_{P}(G)))=\phi(p)K_{P}(G).$ Clearly $\chi$ is one-one
homomorphism. Hence $[P:\phi^{-1}(K_{P}(G))]\leq[P:K_{P}(G)]$ and it
establishes that $J_{P}(G)$ is finite indexed normal subgroup as it is
intersection of finitely many finite indexed normal subgroup of $P$. The full
invariance of $J_{P}(G)$ in $P$ follows from construction. When we talk about
endomorphism in case of profinite group, it means continuous endomorphism. Now
we give analogous theorem of [6, Theorem 2.2].

######  Theorem 2.11.

Let $P$ be a finitely generated group and $\gamma_{1},\gamma_{2}\in P.$ Then
the following are equivalent:

  1. (1)

$i(\gamma_{1})$ and $i(\gamma_{2})$ are endomorphic equivalent in $\hat{P}$.

  2. (2)

If $\mathrm{Hom}_{\gamma_{1},g}(P,G)$ is nonempty if and only if
$\mathrm{Hom}_{\gamma_{2},g}(P,G)$ is nonempty.

  3. (3)

$\gamma_{1}J$ and $\gamma_{2}J$ are endomorphic equivalent in $P/J,$ where
$J=J_{P}(G)$ for every finite group $G.$

  4. (4)

For every $N\unlhd_{\mathrm{f.i.}}P,$ there exists $J\unlhd_{\mathrm{f.i.}}P$
with $J\leq N$ such that $\gamma_{1}J$ and $\gamma_{2}J$ are endomorphic
equivalent in $P/J.$

We notice that for a group $P$, $\mathrm{End}(P)$ is a monoid and
$\mathrm{End}(\hat{P})$ is profinite monoid. More details about profinite
monoid can be found in [1]. We need the following lemma to prove the above
theorem.

######  Lemma 2.12.

Let $P$ be a finitely generated group. Then
$\mathrm{End}(\hat{P})=\varprojlim_{N\unlhd_{\text{f.i.}}P}\mathrm{End}(P/N)$,
where inverse limit is taken over all $N$ such that $N=J_{P}(G)$ for some
finite group $G.$

###### Proof.

Assume that $J_{P}(G_{1})=J_{1}\leq J_{2}=J_{P}(G_{2}),$ and denote
$Q_{i}=P/J_{i}$ for $i=1,2.$ The image of $J_{2}$ in $Q_{1}$ is equal to
$J_{Q_{1}}(G_{2}),$ making this image fully invariant in $Q_{1}.$
Consequently, there exists a well-defined monoid homomorphism
$\mathrm{End}(Q_{1})\longrightarrow\mathrm{End}(Q_{2}).$

Furthermore, for every $J_{1}=J_{P}(G_{1})$ and $J_{2}=J_{P}(G_{2}),$ define
$G_{3}=P/{J_{1}\cap J_{2}}$. Then $J_{3}=J_{P}(G_{3})\leq J_{1},J_{2}.$
Therefore, the inverse limit $\varprojlim_{N\unlhd_{f.i.}P}\mathrm{End}(P/N)$
is well-defined, where the inverse limit is taken over all $N$ such that
$N=J_{P}(G)$ for some finite group $G$ is well-defined inverse system.

By [9, Proposition 3.2.2], for $J\unlhd_{f.i.}P$ we have
$P/J\cong\hat{P}/\overline{i(J)},$ and $i(J_{P}(G))=J_{\hat{P}}(G).$
Therefore, it suffices to show that:

$\mathrm{End}(\hat{P})=\varprojlim_{J}\mathrm{End}(\hat{P}/J),$

where the inverse limits is taken over all subgroups $J\leq\hat{P}$ such that
$J=J_{P}(G)$ for some finite group $G.$ Since $J=J_{\hat{P}}(G)$ is fully
invariant in $\hat{P}$, every endomorphism of $\hat{P}$ induces an
endomorphism of $\hat{P}/J$ which agrees with the inverse system, leading to a
natural continuous homomorphism:

$\omega:\mathrm{End}(\hat{P})\longrightarrow\underset{}{}\varprojlim_{J}\mathrm{End}(\hat{P}/J).$

The map $\omega$ is injective becasuse
$\underset{G\,\mathrm{finite}}{\cap}J_{\hat{P}}(G)=\\{e_{\hat{P}}\\}.$ The map
$\omega$ is surjective because every element of the inverse system
$\varprojlim_{J}\mathrm{End}(\hat{P}/J)$ defines a continuous endomorphism of
$\varprojlim_{J}\hat{P}/J\cong\hat{P}.$ ∎

###### Proof of Theorem 2.11.

$1\implies 2$:

The proof follows from the fact that $\mathrm{Hom}(P,G)$ and $Hom(\hat{P},G)$,
for every finite group $G$ are in one correspondence and we have
$|\mathrm{Hom}_{\gamma,g}(P,G)|=|\mathrm{Hom}_{i(\gamma),g}(\hat{P},G)|.$ And
hence

$|\mathrm{Hom}_{\gamma_{1},g}(P,G)|=|\mathrm{Hom}_{i(\gamma_{1}),g}(\hat{P},G)|=|\mathrm{Hom}_{i(\gamma_{2}),g}(\hat{P},G)|=|\mathrm{Hom}_{\gamma_{2},g}(P,G)|.$

$2\implies 3$:

Let $J=J_{P}(G)$ for some finite group $G$. As
$\mathrm{Hom}_{\gamma_{1},\gamma_{1}K}(P,P/J)\neq\emptyset\implies\mathrm{Hom}_{\gamma_{2},\gamma_{1}K}(P,P/J)\neq\emptyset$.
Let $f_{1}\in\mathrm{Hom}_{\gamma_{2},\gamma_{1}K}(P,P/J).$ Then $J\subseteq
ker(f_{1})$ because of the choice of $J.$ Hence there exist an induced map
$\tilde{f}:P/J\longrightarrow P/J$ such that
$(\tilde{f}_{1})(\gamma_{2}J)=\gamma_{1}J.$ Similarly we can do the other way.
Hence $\gamma_{1}K,\gamma_{2}K$ are endomorphic equivalent in $P/K$.

$3\implies 4$:

For every $N\unlhd_{f.i.}P,$ we can take $J=J_{P}(P/N).$

$4\implies 3$:

Let $J=J_{P}(G)$ for some finite group $G.$ By assumption, there exists a
subgroup $J^{\prime}\leq J$ such that $J^{\prime}\unlhd_{f.i.}P$ with
$\gamma_{1}J^{\prime}$ and $\gamma_{2}J^{\prime}$ endomorphic equivalent. But
the image of $J$ in $P/J^{\prime}$ is $J_{P/J^{\prime}}(G),$ so this image is
fully invariant, hence every endomorphism of $P/J^{\prime}$ induces an
endomorphism in $P/J$. Hence we conclude that $\gamma_{1}J,\gamma_{2}J$ are
endomorphic equivalent in $P/J.$

$3\implies 1$:

For every $J$ in the inverse system, we get that there exists an endomorphism
from $\gamma_{1}J$ to $\gamma_{2}J$. By standard compactness argument, there
exists an element in
$\varprojlim_{J\unlhd_{f.i.}P,J=J_{P}(G)}\mathrm{End}(P/J)$ mapping
$\gamma_{1}J$ to $\gamma_{2}J$. Similarly we can prove the vice versa. This
proves our result.

∎

The above theorem is the reason to generalise the definition 2.4.

######  Definition 2.13.

Let $P$ be a finitely generated group. An element $\gamma_{1}\in P$ is said to
be weakly profinitely rigid, whenever $\gamma_{2}\in P$ satisfies
$i(\gamma_{1})$ and $i(\gamma_{2})$ are endomorphically equivalent in
$\hat{P}$, then $\gamma_{1},\gamma_{2}$ are endomorphically equivalent in $P.$

##  3\. Test Elements in Free Profinite Group

An element $g\in G$ is said to be test element if any endomorphism $\phi$ of
the group $G$ that satisfies $\phi(g)=g$ is an automorphism. In this section,
we will disucss words in finitely generated free group that are test words in
their profinite completion.

######  Definition 3.1.

A subgroup of a group $G$ is said to be retract $G$ if there exists a
homomorphism $\phi:G\longrightarrow H$ such that $\phi|_{H}$ is identity map.

Turner has classified test words in free groups. The next lemma is a corollary
of a result of Turner.

######  Theorem 3.2.

[13] The test words in a free group $F_{n}$ are words not contained in any
proper retract.

Snopce and Tanushevski have proved the analogous result for finitely generated
profinite group.

######  Theorem 3.3.

[12, Theorem 3.5] The test elements of a finitely generated profinite group
are exactly the elements not contained in any proper retracts.

######  Proposition 3.4.

A word $w\in F_{n}$ is a test element if and only if $i(w)$ is a test element
in $\hat{F_{n}}$.

###### Proof.

Suppose $w$ is not a test element in $F_{n}$. By Theorem 3.5, there exists a
proper retract $R$ of $F_{n}$ containing $w$. Let $r:F_{n}\longrightarrow R$
be the retraction. Take the induced map on its profinite completion we get
$\hat{r}:\hat{F_{n}}\longrightarrow\hat{R}$, which will again be a proper
retraction, since the rank of proper retract of $F_{n}$ is stricly less than
$n$. Hence $i(w)$ is not a test word in $\hat{F_{n}}$ by Theorem 3.3.

Suppose $i(w)$ is not a test word. By Theorem 3.3, there exists a proper
retract $S$ of $\hat{F_{n}}$ containing $i(w)$. Consider $R=S\cap F_{n}$,
where $F_{n}$ and $i(F_{n})$ are identified. $R$ is clearly a proper retract
of $F_{n}$ containing $w$. Hence $w$ is not a test word. ∎

The next lemma is the explicit description of test words in $F_{2}$ given by
Turner.

######  Lemma 3.5.

[13] Let $w\in F_{2}\setminus\\{e\\}$. Write
$w=u^{m},m\in{\mathbb{Z}}\setminus\\{0\\}$ such that $u$ is not a proper power
word. Suppose $u=x_{1}^{m_{1}}x_{2}^{m_{2}}c,$ where $c\in F_{2}^{\prime}$. If
$gcd(m_{1},m_{2})\neq 1$, then $w$ is test word.

##  4\. Main Results

We begin this section by following result.

######  Theorem 4.1.

Let $F_{n}$ be a free group. If $w\in F_{n}$ is weakly profinitely rigid, then
$w^{d}|d\in{\mathbb{Z}}$ is also weakly profinitely rigid.

######  Lemma 4.2.

[6] Let $w\in F_{n}$ such that every image of $w$ on symmetric group $S_{n}$
is $dth$ power for some $d\in\mathbb{Z}$, then there exists $v\in F_{n}$ such
that $w=v^{d}$.

######  Lemma 4.3.

[6] Every root of $w\in F_{n}$ in $\hat{F_{n}}$ belongs to $F_{n}$.

###### Proof of Theorem 4.1.

Lemma 4.2 suggests that any word which have same image as
$w^{d}|d\in{\mathbb{Z}}$ on every finite group will be of the form $v^{d}$ for
some $v\in F_{n}$. By Theorem 2.11, we get that there exist endomorphisms
$\phi_{1},\phi_{2}\in End(\hat{F_{n}})$ such that
$\phi_{1}(i(w^{d}))=i(v^{d})$ and $\phi_{2}(i(v^{d}))=i(w^{d})$. Lemma 4.3
says that every $d$ th root of $\phi_{1}(i(w))^{d}$ belongs to $F_{n}$ so we
get that $\phi_{1}(i(w))=i(v).$ Since $w$ is weakly profinitely rigid, $w$ and
$v$ are endomorphic equivalent. And hence $w^{d}$ and $v^{d}$ are endomorphic
equivalent proving $w^{d}$ is weakly profinitely rigid. ∎

######  Definition 4.4.

A word $w$ is said to be surjective, if $G_{w}=G$ for every group $G$.

######  Theorem 4.5.

Every surjective word in any finitely generated free group is weakly
profinitely rigid.

###### Proof.

Let $w\in F_{d}$. Express $w=x_{1}^{t_{1}}x_{2}^{t_{2}}\ldots x_{d}^{t_{d}}c$,
where $c$ lies in commutator subgroup of $F_{d}$. It is enough to prove that
$w$ is surjective as a word map on every finite group if and only if
$gcd\\{t_{1},t_{2},\ldots,t_{d}\\}=1$. Suppose on the contrary that
$gcd\\{t_{1},t_{2},\ldots,t_{d}\\}=r\neq 1$. If $r\geq 2,$ then for
$G={\mathbb{Z}}_{d}$, we get $G_{w}$ to be trivial which is not surjective.
Similarly if $t_{i}=0$ for $i\in\\{1,2,\ldots,d$, then for any abelian finite
group $G$, $G_{w}$ will be trivial.

If $r=1$, then there exists $n_{i}\in{\mathbb{Z}}$ such that $\sum
n_{i}t_{i}=1$. For any $g\in G$, send $x_{i}$ to $g^{n_{i}}$, we get
$w(g^{n_{1}},g^{n_{2}},\ldots,g^{n_{d}})=g$ and hence $G_{w}=G.$ ∎

Every primitive word is weakly profinitely rigid as well as profinitely rigid
[8]. We want to mention that profinite rigidity of non primitive surjective
words is not known.

######  Corollary 4.6.

Let $w$ be any surjective word. Then $w^{d}\,|\,d\in{\mathbb{Z}}$ is weakly
profinitely rigid.

###### Proof.

It directly follows from Theorems 4.5,4.1. ∎

######  Theorem 4.7.

For a test word $w_{1}\in F_{n}$, word map corresponding to another word
$w_{2}\in F_{n}$ having same image as $w_{1}$ on every finite group is enough
to ensure that they induce same probability measure on every finite group.

###### Proof.

By Theorem 2.11, we deduce that $i(w_{1})$ and $i(w_{2})$ are endomorphic
equivalent. By Proposition 3.4, we get that $i(w_{1})$ and $i(w_{2})$ are
automorphic. Hence by [6, Theorem 2.2], we get that $w_{1}$ and $w_{2}$ induce
same probability measure. ∎

######  Theorem 4.8.

A test word in $F_{n}$ is weakly profinitely rigid if and only if it is
profinitely rigid.

###### Proof.

Let $w_{1}$ be a test word in $F_{n}$ which is also weakly profinitely rigid.
Let $w_{2}$ be another word in $F_{n}$ such that it induces same measure as
the word $w_{1}$ on every finite group. By Theorem 2.11, $i(w_{1})$ is
endomorphic equivalent to $i(w_{2})$. Weak profinite rigidity of $w_{1}$
implies $w_{1}$ is endomorphic equivalent to $w_{2}$. Suppose $\sigma_{1}$ and
$\sigma_{2}$ be two endomorphism of $F_{n}$ such that
$\sigma_{1}(w_{1})=w_{2}$ and $\sigma_{2}(w_{2})=w_{1}$. The composition
$\sigma_{2}\circ\sigma_{1}$ fixes the test word $w_{1}$ and therefore
$\sigma_{2}\circ\sigma_{1}$ is an automorphism. Hence $\sigma_{2}$ is
surjective and free groups are hopfian implies $\sigma_{2}$ is an automorphism
and $\sigma_{1}$, a composition of two automorphism, is also an automorphism.
Hence, $w_{1}$ is profinitely rigid.

Now suppose $w_{1}\in F_{n}$ is profinitely rigid. Let $w_{2}\in F_{n}$ be
another word such that it has same image on every finite group as the word
$w_{1}.$ By Theorem 4.7, we get that $w_{1}$ and $w_{2}$ induce same measure
on every finite group. Profinite rigidity of $w_{1}$ implies that $w_{1}$ and
$w_{2}$ are automorphic. Hence they are endomorphic equivalent. Hence $w_{1}$
is weakly profinitely rigid. ∎

######  Corollary 4.9.

Word $[x,y]^{d}\in F_{2}$ for $d\in{\mathbb{Z}}$, is weakly profinitely rigid.

###### Proof.

As word $[x,y]^{d}$ is profinitely rigid [6], the result follows from previous
theorem. ∎

A word in $F_{2}$ can be either a power of a surjective word or a test word by
Lemma 3.5. The following corollary can also be seen as another
characterisation of test words in $F_{2}$.

######  Corollary 4.10.

Let $w\in F_{2}\setminus\\{e\\}$ be a word. If, for any word $u\in F_{2}$, $w$
and $u$ having same image on every finite group ensures that $w$ and $u$
induce same measure on every finite group, then $w$ is a test word.

###### Proof.

It is enough to worry about non-trivial non-test word by Theorem 4.7. If $w$
is a non-trivial non-test word, then it is a power of some surjective word
$w_{1}$, say $w=w_{1}^{d}$. Then we also have a surjective word $u$ which is
not automorphic to $w_{1}$, then $w=w_{1}^{d}$ and $u^{d}$ will have same
image on every finite group but there exists a finite group $G$ on which they
will not induce same measure [6]. ∎

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