A nilpotency criterion for some verbal subgroups

TL;DR

This paper establishes a criterion for nilpotency of certain verbal subgroups in finite and residually finite groups based on the order properties of their $w$-values, specifically for simple commutator words.

Contribution

It provides a new nilpotency criterion for verbal subgroups generated by specific simple commutator words in finite and residually finite groups.

Findings

Verbal subgroup is nilpotent iff $|ab|=|a||b|$ for coprime $w$-values.

Criterion applies when $i_1 eq i_j$ for all $j eq 1$ in the word.

Extension of results to residually finite groups with finite $w$-values.

Abstract

The word $w=[x_{i_1},x_{i_2},\dots,x_{i_k}]$ is a simple commutator word if $k\geq 2, i_1\neq i_2$ and $i_j\in \{1,\dots,m\}$, for some $m>1$. For a finite group $G$, we prove that if $i_{1} \neq i_j$ for every $j\neq 1$, then the verbal subgroup corresponding to $w$ is nilpotent if and only if $|ab|=|a||b|$ for any $w$-values $a,b\in G$ of coprime orders. We also extend the result to a residually finite group $G$, provided that the set of all $w$-values in $G$ is finite.

Equations

$$[x_{i_1},x_{i_2},\dots ,x_{i_k}]=[\dots [[x_{i_1},x_{i_2}],x_{i_3}],\dots ,x_{i_k}]$$

$$[x,{}_{n}y]=[x,\underset{n}{\underbrace{y,\dots ,y}}].$$

$$[X,{}_{n}Y]=[[X,{}_{n-1}Y],Y]$$

$$[X,{}_{n}y]=[X,{}_n⟨y⟩]$$

$$[P,x]=[P,x,x].$$

$$[P,x^{-1}]=[P,{}_{k-1}{x}^{-1}];$$

$$|[g,{}_{k-1}{x}^{-1}]x|=|[g,{}_{k-1}{x}^{-1}]||x|.$$

$$[g,{}_{k-1}{x}^{-1}]x=[g,{}_{k-2}{x}^{-1}{]}^{-1}x[g,{}_{k-2}{x}^{-1}]$$

$$[a_{i_1}{b}_{i_1},\dots ,a_{i_k}{b}_{i_k}]=[a_{i_1},\dots ,a_{i_k}][b_{i_1},\dots ,b_{i_k}]$$

$$P_1{P}_2{P}_3\cdots P_h$$

$$P_3=[P_3,[P_2,P_1]].$$

$$[P_2,x]=[P_2,{}_{k-1}x]]$$

$$[P_3,[P_2,P_1]]=1,$$

$$1\ne [H,a]=[H,{}_{k-1}a]\le N_p,$$

$$H/Z(G)=A/Z(G)⋊T/Z(G)$$

$$A/Z(G)=[A/Z(G),{}_{k-1}T/Z(G)].$$

$$1=[x^2,g]=[x,g{]}^x[x,g],$$

$$w=[x_{i_1},\dots ,x_{i_m},{}_{n-m}{x}_{i_1}]$$

$$[g,h,h,h]=[(h^{-1}{)}^{[g,h]},h{]}^h=[(h^{-1}{)}^{(h^{-1}{)}^{g}h},h{]}^h=[(h^{-1}{)}^k,h{]}^{h^{-1}},$$

$$[(h^{-1}{)}^k,h{]}^{-(h^{-1}{)}^k{h}^{-1}{k}^{-1}}=[((h^{-1}{)}^k{)}^{-1},h^{-1}{]}^{-k^{-1}}=[(h^{-1}{)}^{k^{-1}},h];$$

$$[(h^{-1}{)}^k,h{]}^{h^k{k}^{-1}}=[(h^k{)}^{-1},h{]}^{h^k{k}^{-1}}=[h^k,h{]}^{-k^{-1}}=[h^{k^{-1}},h].$$

$$[(h^{-1}{)}^k,h]=[(acb{)}^{(aef)},(abc)]=[(bec),(abc)]=(ac)(be).$$

Full text

2010 Mathematics Subject Classification: 20F18, 20F45

Keywords: nilpotent, Engel word, verbal subgroup

1 Introduction

Let $F$ be the free group on free generators $x_{1},\dots,x_{m}$, for some $m>1$. A group-word is any nontrivial element of $F$, that is, a product of finitely many $x_{i}$’s and their inverses. The elements of the commutator subgroup of $F$ are called commutator words. We say that the commutator word

[TABLE]

is a simple commutator word if $k\geq 2,i_{1}\neq i_{2}$ and $i_{j}\in\{1,\dots,m\}$ for every $j\in\{1,\dots,k\}$. Examples of simple commutator words are the lower central words and the $n$-Engel word

[TABLE]

Let $w=w(x_{1},\dots,x_{k})$ be a group-word in the variables $x_{1},\dots,x_{k}$. For any group $G$ and arbitrary $g_{1},\dots,g_{k}\in G$, the elements of the form $w(g_{1},\dots,g_{k})$ are called the $w$-values in $G$. We denote by $G_{w}$ the set of all $w$-values in $G$. The verbal subgroup of $G$ corresponding to $w$ is the (normal) subgroup $w(G)$ of $G$ generated by $G_{w}$. If $w(G)=1$, then $w$ is said to be a law in $G$.

Recently the following question has been considered ([2], see also [3]):

Let $w$ be a commutator word and let $G$ be a finite group with the property that $|ab|=|a||b|$ for any $a,b\in G_{w}$ of coprime orders. Is the verbal subgroup $w(G)$ nilpotent? Here $|x|$ stands for the order of the element $x\in G$.

As remarked in [2], by a result of Kassabov and Nikolov [9] this is not true in general (see Example 4.3). Two easier counterexamples are given in Section 4. On the other hand, the answer to the above question is positive when $w$ is a lower central word [2, 1]. Motivated by this, in the present paper we prove the following nilpotency criterion for $w(G)$, where $w$ is a simple commutator word without any repetition of the first variable.

Theorem 1.1**.**

Let $w=[x_{i_{1}},\ldots,x_{i_{k}}]$ be a simple commutator word with $i_{1}\neq i_{j}$ for every $j\in\{2,\ldots,k\}$, and let $G$ be a finite group. Then $w(G)$ is nilpotent if and only if $|ab|=|a||b|$ for any $a,b\in G_{w}$ of coprime orders.

We also extend Theorem 1.1 to a residually finite group $G$, provided that the set of all $w$-values in $G$ is finite.

Corollary 1.2**.**

Let $w=[x_{i_{1}},\ldots,x_{i_{k}}]$ be a simple commutator word with $i_{1}\neq i_{j}$ for every $j\in\{2,\ldots,k\}$, and let $G$ be a residually finite group in which $G_{w}$ is finite. Then $w(G)$ is finite, and it is nilpotent if and only if $|ab|=|a||b|$ for any $a,b\in G_{w}$ of coprime orders.

Recall that a group is residually finite if the intersection of its subgroups of finite index is trivial. Notice also that in Corollary 1.2 the finiteness of $G_{w}$ depends on the conciseness of the word $w$ in the class of residually finite groups (see Section 3). This is related to a question of P. Hall on words assuming only finitely many values in a group (see [10, Part 1, p. 119] for an account).

2 Proof of Theorem 1.1

The aim of this section is to prove the “if” part of Theorem 1.1, being the “only if” part clear. We will split the proof in two cases, depending on whether the finite group $G$ is soluble or not. In particular, we will show that only the soluble case can occur.

2.1 The soluble case

If $G$ is a finite soluble group, the Fitting height of $G$ is the least integer $h$ such that $F_{h}(G)=G$, where $F_{0}(G)=1$ and $F_{i}(G)/F_{i-1}(G)=F(G/F_{i-1}(G))$ is the Fitting subgroup of $G/F_{i-1}(G)$ for every $i\geq 1$. A finite soluble group with Fitting height at most $2$ is said to be metanilpotent.

The following lemma is well-known (see [1, Lemma 3] for a proof).

Lemma 2.1**.**

Let $G$ be a finite metanilpotent group with Fitting subgroup $F(G)$. For $p$ a prime, denote by $O_{p^{\prime}}(G)$ the maximal normal subgroup of $G$ of order coprime to $p$. If $x\in G$ is a $p$-element such that $[O_{p^{\prime}}(F(G)),x]=1$, then $x\in F(G)$.

A subgroup $H$ of a finite group $G$ is called a tower of height $h$ if $H$ can be written as a product $H=P_{1}\cdots P_{h}$, where

1. (1)

$P_{i}$ is a $p_{i}$-group ($p_{i}$ a prime) for $i=1,\dots,h$; 2. (2)

$P_{i}$ normalizes $P_{j}$ for $i<j$; 3. (3)

$[P_{i},P_{i-1}]=P_{i}$ for $i=2,\dots,h$.

It follows from (3) that $p_{i}\neq p_{i+1}$ for $i=1,\dots,h-1$.

The next lemma is taken from [14, Lemma 1.9].

Lemma 2.2**.**

A finite soluble group $G$ has Fitting height at least $h$ if and only if $G$ has a tower of height $h$.

Given two nonempty subsets $X$ and $Y$ of a group $G$, let

[TABLE]

where $n\geq 2$ and $[X,Y]$ is the commutator subgroup of $X$ and $Y$. We write $[X,\,_{n}y]$ when $Y=\{y\}$. Then, assuming that $X$ is normalized by $y$, it easy to see that

[TABLE]

for every $n\geq 1$.

As a straightforward corollary of [6, Theorem 5.3.6], we have:

Lemma 2.3**.**

For $p$ a prime, let $P$ be a $p$-subgroup of a finite group $G$. Suppose that $P$ is normalized by an element $x\in G$ of $p^{\prime}$-order. Then

[TABLE]

Lemma 2.4**.**

Let $w=[x_{i_{1}},\ldots,x_{i_{k}}]$ be a simple commutator word with $i_{1}\neq i_{j}$ for every $j\in\{2,\ldots,k\}$, and let $G$ be a finite group in which $|ab|=|a||b|$ for any $w$-values $a,b\in G$ of coprime orders. For $p$ a prime, let $P$ be a $p$-subgroup of $G$ normalized by a $w$-value $x\in G$ of $p^{\prime}$-order. Then $[P,x]=1$.

Proof.

By Lemma 2.3,

[TABLE]

thus the result will follow once it is shown that $N=[P,\,_{k-1}x^{-1}]=1$.

Let $[g,\,_{k-1}x^{-1}]\in N$, for some $g\in P$. Of course the orders of the $w$-values $x$ and $[g,\,_{k-1}x^{-1}]$ are coprime. Then, by hypothesis,

[TABLE]

However

[TABLE]

is a conjugate of $x$. So $|[g,\,_{k-1}x^{-1}]x|=|x|$ and consequently $[g,\,_{k-1}x^{-1}]=~{}1$. ∎

Lemma 2.5**.**

Let $w=[x_{i_{1}},\ldots,x_{i_{k}}]$ be a simple commutator word and let $G=A\times B$ be an arbitrary group. Then $w(G)=w(A)\times w(B)$.

Proof.

By induction on $n$, we get

[TABLE]

for every $a_{i_{1}},\ldots,a_{i_{k}}\in A$, and every $b_{i_{1}},\ldots,b_{i_{k}}\in B$. ∎

We are now able to prove the announced result, for soluble groups.

Proposition 2.6**.**

Let $w=[x_{i_{1}},\ldots,x_{i_{k}}]$ be a simple commutator word with $i_{1}\neq i_{j}$ for every $j\in\{2,\ldots,k\}$, and let $G$ be a finite soluble group in which $|ab|=|a||b|$ for any $a,b\in G_{w}$ of coprime orders. Then $w(G)$ is nilpotent.

Proof.

Let $h$ be the Fitting height of $G$. Firstly, we show that $h\leq 2$. Suppose by way of contradiction that $h\geq 3$. Then, by Lemma 2.2, there exists a tower

[TABLE]

of height $h$ in $G$. Since $P_{2}=[P_{2},P_{1}]$ and $P_{3}=[P_{3},P_{2}]$, we have

[TABLE]

Furthermore, by Lemma 2.3

[TABLE]

for every $x\in P_{1}$. Hence $[P_{2},x]$ is generated by $w$-values of $p_{2}$-orders. Applying Lemma 2.4, we deduce that $P_{3}$ commutes with $[P_{2},x]$. Thus

[TABLE]

which is impossible.

Now let $h=2$, the case $h=1$ being obvious. Denote by $F$ the Fitting subgroup of $G$. If $w(G)\leq F$, we are done. Suppose that $w(G)$ is not contained in $F$. Since $G/F$ is nilpotent, by Lemma 2.5 there exists a Sylow $p$-subgroup $P$ of $G$ such that $w(P/F)$ = $w(P)F/F$ is nontrivial. Let $x\in w(P)$ be a $w$-value which does not belong to $F$. Then $[O_{p^{\prime}}(F),x]=1$, by Lemma 2.4, from which it follows that $x\in F$, by Lemma 2.1: a contradiction. ∎

2.2 The general case

The following is a well-known consequence of the Baer-Suzuki Theorem (see, for instance, [8, Theorem 2.13]).

Lemma 2.7**.**

Let $G$ be a finite non-abelian simple group. If $x$ is an element of $G$ of order $2$, then there exists $g\in G$ such that $[x,g]$ has odd prime order.

In the sequel we will require a property of finite simple groups whose proper subgroups are soluble. These groups have been classified by Thompson in [13], and they are known as finite minimal simple groups.

Proposition 2.8**.**

Let $G$ be a finite minimal simple group. Then $G$ contains a subgroup $H=A\rtimes T$ where $A$ is an elementary abelian $2$-group and $T$ is a subgroup of odd order such that $C_{A}(T)=1$. Further, $A=[A,T]$.

Proof.

According to Thompson’s classification [13, Corollary 1], the group $G$ is isomorphic to one of the following groups:

1. (1)

$\operatorname*{PSL}(2,2^{p})$, where $p$ is any prime; 2. (2)

$\operatorname*{PSL}(2,3^{p})$, where $p$ is any odd prime; 3. (3)

$\operatorname*{PSL}(2,p)$, where $p>3$ is any prime such that $p^{2}+1\equiv 0\mod 5$; 4. (4)

$\operatorname*{PSL}(3,3)$; 5. (5)

$\operatorname*{Sz}(2^{p})$, where $p$ is any odd prime.

Since the groups in (2), (3) and (4) have a subgroup isomorphic to the alternating group of degree $4$ (see, for instance, [11, Theorem 6.26] and [4, Theorem 7.1(2)]), we may consider the other two cases.

If $G$ is isomorphic to $\operatorname*{PSL}(2,2^{p})$, then Theorem 6.25 of [11] shows that $G$ contains a Frobenius group $H=A\rtimes T$ where $A$ is an elementary abelian $2$-group of order $q$, and $T$ is a cyclic group of order $q-1$.

If $G$ is isomorphic to the Suzuki group $\operatorname*{Sz}(2^{p})$, then $G$ contains a Frobenius group $F=Q\rtimes T$ where $Q$ is a Sylow $2$-subgroup of $G$ of order $2^{2p}$ and $T$ is a cyclic subgroup of order $2^{p}-1$ (see [12, Theorem 9]). Thus, taking $A$ to be a minimal normal subgroup of $F$ contained in $Q$, the subgroup $H=A\rtimes T$ is as required.

Finally, notice that in both cases we have $A=[A,T]$ by [6, Theorem 5.2.3]. ∎

Lemma 2.9**.**

Let $w=[x_{i_{1}},\ldots,x_{i_{k}}]$ be a simple commutator word with $i_{1}\neq i_{j}$ for every $j\in\{2,\ldots,k\}$, and let $G$ be a finite group such that $G=G^{\prime}$. If $q\in\pi(G)$, then $G$ is generated by $w$-values of $p$-power order for primes $p\neq q$.

Proof.

For each prime $p\in\pi(G)\setminus\{q\}$, denote by $N_{p}$ the subgroup of $G$ generated by all $w$-values of $p$-power order. Let us show that each Sylow $p$-subgroup of $G$ is contained in $N_{p}$. Suppose this is false and choose $p$ such that a Sylow $p$-subgroup of $G$ is not contained in $N_{p}$. Of course, $N_{p}$ is a normal subgroup of $G$. We may pass to the quotient $G/N_{p}$ and assume that $N_{p}=1$. Since $G=G^{\prime}$, it is clear that $G$ does not possess a normal $p$-complement. Thus the Frobenius Theorem (see [6, Theorem 7.4.5]) implies that $G$ has a $p$-subgroup $H$ and a $p^{\prime}$-element $a\in N_{G}(H)$ such that $[H,a]\neq 1$. By Lemma 2.3, we have

[TABLE]

a contradiction. Hence $N_{p}$ contains the Sylow $p$-subgroups of $G$. Let $T$ be the product of all subgroups $N_{p}$, with $p\neq q$. Then $G/T$ is a $q$-group and, since $G=G^{\prime}$, we conclude that $G=T$. It follows that $G$ can be generated by $w$-values of $p$-power order for $p\neq q$. ∎

In order to complete the proof of Theorem 1.1, we recall that if a simple commutator word is a law in a finite group $G$, then $G$ is nilpotent [7].

Proposition 2.10**.**

Let $w=[x_{i_{1}},\ldots,x_{i_{k}}]$ be a simple commutator word with $i_{1}\neq i_{j}$ for every $j\in\{2,\ldots,k\}$, and let $G$ be a finite group in which $|ab|=|a||b|$ for any $a,b\in G_{w}$ of coprime orders. Then $G$ is soluble, and $w(G)$ is nilpotent.

Proof.

By Proposition 2.6, it is enough to show that $G$ is soluble. Suppose that $G$ is not soluble. Of course, we may assume that $G$ is a counterexample of minimal order. Then every proper subgroup $K$ of $G$ is soluble: indeed, $K/w(K)$ is nilpotent [7, 6.1 Satz] and so is $w(K)$ by Proposition 2.6. It follows that $G=G^{\prime}$.

Let $R$ be the soluble radical of $G$, that is, the subgroup of $G$ generated by all normal soluble subgroups of $G$. Then $G/R$ is a nonabelian simple group and, by Proposition 2.6, $w(R)$ is nilpotent. We claim that $R=Z(G)$. Choose $q\in\pi(F(G))$ and let $Q$ be the Sylow $q$-subgroup of $F(G)$. According to Lemma 2.9, the group $G$ is generated by $w$-values of $p$-power order for primes $p\neq q$. Also, by Lemma 2.4, $[Q,x]=1$ for every $w$-value $x$ of $q^{\prime}$-order. Thus $Q\leq Z(G)$. This happens for each choice of $q\in\pi(F(G))$, so that $F(G)=Z(G)$. Since $w(R)\leq F(G)$, we have $[x_{i_{1}},\ldots,x_{i_{k}},y]=1$ for every $x_{i_{1}},\ldots,x_{i_{k}},y\in R$. Hence $R$ is nilpotent [7, 6.1 Satz] and therefore $R\leq F(G)$. In particular, $R=Z(G)$.

Next, we prove that $G$ contains a $w$-value $x$ such that $x$ is a 2-element of order $2$ modulo $Z(G)$. First, notice that $G/Z(G)$ is a finite minimal simple group. Then, by Proposition 2.8, $G/Z(G)$ has a subgroup

[TABLE]

where $A/Z(G)$ is an elementary abelian $2$-group and $T/Z(G)$ is a group of odd order such that

[TABLE]

Let $P$ be the Sylow $2$-subgroup of $A$. Thus $x=[a,\,_{k-1}t]$, for some $a\in P$ and $t\in T$, is a $w$-value as desired.

Now, take $x\in G_{w}$ with the above properties. By Lemma 2.7, there exists an element $g\in G$ such that the order of $[x,g]$ is an odd prime. Since

[TABLE]

$x$ inverts $[x,g]$ and so, by Lemma 2.4, $[\langle[x,g]\rangle,x]=1$. This gives $[x,g]=1$, which is a contradiction. ∎

3 Proof of Corollary 1.2

Following [5], we say that a word $w$ implies virtual nilpotency if every finitely generated metabelian group, where $w$ is a law, has a nilpotent subgroup of finite index. Since finitely generated $n$-Engel groups are nilpotent (see [10, Part 2, Theorem 7.3.5]), the Engel words imply virtual nilpotency. More generally, this is true for simple commutator words.

Lemma 3.1**.**

Let $w=[x_{i_{1}},\ldots,x_{i_{n}}]$ be a simple commutator word and let $G$ be a metabelian group such that $w(G)=1$. Then $G$ is $n$-Engel.

Proof.

Since $G$ is metabelian, we have $[c,x_{i_{j}},x_{i_{k}}]=[c,x_{i_{k}},x_{i_{j}}]$ for every $c\in G^{\prime}$. Then, without loss of generality, we may assume that

[TABLE]

where $1<m<n$ and $i_{1}\neq i_{j}$ for every $j\in\{2,\dots,m\}$. Thus, for any $x,y\in G$, taking $x_{i_{1}}=y[x,y]$ and $x_{i_{2}}=\dots=x_{i_{m}}=y$, we have $[x,\,_{n}y]=1$ and therefore $G$ is $n$-Engel. ∎

Corollary 3.2**.**

Every simple commutator word implies virtual nilpotency.

A word $w$ is said to be boundedly concise in a class of groups $\mathcal{C}$ if for every integer $m$ there exists a number $\nu=\nu(\mathcal{C},w,m)$ such that whenever $|G_{w}|\leq m$ for a group $G\in\mathcal{C}$ it always follows that $|w(G)|\leq\nu$. According to Theorem 1.2 of [5], words implying virtual nilpotency are boundedly concise in residually finite groups. This, together with Corollary 3.2, yields the following corollary.

Corollary 3.3**.**

Every simple commutator word is boundedly concise in the class of residually finite groups.

We are finally ready to prove Corollary 1.2: Let $w=[x_{i_{1}},\ldots,x_{i_{k}}]$ be a simple commutator word with $i_{1}\neq i_{j}$ for every $j\in\{2,\ldots,k\}$, and let $G$ be a residually finite group in which $G_{w}$ is finite. Then $w(G)$ is finite, and it is nilpotent if and only if $|ab|=|a||b|$ for any $a,b\in G_{w}$ of coprime orders.

Proof of Corollary 1.2.

Of course $w(G)$ is finite, by Corollary 3.2. Let us show that $w(G)$ is nilpotent, whenever $|ab|=|a||b|$ for any $a,b\in G_{w}$ of coprime orders. The converse is clear.

Since $G$ is residually finite, there exists a normal subgroup $N$ of $G$ such that $N\cap w(G)=1$ and $G/N$ is finite. Notice that for any $w$-value $xN\in G/N$, we have $x\in G_{w}$ and $|xN|=|x|$. It follows that $G/N$ satisfies the hypotheses of Theorem 1.1 and therefore $w(G/N)\simeq w(G)$ is nilpotent. ∎

4 Examples

In this section we collect some examples showing that Theorem 1.1 does not hold for an arbitrary commutator word.

Example 4.1**.**

Let $w=[x,y]^{3}$ and let $G=(S\times S)\rtimes C$ where $S$ is the symmetric group of degree $3$, $C=\langle g\rangle$ is the cyclic group of order $2$, and the action is given by $(a,b)^{g}=(b,a)$ for every $(a,b)\in S\times S$. Then every nontrivial $w$-value has order $2$, and $w(G)=S\times S$.

Example 4.2**.**

Let $w=[x,y^{10},y^{10},y^{10}]$ and let $G$ be the alternating group of degree $5$. Then $G_{w}$ consists of the identity and all products of two transpositions. In particular, $w(G)=G$.

Proof.

Of course $G_{w}$ is the set of all commutators $[g,h,h,h]$, where $g,h\in G$ and $h$ is a 3-cycle. Since $[g,h,h]=(h^{-1})^{[g,h]}h$, we have

[TABLE]

where $k=(h^{-1})^{g}$ is a $3$-cycle. For any $3$-cycles $h,k\in G$, we claim that $[(h^{-1})^{k},h]$ is either trivial or a product of two transpositions, from which it follows that so is $[(h^{-1})^{k},h]^{h^{-1}}$.

Let $h=(a\ b\ c)$ and $k=(d\ e\ f)$. Clearly, we may assume $a=d$. Furthermore, it is enough to consider the cases:

1. (1)

$b=e$ and $c\neq f$; 2. (2)

$e,f\not\in\{b,c\}$.

In fact, the other (nontrivial) cases can be deduced applying the following identities:

[TABLE]

[TABLE]

Now, in the first case, $h$ and $k$ belong to the alternating group of degree $4$ and therefore $[(h^{-1})^{k},h]$ is the product of two transpositions. In the second case, we have

[TABLE]

This proves our claim. Also, it implies that $G_{w}$ contains all products of two transpositions. ∎

Example 4.3** (see Theorem 1.2 of [9]).**

For every $n\geq 7$, there exists a commutator word $v$ such that, for $w=v^{10}$, the set of $w$-values of the alternating group of degree $n$ consists of the identity and all $3$-cycles.