On the multiple holomorph of groups of squarefree or odd prime power order

TL;DR

This paper investigates the structure of the quotient group T(G) related to the holomorph of finite groups, showing it is elementary 2-abelian for squarefree order groups and not a 2-group for certain p-groups of small nilpotency class.

Contribution

It proves T(G) is elementary 2-abelian for all finite groups of squarefree order and identifies cases where T(G) is not a 2-group among certain p-groups.

Findings

T(G) is elementary 2-abelian for all finite squarefree order groups.

T(G) is not a 2-group for some p-groups with nilpotency class at most p-1.

Results extend understanding of the multiple holomorph structure for specific group classes.

Abstract

Let $G$ be a group and write $\mbox{Perm}(G)$ for its symmetric group. Define $\mbox{Hol}(G)$ to be the holomorph of $G$, regarded as a subgroup of $\mbox{Perm}(G)$, and let $\mbox{NHol}(G)$ denote its normalizer. The quotient $T(G) = \mbox{NHol}(G)/\mbox{Hol}(G)$ has been computed for various families of groups $G$, and in most of the known cases, it turns out to be elementary $2$-abelian, except for two groups of order $16$ and some groups of odd prime power order and nilpotency class two. In this paper, we shall show that $T(G)$ is elementary $2$-abelian for all finite groups $G$ of squarefree order, and that $T(G)$ is not a $2$-group for certain finite $p$-groups $G$ of nilpotency class at most $p-1$.

Tables

Family	Number of groups	$n_c$	$(r,t)$	$r\ge t+1?$
$3$	$13$	$3$	$(1,1)$	No
$6$	$p+7$	$3$	$(1,1)$	No
$7$	$5$	$3$	$(1,1)$	No
$8$	$1$	$3$	$(2,1)$	Yes
$9$	$\gcd (p-1,3)+2$	$4$	$(1,1)$	No
$10$	$\gcd (p-1,3)+\gcd (p-1,4)+1$	$4$	$(1,1)$	No

Family	Number of groups	$n_c$	$(r,t)$	$r\ge t+1?$
$3$	$32$	$3$	$(1,1)$	No
$6$	$2p+21$	$3$	$(1,1)$	No
$7$	$21$	$3$	$(1,1)$	No
$8$	$p+5$	$3$	$(2,1)$	Yes
$9$	$3\gcd (p-1,3)+7$	$4$	$(1,1)$	No
$10$	$3\gcd (p-1,3)+3\gcd (p-1,4)+4$	$4$	$(1,1)$	No
$16$	$p+\gcd (p-1,3)+12$	$3$	$(1,1)$	No
$17$	$4p+\gcd (p-1,3)+30$	$3$	$(1,1)$	No
$18$	$3p+\gcd (p-1,3)+\gcd (p-1,4)+9$	$3$	$(1,1)$	No
$19$	$\left(3p^2+10p+21\right)/2$	$3$	$(1,1)$	No
$20$	$5p+\gcd (p-1,3)+\gcd (p-1,4)+13$	$3$	$(1,1)$	No
$21$	$\left(3p^2+4p+5\right)/2$	$3$	$(1,1)$	No
$22$	$7$	$3$	$(1,1)$	No
$23$	$p+4\gcd \left(p-1\right)+\gcd (p-1,4)+5$	$3$	$(1,1)$	No
$24$	$\gcd (p-1,3)+3$	$3$	$(1,1)$	No
$25$	$\left(p+3\right)/2$	$3$	$(2,1)$	Yes
$26$	$\left(p+3\right)/2$	$3$	$(2,1)$	Yes
$27$	$\gcd (p-1,3)+\gcd (p-1,4)+3$	$3$	$(1,1)$	No
$28$	$p$	$3$	$(2,1)$	Yes
$29$	$p$	$3$	$(2,1)$	Yes
$30$	$2\gcd (p-1,3)+4$	$3$	$(1,1)$	No
$31$	$7$	$3$	$(1,1)$	No
$32$	$5$	$3$	$(1,1)$	No
$33$	$6$	$3$	$(1,1)$	No
$34$	$3$	$3$	$(2,1)$	Yes
$35$	$\gcd (p-1,4)+2$	$3$	$(1,1)$	No
$36$	$\gcd (p-1,4)+\gcd (p-1,6)+1$	$3$	$(1,1)$	No
$37$	$\gcd (p-1,4)+4$	$3$	$(1,1)$	No
$38$	$p+\gcd (p-1,4)+\gcd (p-1,5)$	$3$	$(1,1)$	No
$39$	$p+\gcd (p-1,5)+\gcd (p-1,6)$	$3$	$(1,1)$	No
$40$	$\gcd (p-1,3)+2$	$3$	$(1,1)$	No
$41$	$\gcd (p-1,3)+1$	$3$	$(1,1)$	No
$42$	$p+1$	$3$	$(2,1)$	Yes
$43$	$p$	$3$	$(2,1)$	Yes

Equations

$${\xi }_N:N⟶G;{\xi }_N(\eta )=\eta (1_G)$$

$$\{\begin{array}{l}\lambda :G⟶\mathrm{Perm}(G);\lambda (\sigma )=(x\mapsto \sigma x)\\ \rho :G⟶\mathrm{Perm}(G);\rho (\sigma )=(x\mapsto x{\sigma }^{-1})\end{array}$$

$$\mathrm{Hol}(G)=\rho (G)⋊\mathrm{Aut}(G).$$

$${\mathrm{Norm}}_{\mathrm{Perm}(G)}(\lambda (G))=\mathrm{Hol}(G)={\mathrm{Norm}}_{\mathrm{Perm}(G)}(\rho (G)).$$

$$\mathcal{H}_{0}(G)=\left\{\begin{array}[]{c}\mbox{regular subgroups $N$ of $\mathrm{Perm}(G)$ isomorphic to $G$}\\ \mbox{and such that $\mathrm{Norm}_{\mathrm{Perm}(G)}(N)=\mathrm{Hol}(G)$}\end{array}\right\}.$$

$$\mathrm{NHol}(G)={\mathrm{Norm}}_{\mathrm{Perm}(G)}(\mathrm{Hol}(G)),$$

$$T(G)=\frac{\mathrm{NHol}(G)}{\mathrm{Hol}(G)}$$

$${\mathcal{H}}_0(G)=\{\pi \lambda (G){\pi }^{-1}:\pi \in \mathrm{NHol}(G)\},$$

$$\mathcal{H}_{0}(G)=\{N\lhd\mathrm{Hol}(G):N\simeq G\mbox{ and $N$ is regular}\}.$$

$$(n,i)\in \{(48,12),(48,14),(63,1),(80,12),(80,14)\},$$

$$G=G(d,e,k)=⟨\sigma ,\tau :{\sigma }^e=1_G,{\tau }^d=1_G,\tau \sigma {\tau }^{-1}={\sigma }^k⟩,$$

$$\gcd (d,e)=1\text{mbox}andbothd,e\text{mbox}aresquarefree.$$

$$|G|=de=d(gz),\text{mbox}wherez=\gcd (e,k-1)\text{mbox}andg=e/z.$$

$$k\equiv 1\pmod{z},\mbox{ and }k\not\equiv 1\pmod{q}\mbox{ for all primes $q\mid g$}.$$

$$\mathfrak{o}(G)=\{(f_1,f_2)\in {\mathbb{N}}^2:g=f_1{f}_2\text{mbox}and\gcd (d_{f_1},d_{f_2})\le 2\}.$$

$$|\mathrm{Aut}(G)|=g\phi (e)\text{mbox}and\mathrm{Aut}(G)=⟨\theta ⟩⋊\{{\varphi }_s{\}}_{s\in U(e)},$$

$$\theta (\sigma )=\sigma ,\theta (\tau )={\sigma }^z\tau ,{\varphi }_s(\sigma )={\sigma }^s,{\varphi }_s(\tau )=\tau .$$

$${\theta }^g={\mathrm{Id}}_G,{\varphi }_s{\varphi }_t={\varphi }_{st},{\varphi }_s\theta {\varphi }_s^{-1}={\theta }^s.$$

$$\displaystyle\mbox{in $G$:}\hskip 29.87547pt\tau^{b}\sigma^{a}$$

$$\displaystyle\mbox{in $\mathrm{Aut}(G)$:}\hskip 31.2982pt\phi_{s}\theta^{c}$$

$$\displaystyle\mbox{in $\mathrm{Hol}(G)$:}\hskip 21.33955pt\pi\rho(x)$$

$$S(h,i)=\sum _{\ell =0}^{i-1}{h}^{\ell }=1+h+\cdots +h^{i-1},$$

$$({\sigma }^a\tau {)}^b={\sigma }^{aS(k,b)}{\tau }^b\text{mbox}foranya\in \mathbb{Z},b\in {\mathbb{N}}_{\ge 0}.$$

$$({\theta }^c{\varphi }_s)({\sigma }^a{\tau }^b)={\sigma }^{as+czS(k,b)}{\tau }^b\text{mbox}foranya,c\in \mathbb{Z},b\in {\mathbb{N}}_{\ge 0},s\in U(e).$$

$$\Phi =\rho ({\sigma }^a{\tau }^b){\theta }^c{\varphi }_s(a,c\in \mathbb{Z},b\in {\mathbb{N}}_{\ge 0},s\in U(e))$$

$${\Phi }^i=\rho ({\sigma }^{A_{\Phi }(i)}{\tau }^{bi}){\theta }^{cS(s,i)}{\varphi }_{s^i},$$

$$A_{\Phi }(i)=aS(s{k}^b,i)+czS(k,b)\sum _{\ell =1}^{i-1}S(s,\ell )k^{b\ell }.$$

$${\Phi }^{i+1}$$

$$=\rho ({\sigma }^{A_{\Phi }(i)}{\tau }^{bi}\cdot {\sigma }^{a{s}^i+cS(s,i)zS(k,b)}{\tau }^b)\cdot {\theta }^{cS(s,i)}{\varphi }_{s^i}\cdot {\theta }^c{\varphi }_s$$

$$=\rho ({\sigma }^{A_{\Phi }(i)+(a{s}^i+czS(k,b)S(s,i))k^{bi}}{\tau }^{b(i+1)})\cdot {\theta }^{c(S(s,i)+s^i)}{\varphi }_{s^{i+1}}.$$

$$a(S(s{k}^b,i)+(s{k}^b{)}^i)+czS(k,b)\left(\sum _{\ell =1}^{i-1}S(s,\ell )k^{b\ell }+S(s,i)k^{bi}\right),$$

$$\#\{{\Phi }^i(1_G):i\in {\mathbb{N}}_{\ge 0}\}=e$$

Full text

1. 1 Introduction

2. 2 Groups of squarefree order

3. 2.1 Preliminaries

4. 2.2 Congruence criteria

5. 2.3 Proof of Theorem 2.1

6. 3 Groups of odd prime power order

7. 3.1 Some commutator calculations

8. 3.2 Proof of Theorem 3.1

9. 3.3 Further questions

10. 3.3.1 Applicability

11. 3.3.2 Large nilpotency class

12. 4 Acknowledgments

1. Introduction

Let $G$ be a group and write $\mathrm{Perm}(G)$ for its symmetric group. Recall that a subgroup $N$ of $\mathrm{Perm}(G)$ is said to be regular if the map

[TABLE]

is bijective, or equivalently, if the $N$-action on $G$ is both transitive and free. For example, both $\lambda(G)$ and $\rho(G)$ are regular subgroups of $\mathrm{Perm}(G)$, where

[TABLE]

denote the left and right regular representations of $G$, respectively. Further, recall that the holomorph of $G$ is defined to be

[TABLE]

Alternatively, it is not hard to verify that

[TABLE]

Then, it seems natural to ask whether $\mathrm{Perm}(G)$ has other regular subgroups whose normalizer is also equal to $\mathrm{Hol}(G)$. This problem was first considered by G. A. Miller [15]. More specifically, put

[TABLE]

In [15], he defined the multiple holomorph of $G$ to be

[TABLE]

which clearly acts on $\mathcal{H}_{0}(G)$ via conjugation, and he showed that this action is transitive. Hence, the quotient group

[TABLE]

acts regularly on and so has the same cardinality as $\mathcal{H}_{0}(G)$. In fact, we have

[TABLE]

and in the case that $G$ is finite, we also have

[TABLE]

These facts are easy to show, and a proof may be found in [20, Section 2] or [13, Section 1], for example. Let us mention in passing that regular subgroups of the holomorph, not just the ones which are normal, have close connections with Hopf-Galois structures on field extensions and set-theoretic solutions to the Yang-Baxter equation; see [8, Chapter 2] and [9].

The study of $T(G)$ did not attract much attention initially other than [15] and [14]. But recently, T. Kohl, who was originally interested in Hopf-Galois structures, revitalized this topic of research in [13]. His work then motivated the calculation of $T(G)$ for other groups $G$. Note that for non-abelian $G$, we have $T(G)\neq 1$ because the element $\pi_{-1}\mathrm{Hol}(G)$ is non-trivial, where $\pi_{-1}$ is the involution permutation $x\mapsto x^{-1}$ on $G$. It turns out that $T(G)$ is elementary $2$-abelian (including the trivial group) in all of the following five cases.

• •

finitely generated abelian groups $G$; see [15, 14, 5].

• •

finite dihedral groups $G$ of order at least $6$; see [13].

• •

finite dicyclic groups $G$ of order at least $12$; see [13].

• •

finite perfect groups $G$ with trivial center; see [6].

• •

finite quasisimple or almost simple groups $G$; see [20].

Nevertheless, there are examples of groups $G$ for which $T(G)$ is not elementary $2$-abelian. In [13, Section 3], it was noted that $T(G)$ is non-abelian (but still a $2$-group) for two groups $G$ of order $16$. Also, in [7, Proposition 3.1], it was shown that $T(G)$ is not even a $2$-group for a large subfamily of

• •

finite groups of odd prime power order and nilpotency class two.

In this paper, we shall consider two new families of finite groups $G$. First, in Section 2, we shall show that $T(G)$ is elementary $2$-abelian for

• •

finite groups $G$ of squarefree order.

This was motivated by [4], in which all cyclic regular subgroups of $\mathrm{Hol}(G)$ for $G$ of squarefree order were enumerated, and these correspond to Hopf-Galois structures on cyclic field extensions of squarefree degree. Then, in Section 3, we shall show that $T(G)$ is not a $2$-group for a subfamily of

• •

finite $p$-groups $G$ of nilpotency class at most $p-1$.

We shall state the precise statements of our results in Theorems 2.1 and 3.1 below. We chose not to include them here because more notation would have to be introduced.

In view of the known results so far, it seems natural to ask whether there is a finite non-nilpotent group $G$ for which $T(G)$ is not elementary $2$-abelian or not even a $2$-group. By calculating the size of the set in (1.4), which equals the order of $T(G)$, we have checked in Magma [3] that such a group $G$ does exist. However, among the non-nilpotent groups $G$ of order at most $95$, the quotient $T(G)$ is not a $2$-group only for $G=\mbox{{\sc SmallGroup}}(n,i)$ [2], where

[TABLE]

and they are all solvable. Let us end with the following question.

Question 1.1**.**

Is $T(G)$ a $2$-group for all finite insolvable groups $G$?

It would also be interesting to find out whether $T(G)$ is a $2$-group for most finite solvable (non-nilpotent) groups $G$, and if so, to give an explanation.

2. Groups of squarefree order

Throughout this section, let $G$ denote a finite group of squarefree order. It is known, by [16, Lemma 3.5], that $G$ admits a presentation

[TABLE]

where $d,e\in\mathbb{N}$, and $k\in\mathbb{N}$ is coprime to $e$ whose multiplicative order mod $e$ is equal to $d$. Note that $|G|=de$, which is squarefree by assumption, so

[TABLE]

Following [4, Proposition 3.5], let us write

[TABLE]

Observe that by definition and the fact that $\gcd(g,z)=1$, we have

[TABLE]

For each divisor $f$ of $g$, define $d_{f}$ to be the multiplicative order of $k$ mod $f$. Since $\gcd(g,z)=1$ and $z$ divides $k-1$, we have $d=d_{g}$. Also, let us define

[TABLE]

Note that this set depends only on the values of $g$ and $k$. We shall prove:

Theorem 2.1**.**

The group $T(G)$ is elementary $2$-abelian of order $\#\mathfrak{o}(G)$.

Remark 2.2*.*

The number $\#\mathfrak{o}(G)$ is indeed a power of $2$. To see why, define a graph $\Gamma(G)$ as follows.

• •

The vertex set of $\Gamma(G)$ is the set of distinct prime factors of $g$.

• •

Two vertices $q,q^{\prime}$ are joined by an edge if and only if $\gcd(d_{q},d_{q^{\prime}})\geq 3$.

Since $g$ is squarefree, for any divisor $f$ of $g$, we have $d_{f}=\operatorname*{lcm}_{q\mid f}d_{q}$. Hence, a pair $(f_{1},f_{2})\in\mathbb{N}^{2}$ with $g=f_{1}f_{2}$ lies in $\mathfrak{o}(G)$ if and only if for each connected component $\Gamma_{0}$ of $\Gamma(G)$, the primes in $\Gamma_{0}$ either all divide $f_{1}$ or all divide $f_{2}$. We then deduce that $\#\mathfrak{o}(G)=2^{\mathfrak{c}(G)}$, where $\mathfrak{c}(G)$ is the number of connected components in $\Gamma(G)$.

To prove Theorem 2.1, recall that $T(G)$ has the same size as $\mathcal{H}_{0}(G)$. Below, we shall use (1.4) to give a description of the elements of $\mathcal{H}_{0}(G)$ in terms of certain congruence conditions.

2.1. Preliminaries

Let $U(e)$ denote the set of integers $s\in\mathbb{Z}$ coprime to $e$ and write $\varphi(\cdot)$ for Euler’s totient function.

Lemma 2.3**.**

We have

[TABLE]

where $\theta$ and $\phi_{s}$ are the automorphisms on $G$ determined by

[TABLE]

Moreover, we have the relations

[TABLE]

Proof.

See the proof of [4, Lemma 4.1]. ∎

It is easy to see that we have the relations

[TABLE]

where $k^{-1}$ is to be interpreted as the multiplicative inverse of $k$ mod $e$ for $b$ negative. Following [4, (5)], for $h\in\mathbb{Z}$ and $i\in\mathbb{N}_{\geq 0}$, let us define

[TABLE]

with the empty sum $S(h,0)$ representing zero. Then, as in [4, (6)], we have

[TABLE]

Using this, a simple induction on $c$ shows that

[TABLE]

We shall frequently use above relations without referring to them explicitly. In what follows, consider an arbitrary element

[TABLE]

of $\mathrm{Hol}(G)$. The next lemma is essentially [4, Lemma 4.2].

Lemma 2.4**.**

For any $i\in\mathbb{N}_{\geq 0}$, we have

[TABLE]

where we define

[TABLE]

Proof.

We shall use induction. The case $i=0$ is clear. Suppose now that the claim holds for $i$. Then, we have

[TABLE]

Plainly, we have $S(s,i)+s^{i}=S(s,i+1)$, and the exponent of $\sigma$ simplifies to

[TABLE]

which is clearly equal to $A_{\Phi}(i+1)$. This proves the claim. ∎

Lemma 2.5**.**

The element $\Phi$ has order $e$ and satisfies

[TABLE]

precisely when

[TABLE]

Proof.

We may assume $b\equiv 0$ (mod $d$), for otherwise $\Phi^{e}\neq\mathrm{Id}_{G}$ by Lemma 2.4 because $\gcd(d,e)=1$. In this case, for any $i\in\mathbb{N}_{\geq 0}$, we have

[TABLE]

Suppose that $\Phi$ has order $e$ and (2.1) holds. For any $i,i^{\prime}\in\mathbb{N}_{\geq 0}$, we then have

[TABLE]

This clearly yields $\gcd(a,e)=1$. The above also implies that $s\equiv 1$ (mod $e$). To see why, consider the set

[TABLE]

where $i_{p}$ denotes the multiplicative order of $s$ mod $p$. Also, let $i$ denote the multiplicative order of $s$ mod $e$, and note that $S(s,i)\equiv 0$ (mod $p$) for $p\in\mathcal{P}$. Since $e$ is squarefree, we then deduce that

[TABLE]

It follows from (2.2) that

[TABLE]

But we also know that

[TABLE]

This leads to a contradiction unless $\mathcal{P}$ is empty. It follows that $i_{p}=1$ for all prime divisors $p$ of $e$. Since $e$ is squarefree, this implies that $s\equiv 1$ (mod $e$), as claimed. The converse is easily verified. ∎

Lemma 2.6**.**

Assume that $b$ is coprime to $d$. Then, the element $\Phi$ has order $d$ precisely when

[TABLE]

Proof.

By Lemma 2.4, the stated conditions are equivalent to $\Phi^{d}=\mbox{Id}_{G}$. In this case, the order of $\Phi$ is exactly $d$ because for any $i\in\mathbb{N}_{\geq 0}$, the exponent of $\tau$ in the expression for $\Phi^{i}$ in Lemma 2.4 is $bi$, and $b$ is coprime to $d$.∎

2.2. Congruence criteria

In this subsection, let us fix two elements

[TABLE]

of $\mathrm{Hol}(G)$. In order to compute $\mathcal{H}_{0}(G)$, we shall determine exactly when

[TABLE]

are all satisfied. By Lemma 2.5, we may and shall assume that

[TABLE]

Then, for any $i,j\in\mathbb{N}_{\geq 0}$, using Lemma 2.4, we compute that

[TABLE]

This expression shall be used repeatedly in the subsequent calculations.

Lemma 2.7**.**

We have $\Phi_{2}\Phi_{1}\Phi_{2}^{-1}=\Phi_{1}^{k}$ precisely when

[TABLE]

Proof.

On the one hand, we have

[TABLE]

On the other hand, from (2.4) and the fact that $A_{\Phi_{2}}(1)=a_{2}$, we have

[TABLE]

The claim then follows by equating the exponents of $\sigma$ and $\theta$. ∎

Lemma 2.8**.**

Assume that $\Phi_{2}$ has order $d$. Then, we have

[TABLE]

precisely when

[TABLE]

Proof.

For any $i,j\in\mathbb{N}_{\geq 0}$, from (2.4) we know that

[TABLE]

Thus, for (2.5) to hold, necessarily $\gcd(b_{2},d)=1$. In this case, since $\Phi_{2}$ has order $d$, the equality (2.5) holds if and only if for each $j\in\mathbb{N}_{\geq 0}$, the set

[TABLE]

runs over all residue classes mod $e$, which is equivalent to the coefficient of $i$ being invertible mod $e$. This proves the claim. ∎

Lemma 2.9**.**

Assume that $b_{2}$ is coprime to $d$, that $\Phi_{2}$ has order $d$, and that

[TABLE]

Then, we have $\langle\Phi_{1},\Phi_{2}\rangle$ is normal in $\mathrm{Hol}(G)$ precisely when the conditions

1. (1)

There exists $i\in\mathbb{N}_{\geq 0}$ such that

[TABLE] 2. (2)

There exists $i\in\mathbb{N}_{\geq 0}$ such that

[TABLE] 3. (3)

There exists $i\in\mathbb{N}_{\geq 0}$ such that

[TABLE] 4. (4)

There exists $i\in\mathbb{N}_{\geq 0}$ such that

[TABLE] 5. (5)

For every $s\in U(e)$, there exists $i\in\mathbb{N}_{\geq 0}$ such that

[TABLE]

are all satisfied. Here, the $i\in\mathbb{N}_{\geq 0}$ are independent of each other.

Proof.

By (2.6), the subgroup $\langle\Phi_{1},\Phi_{2}\rangle$ is normal in $\mathrm{Hol}(G)$ if and only if for each $\Psi$ in some set of generators of $\mathrm{Hol}(G)$, we have

[TABLE]

Now, clearly $\mathrm{Hol}(G)$ is generated by $\rho(\sigma),\rho(\tau),\theta$, and $\phi_{s}$ for $s\in U(e)$. Since

[TABLE]

these three equations impose no condition. Next, we compute that

[TABLE]

as well as that

[TABLE]

Since $\gcd(b_{2},d)=1$, by comparing the exponents of $\tau$, we see from (2.4) that

[TABLE]

necessarily. Since $\Phi_{2}$ has order $d$, we may assume that these are equalities in $\mathbb{Z}$. Notice that $A_{\Phi_{2}}(0)=0$ and $A_{\Phi_{2}}(1)=a_{2}$. Hence, the claim now follows by comparing the exponents of $\sigma$ and $\theta$ in the above with those in (2.4). ∎

Let us make a useful observation. Given a prime divisor $q$ of $g$, recall that $k\not\equiv 1$ (mod $q$) and that $d_{q}$ denotes the multiplicative order of $k$ mod $q$, so in particular $S(k,d_{q})\equiv 0$ (mod $q$). For $i,i^{\prime}\in\mathbb{N}_{\geq 0}$, we then have the relation

[TABLE]

We are now ready to give criteria for the conditions in (2.3) to all hold.

Proposition 2.10**.**

The five conditions in $(\ref{conditions})$ simultaneously hold precisely when the conditions

1. (a)

We have $\gcd(a_{1},e)=1$ and $\gcd(b_{2},d)=1$. 2. (b)

*We have $s_{2}\equiv 1\pmod{z}$ and $a_{2}\equiv 0\pmod{z}$. * 3. (c)

For each prime $q\mid g$ and $q\mid c_{1}$, we have

[TABLE] 4. (d)

For each prime $q\mid g$ and $q\nmid c_{1}$, we have

[TABLE]

are all satisfied.

Proof.

First, suppose that all of the conditions in (2.3) hold.

1. (a)

By Lemmas 2.5 and 2.8, we have $\gcd(a_{1},e)=1$ and $\gcd(b_{2},d)=1$. 2. (b)

Since $k\equiv 1$ (mod $z$), we have have $s_{2}\equiv 1$ (mod $z$) by Lemma 2.7. From Lemma 2.6, we then see that

[TABLE]

This yields $a_{2}\equiv 0$ (mod $z$) because $d$ is coprime to $e$ and hence to $z$. 3. (c)

For each prime $q\mid g$ and $q\mid c_{1}$, we have $s_{2}\equiv 1$ (mod $q$) by Lemma 2.9(4). From Lemma 2.6, we then see that

[TABLE]

This yields $c_{2}\equiv 0$ (mod $q$) because $d$ is coprime to $e$ and hence to $g$. By Lemma 2.7, we also have

[TABLE]

which implies that $b_{2}\equiv 1$ (mod $d_{q}$). 4. (d)

For each prime $q\mid g$ and $q\nmid c_{1}$, we have $s_{2}\equiv k$ (mod $q$) by Lemma 2.7. From Lemma 2.9(2), we then deduce that

[TABLE]

which implies that $b_{2}\equiv-1$ (mod $d_{q}$). The other two congruences follow directly from Lemma 2.9(1),(3).

We have thus shown that (a) through (d) are all satisfied.

Conversely, suppose that (a) through (d) are all satisfied. Then, clearly $\Phi_{1}$ has order $e$ by Lemma 2.5. Hence, it remains to verify that the conditions in Lemmas 2.6, 2.7, 2.8, and 2.9 hold. Since $e$ is squarefree, it suffices to check them mod $z$ as well as mod $q$ for each prime divisor $q$ of $g$.

1. (i)

Lemma 2.6: It is obvious that

[TABLE]

Since $S(k,d)\equiv 0$ (mod $q$), we also clearly have

[TABLE]

For $q\nmid c_{1}$, using (2.7), for any $j\in\mathbb{N}_{\geq 0}$ we compute that

[TABLE]

Since $b_{2}$ is coprime to $d$, we have $k^{b_{2}}\not\equiv 1$ (mod $d_{q}$), and the above yields

[TABLE]

We have thus shown that $\Phi_{2}$ has order $d$. 2. (ii)

Lemma 2.7: The conditions there clearly hold so $\Phi_{2}\Phi_{1}\Phi_{2}^{-1}=\Phi_{1}^{k}$. 3. (iii)

Lemma 2.8: For any $j\in\mathbb{N}_{\geq 0}$, observe that

[TABLE]

Since $a_{1}$ is coprime to $e$, it follows that the above expression on the left is coprime to $e$. We have thus shown that $\langle\Phi_{1},\Phi_{2}\rangle$ is regular. 4. (iv)

Lemma 2.9: By (a) and (iii) above, there exist $\widetilde{a_{1}},\widetilde{u}\in\mathbb{Z}$ such that

[TABLE]

To exhibit the existence of $i\in\mathbb{N}_{\geq 0}$ claimed in Lemma 2.9, simply take

[TABLE]

In each part of Lemma 2.9, the first congruence then clearly holds, and for $q\mid c_{1}$, the second congruence also holds. For $q\nmid c_{1}$, observe that

[TABLE]

and so the second congruence holds in parts (1), (2), and (3). Also, note that by (iii) above, we have $\widetilde{u}\equiv\widetilde{a_{1}}k$ (mod $q$). We then compute that

[TABLE]

which verifies the second congruence in part (4), and similarly that

[TABLE]

which verifies the second congruence in part (5) since $z$ is coprime to $g$. We have thus shown that $\langle\Phi_{1},\Phi_{2}\rangle$ is normal in $\mathrm{Hol}(G)$.

This completes the proof of the proposition. ∎

In order to compute the actual group structure of $T(G)$, we shall also need to understand the map $\xi_{\langle\Phi_{1},\Phi_{2}\rangle}$ defined in (1.1).

Proposition 2.11**.**

Assume that the conditions $(a)$ to $(d)$ in Proposition 2.10 are all satisfied. Then, for any $i,j\in\mathbb{N}_{\geq 0}$, we have

[TABLE]

where we define

[TABLE]

and $\widetilde{k}\in\mathbb{Z}$ is such that $k\widetilde{k}\equiv 1\pmod{e}$. Moreover, we have

[TABLE]

Proof.

The first claim follows from (2.4). The congruence mod $z$ is also clear since $k\equiv 1$ (mod $z$) and $A_{\Phi_{2}}(j)\equiv 0$ (mod $z$). For a prime $q\mid g$, recall that

[TABLE]

For $q\mid c_{1}$, we have $A_{\Phi_{2}}(j)\equiv a_{2}S(k,j)$ (mod $q$), and so the claim is clear. As for $q\nmid c_{1}$, observe that

[TABLE]

by (iii) and (i) in the proof of Proposition 2.10. Since $k^{b_{2}}\equiv\widetilde{k}$ (mod $q$) in this case, we deduce that

[TABLE]

as claimed. This proves the proposition. ∎

2.3. Proof of Theorem 2.1

Let us first make a definition.

Definition 2.12**.**

Given a tuple $\mathfrak{m}=(a_{1},c_{1},a_{2},b_{2},c_{2},s_{2})$ of integers, we shall say that $\mathfrak{m}$ is admissible if the conditions (a) to (d) in Proposition 2.10 are all satisfied. The admissibility of $\mathfrak{m}$ depends only on its class mod $\mathbb{M}$, where

[TABLE]

In the case that $\mathfrak{m}$ is admissible, define

[TABLE]

which satisfy (2.3) by Proposition 2.10. Further, define

[TABLE]

An explicit description of $\pi_{\mathfrak{m}}$ was given in Proposition 2.11 and the $\phi_{-1}$ above is only to remove the negative sign in the exponent of $\sigma$ for convenience.

Recall that $T(G)$ has the same size as $\mathcal{H}_{0}(G)$. By (1.4), we plainly have

[TABLE]

From Lemma 2.3 and Proposition 2.10, we then deduce that

[TABLE]

We shall now compute this number.

Proof for the order.

Consider a pair $(f_{1},f_{2})\in\mathbb{N}^{2}$ with $g=f_{1}f_{2}$. Below, let $q$ be an arbitrary prime divisor of $g$, and we shall count the admissible tuples

[TABLE]

under the restriction that

[TABLE]

Since $g$ is squarefree, this is equivalent to

[TABLE]

Since $d$ is squarefree and $d=\operatorname*{lcm}_{q\mid g}d_{q}$, there exists $b_{2}\in\mathbb{Z}$ satisfying

[TABLE]

exactly when $(f_{1},f_{2})\in\mathfrak{o}(G)$. Hence, for $(f_{1},f_{2})\notin\mathfrak{o}(G)$, we get no admissible tuple satisfying (2.9). So let us assume that $(f_{1},f_{2})\in\mathfrak{o}(G)$. Once we impose (2.9), in order for $\mathfrak{m}$ to satisfy the conditions in Proposition 2.10, the classes $s_{2}$ mod $e$ and $b_{2}$ mod $d$ are uniquely determined. Note that mod $z$, we have

• •

$\varphi(z)$ choices of $a_{1}$,

• •

$1$ choice of $a_{2}$.

Similarly, mod $q$ for $q\mid f_{1}$, we have

• •

$1$ choice of $c_{1}$ and $c_{2}$,

• •

$\varphi(q)$ choices of $a_{1}$,

• •

$q$ choices of $a_{2}$.

Finally, mod $q$ for $q\mid f_{2}$, we have

• •

$\varphi(q)$ choices of $c_{1}$ and $c_{1}$ determines $a_{1}$ uniquely,

• •

$q$ choices of $c_{2}$ and $c_{2}$ determines $a_{2}$ uniquely,

because $k\not\equiv 1$ (mod $q$). It follows that mod $\mathbb{M}$, we have

[TABLE]

admissible tuples satisfying (2.9). Therefore, in total, we have $\#\mathfrak{o}(G)\cdot g\varphi(e)$ admissible tuples without any restriction, so indeed $|T(G)|=\#\mathfrak{o}(G)$.∎

A proof of (1.3) was given in [20, Section 2] and the key is that isomorphic regular subgroups of $\mathrm{Perm}(G)$ are always conjugates. In fact, by the proof of [20, Lemma 2.1], if $N_{1}$ and $N_{2}$ are isomorphic regular subgroups of $\mathrm{Perm}(G)$, via the isomorphism $\psi:N_{1}\longrightarrow N_{2}$ say, then

[TABLE]

For any admissible tuple $\mathfrak{m}=(a_{1},c_{1},a_{2},b_{2},c_{2},s_{2})$, by taking

[TABLE]

to be the natural isomorphism, we see that

[TABLE]

Note that $\pi_{\mathfrak{m}}=\phi_{-1}\circ\widetilde{\pi_{\mathfrak{m}}}$ and $\phi_{-1}\in\mathrm{Aut}(G)$. By (1.3), we then have

[TABLE]

We shall now compute the structure of $T(G)$.

Proof for the structure.

It suffices to show that the exponent of $T(G)$ divides two because any group of exponent two is necessarily abelian. Let

[TABLE]

be an admissible tuple and we need to show that $\pi_{\mathfrak{m}}^{2}\in\mathrm{Hol}(G)$. Note that for any $i,j\in\mathbb{N}_{\geq 0}$, by Proposition 2.11, we have

[TABLE]

where the exponent of $\tau$ is simply $j$ because $b_{2}^{2}\equiv 1$ (mod $d$) by admissibility. Here $\mathfrak{a}_{\mathfrak{m}}(i,j)$ is as in (2.8) and $\mathfrak{v}_{\mathfrak{m}}(i,j)$ may be computed explicitly as follows.

First, it is clear that we have

[TABLE]

Next, let $q$ denote an arbitrary prime divisor of $g$. Recall that

[TABLE]

by admissibility. Without loss of generality, let us assume that $0\leq j\leq d-1$.

1. $\bullet$

$q\mid c_{1}:$ Using Proposition 2.11 and (2.7), we compute that

[TABLE] 2. $\bullet$

$q\nmid c_{1}:$ Using Proposition 2.11, we compute that

[TABLE]

Since $e=gz$ is squarefree, we then deduce that

[TABLE]

where $\widetilde{z},u\in\mathbb{Z}$ are such that

[TABLE]

It follows that

[TABLE]

This shows that $\pi_{\mathfrak{m}}^{2}\in\mathrm{Aut}(G)$ so $T(G)$ is indeed elementary $2$-abelian. ∎

3. Groups of odd prime power order

Throughout this section, let $G$ denote a finite group of order a power of an odd prime $p$, and of nilpotency class $n_{c}$. Write $Z(G)$ for its center and let

[TABLE]

be its lower central series, namely

[TABLE]

Note that $\gamma_{i}(G)=1$ for $i\geq n_{c}+1$. Write

[TABLE]

For convenience, let us assume that $r\geq 1$, namely $n_{c}\geq 2$, for otherwise $G$ is abelian and the structure of $T(G)$ is known. Consider the subgroup

[TABLE]

of the cyclic multiplicative group $(\mathbb{Z}/p^{r}\mathbb{Z})^{\times}$. Note that

[TABLE]

Further, for each $n\in\mathbb{Z}$, define

[TABLE]

to be the $n$th power map on $G$. We shall prove:

Theorem 3.1**.**

Assume that $2\leq n_{c}\leq p-1$. Then, the map

[TABLE]

is a well-defined injective homomorphism.

Let us remark that Theorem 3.1 is motivated by [7, Proposition 3.1]. The latter treats the special case when $n_{c}=2$.

To prove Theorem 3.1, rather than (1.3) and (1.4), we shall use the definition (1.2) directly. For each $n\in\mathbb{Z}$ coprime to $p$, clearly $\pi_{n}$ is a bijection that centralizes $\mathrm{Aut}(G)$, whence we have

[TABLE]

For each $\sigma\in G$, let us further define

[TABLE]

Explicitly, for any $x\in G$, we have

[TABLE]

where $\widetilde{n}\in\mathbb{Z}$ is such that $n\widetilde{n}\equiv 1$ (mod $\exp(G)$).

Lemma 3.2**.**

Let $n\in\mathbb{Z}$ be coprime to $p$ and let $\sigma\in G$. Then, we have

[TABLE]

Proof.

This is because $\pi_{n}\circ\rho(\sigma)\circ\pi_{n}^{-1}$ sends $1_{G}$ to $\sigma^{-n}$. ∎

3.1. Some commutator calculations

In this subsection, let $x,y\in G$ be two arbitrary elements. In the case that $n_{c}=2$, as shown in [18, (5.3.5)], for example, for any $n\in\mathbb{N}$, we have the well-known formula

[TABLE]

and using this, it may be verified that (3.2) is a homomorphism. This was in fact the key idea to the proof of [7, Proposition 3.1]. In general, although (3.3) might not hold, the difference $(xy)^{-n}x^{n}y^{n}$ is still a product of commutators. This difference was first studied by P. Hall in [11]. For our purpose, we shall use the following so-called Hall-Petresco formula for its description.

Lemma 3.3**.**

For any $n\in\mathbb{N}$, we have the formula

[TABLE]

where $c_{2}(x,y)=[x,y][[x,y],y]$ and $c_{i}(x,y)\in\gamma_{i}(\langle x,y\rangle)$ for each $i=2,\dots,n$.

Proof.

See [19, Theorem 3.5] and the discussion after it. ∎

Our idea is to impose conditions on $n\in\mathbb{N}$ and the nilpotency class $n_{c}$, so that we may reduce the formula in Lemma 3.3 to (3.3), even when $n_{c}\geq 3$.

Lemma 3.4**.**

Assume that $2\leq n_{c}\leq p-1$. Then, for any $n\in\mathbb{N}$ such that

[TABLE]

*the formula $(\ref{c2 formula})$ holds. *

Proof.

We may assume that $n_{c}\geq 3$, which means that $t\geq 1$, whence $p\leq n$. Hence, we have $n_{c}<p\leq n$, and since $\gamma_{i}(G)=1$ for $i\geq n_{c}+1$, the formula in Lemma 3.3 may be rewritten as

[TABLE]

For each $2\leq i\leq n_{c}$, observe that $p^{t}$ divides $n-\epsilon$ and hence divides

[TABLE]

because $i\leq n_{c}<p$ implies $i!$ has no factor of $p$. Since $\gamma_{3}(G)$ has exponent $p^{t}$ by definition, we then deduce that

[TABLE]

Put $N={n\choose 2}$, which is divisible by $p^{t}$. We may then apply the above formula again to the elements $[x,y]$ and $[[x,y],y]$ in $c_{2}(x,y)$ to obtain

[TABLE]

Again, since $\gamma_{3}(G)$ has exponent $p^{t}$, we have

[TABLE]

This proves the claim. ∎

The next lemma is well-known in the special case $n_{c}=2$. Recall that $G$ is said to be regular (as a $p$-group) if for every $\sigma,\tau\in G$ and $i\in\mathbb{N}$, we have

[TABLE]

It is known, by Lemma 3.3, that $G$ is regular when $n_{c}\leq p-1$. We thank the referee for pointing out that Lemma 3.5 below is true as long as $G$ is regular and for giving the following simple proof.

Lemma 3.5**.**

Assume that $G$ is regular (as a $p$-group). Then, for any $n\in\mathbb{N}$ which is a multiple of $p^{t}$, we have

1. (a)

$[x,y]^{n}\in Z(G)$; 2. (b)

$[\sigma,xy]^{n}=[\sigma,x]^{n}[\sigma,y]^{n}$* for all $\sigma\in G$;* 3. (c)

$[x,y]^{n}=[x,y^{n}]$.

Proof.

Let $n=mp^{t}$ with $m\in\mathbb{N}$. For any $z\in G$, by [10, Theorem 12.4.3 (1)] and regularity, we have the equivalence

[TABLE]

The latter holds because $\gamma_{3}(G)$ has exponent $p^{t}$. This yields $[x,y]^{p^{t}}\in Z(G)$, which clearly implies part (a). For any $\sigma\in G$, we have the identity

[TABLE]

By regularity and the fact that $\gamma_{3}(G)$ has exponent $p^{t}$, we obtain

[TABLE]

where $\mu_{1}$ and $\mu_{2}$, respectively, belong to the subgroups

[TABLE]

Both of them lie in $\gamma_{3}(G)$ and thus $\mu_{1}^{p^{t}},\mu_{2}^{p^{t}}=1_{G}$. Since $[\sigma,x]^{p^{t}},[\sigma,y]^{p^{t}}\in Z(G)$ by part (a), raising (3.4) to the $m$th power gives part (b). Now, by regularity, we also have

[TABLE]

where $\mu$ belongs to the subgroup

[TABLE]

This lies in $\gamma_{3}(G)$ and so $\mu^{p^{t}}=1_{G}$. Replacing $y$ by $y^{m}$ in (3.5) then yields

[TABLE]

But by part (b), we know that

[TABLE]

and this proves part (c). ∎

3.2. Proof of Theorem 3.1

Assume that $2\leq n_{c}\leq p-1$. In the following, let $n\in\mathbb{N}$ be such that $p\nmid n$ and $n\equiv 1$ (mod $p^{t}$). Let $\widetilde{n}\in\mathbb{N}$ be as after (3.2), and we have $\widetilde{n}\equiv 1$ (mod $p^{t})$. Also, note that $p^{t}$ divides both ${n\choose 2}$ and ${\widetilde{n}\choose 2}$.

First, let $\sigma\in G$ and consider the map $\zeta_{n,\sigma}$ defined in (3.2). By Lemmas 3.4 and 3.5(a), for any $x\in G$, we may rewrite

[TABLE]

Applying Lemmas 3.4 and 3.5(a),(b) again, for any $x,y\in G$, we then have

[TABLE]

and this proves that $\zeta_{n,\sigma}$ is a homomorphism. By Lemma 3.2 and (3.1), this in turn implies that $\pi_{n}\in\mathrm{NHol}(G)$, so that $\pi_{n}\mathrm{Hol}(G)\in T(G)$.

Next, observe that since $\pi_{n}$ sends $1_{G}$ to $1_{G}$, we have

[TABLE]

From Lemmas 3.4 and 3.5(c), we then deduce that

[TABLE]

where the last equivalence holds because $p\nmid n$, which means that the highest power of $p$ dividing $n\choose 2$ equals that of $n-1$. This shows that $\kappa$ is well-defined and injective. Clearly $\kappa$ is a homomorphism and the theorem now follows.

3.3. Further questions

Theorem 3.1 raises two natural questions:

How often Theorem 3.1 may be applied to show that $T(G)$ is not elementary $2$-abelian or not even a $2$-group? 2. 2.

Can $T(G)$ be a non-$2$-group or can we generalize Theorem 3.1 to the case when $G$ has nilpotency class $n_{c}\geq p$?

Let us briefly discuss these two questions.

3.3.1. Applicability

In the case that $n_{c}=2$, namely $r\geq 1$ and $t=0$, from Theorem 3.1 we see that $T(G)$ has a cyclic subgroup of order $(p-1)p^{r-1}$, so in particular

• •

$T(G)$ is not elementary $2$-abelian when $p\geq 5$;

• •

$T(G)$ is not a $2$-group when $p-1$ is not a power of two or $r\geq 2$.

This was first shown in [7, Proposition 3.1] and some specific examples were given in [7, Section 5].

In the case that $3\leq n_{c}\leq p-1$, we have $t\geq 1$, and from Theorem 3.1 we see that $T(G)$ has a cyclic subgroup of order $\max\{1,p^{r-t}\}$, whence

• •

$T(G)$ is not a $2$-group when $r\geq t+1$.

However, the condition $r\geq t+1$ seems quite restrictive, and in particular it imposes that $r\geq 2$ for $t\geq 1$. Notice that $G$ has order at least $p^{n_{c}+1}$, and $G$ is said to be of maximal class if its order is precisely $p^{n_{c}+1}$.

Lemma 3.6**.**

Assume that $G$ is of maximal class and that $n_{c}\leq p$. Then, the quotient $G/Z(G)$ has exponent $p$, namely $r=1$.

Proof.

See [1, Theorem 9.5].∎

All groups of order $p,p^{2},p^{3}$ have nilpotency class at most $2$. For groups $G$ of order $p^{4}$, none of them can satisfy $n_{c}=3$ and $r\geq t+1$ by Lemma 3.6. As for groups $G$ of order $p^{5},p^{6}$, it is known that they may be separated into $10$ and $43$ so-called isoclinism families, respectively. In the two tables below, let us summarize the information of interest to us, which is found in [12, Section 4]. The list of groups of order $p^{6}$ given in [12] is incomplete and the number of groups in each family stated below is taken from [17, Table 2].

The isoclinism families of groups of order $p^{5}$ and nilpotency class $n_{c}\geq 3$ for $p\geq 5$.

The isoclinism families of groups of order $p^{6}$ and nilpotency class $n_{c}\geq 3$ for $p\geq 5$.

Hence, for $p\geq 5$, there are exactly one and $6p+12$, respectively, groups $G$ of order $p^{5}$ and $p^{6}$ with $n_{c}\geq 3$ and $r\geq t+1$. For both $i=5,6$, the ratio

[TABLE]

is quite small and tends to zero as $p$ increases. But for $i=6$, the numerator tends to infinity as $p$ increases, and so Theorem 3.1 gives infinitely many new examples $G$ (with $3\leq n_{c}\leq p-1$) for which $T(G)$ is not a $2$-group.

3.3.2. Large nilpotency class

The key ingredient to the proof of Theorem 3.1 is Lemmas 3.4 and 3.5. In view of Lemma 3.5, it seems natural to ask whether Theorem 3.1 may be extended to the case when $G$ is only regular but $n_{c}\geq p$. We thank the referee for suggesting this. However, if $G$ is regular but $n_{c}\geq p$, then $\pi_{n}$ is reasonably well-behaved only when $n$ is a power of $p$, in which case $\pi_{n}$ is not bijective. In fact, by using the SmallGroup library [2] in Magma [3], we found a counterexample.

Example 3.7**.**

Let $G=\mbox{{\sc SmallGroup}}(729,22)$. The following information may be easily calculated in Magma.

• •

$G$ has exponent $27$.

• •

$G$ has nilpotency class $3$, namely $n_{c}=3$.

• •

$G/Z(G)$ has exponent $9$, namely $r=2$.

• •

$\gamma_{3}(G)$ has exponent $3$, namely $t=1$.

• •

$\gamma_{2}(G)$ has order $9$.

• •

$\gamma_{2}(G)$ is a cyclic group.

The last fact implies that $G$ is regular as a $3$-group, by [19, (3.13)]. However, running the code below in Magma yields the output $[1,8,10,17,19,26]$.

⬇

G:=SmallGroup(729,22);

X:=[x:x in G];

s:=X[82]; // a choice of $\sigma\in G$

x:=X[244]; // a choice of $x\in G$

y:=x^2;

L:=[];

for n in [n:n in [1..27]|GreatestCommonDivisor(n,3) eq 1] do

nt:=InverseMod(n,27); // the $\widetilde{n}\in\mathbb{N}$ in (3.2)

Zx:=(x^(nt)*s^(-1))^n*s^n; // the value of $\zeta_{n,\sigma}(x)$ as in (3.2)

Zy:=(y^(nt)*s^(-1))^n*s^n; // the value of $\zeta_{n,\sigma}(x^{2})$ as in (3.2)

if Zy eq (Zx)^2 then // test if $\zeta_{n,\sigma}(x^{2})=\zeta_{n,\sigma}(x)^{2}$ holds

  Append(~L,n);

end if;

end for;

L; // list of $n$ mod $27$ for which $\zeta_{n,\sigma}$ might be in $\mathrm{Aut}(G)$

This implies that there exists $\sigma\in G$ such that $\zeta_{n,\sigma}\not\in\mathrm{Aut}(G)$ for all $n\in\mathbb{Z}$ coprime to $3$ with $n\not\equiv\pm 1$ (mod $9$). By (3.1) and Lemma 3.2, it follows that $\pi_{n}\notin\mathrm{NHol}(G)$ for these $n$, whence $\kappa$ is not well-defined.

Let $m\in\mathbb{Z}$. For any $x,y\in G$, by regularity we have

[TABLE]

and $\mu^{9}=1_{G}$ because $\gamma_{2}(G)$ has order $9$. Also, notice that $x^{9},y^{9}\in Z(G)$ since $G/Z(G)$ has exponent $9$. It follows that

[TABLE]

This shows that

[TABLE]

Therefore, the power maps $\pi_{n}$ with $n\in\mathbb{Z}$ coprime to $3$ only give rise to the subgroup $\langle\pi_{-1}\mathrm{Hol}(G)\rangle$ of order two in $T(G)$.

Nonetheless, by computing the size of (1.4), which is the order of $T(G)$, we found in Magma that $T(G)$ has order $18$ and hence is not a $2$-group.

Example 3.7 shows that Theorem 3.1 may not be generalized to all regular $p$-groups. It also shows that there is a (regular) $p$-group $G$ of nilpotency class at least $p$ for which $T(G)$ is not a $2$-group, and the elements in $T(G)$ of odd order do not arise from the power maps. It leads to the following questions.

Question 3.8**.**

Give more examples of $p$-groups $G$ of nilpotency class $n_{c}\geq p$ for which $T(G)$ is not a $2$-group.

Question 3.9**.**

Other than the power maps $\pi_{n}$ for $n$ coprime to $p$, what are other ways to construct elements in $T(G)$ of odd order?

A. Caranti [7] used bilinear maps but his method is specific for $p$-groups of nilpotency class two. It would be interesting to see if there are other methods which work more generally, and in particular, figure out where the elements in $T(G)$ of odd order come from for $G$ with SmallGroup ID equal to $(729,22)$, as well as those in (1.5).

4. Acknowledgments

The author thanks the referee for helpful suggestions which led to the discussion in Subsection 3.3.2.