On connected quandles of prime power order

Abstract

We develop some general ideas to study connected quandles of prime power size and we classify non-affine connected quandles of size $p^3$ for $p>3$, using a combination of group theoretical and universal algebraic tools. As a byproduct we obtain a classification of Bruck loops of the same size and that they are in one-to-one correspondence with commutative automorphic loops.

Appendix

In the present Appendix we illustrate the computations that lead us to determine the conjugacy classes of the automorphisms of the relevant groups which give rise to the isomorphism classes listed in Tables 1 and 2. To this end, for the relevent groups of size $p^4$ we start with a presentation of the generic automorphism of the group, which can be found in Girnat (2018), and then proceed to impose the necessary and sufficient conditions which we have stated in Proposition 5.7 and the discussion thereof.

1.1 Conjugacy classes of automorphisms of ${\mathbb {Z}}_p^2\rtimes {\mathbb {Z}}_p$

In this section we will use the power-commutator presentation of the group $G={\mathbb {Z}}_p^2\rtimes {\mathbb {Z}}_p$ as

$$\begin{aligned} G=\langle g_1, g_2, g_3,\, | \, [g_2,g_1]=g_3\rangle . \end{aligned}$$

In particular, $Z(G)=\gamma _1(G)=\langle g_3\rangle$. The automorphisms of G are given by the mappings

$$\begin{aligned} f= \left\{ \begin{array}{l} g_1\mapsto g_1^a g_2^c g_3^\alpha \\ g_2\mapsto g_1^b g_2^d g_3^\beta ,\\ g_3\mapsto g_3^{ad-bc} \end{array}\right. \end{aligned}$$

where $a,b,c,d,\alpha ,\beta \in \{0,\ldots ,p-1\}$ and $ad-bc\ne 0$. We will denote such automorphism as $f=f(a,b,c,d,\alpha ,\beta )$.

Proposition 7.1

A set of representatives of conjugacy classes of ${\text {Aut}}\left( {G}\right)$ with no fixed points is:

• (i) $f_{\lambda , \mu } = f(\lambda , 0,0,\mu ,0,0)$, where $\lambda , \mu \in \{2,\ldots , p-1\}, \lambda \mu \ne 1$ and $\lambda \le \mu$.

• (ii) $h_{\lambda } = f(\lambda ,0,1,\lambda ,0,0)$, where $\lambda \in \{2,\ldots , p-2\}$.

• (iii) $f_{q} = f(0, 1,-a,-b,0,0)$, where $q(x)=x^2+bx+a$ is an irreducible polynomial and $a\ne 1$.

The number of such conjugacy classes is $p^2-2p-1$.

Proof

Let $f\in {\text {Aut}}\left( {G}\right)$ with no fixed points and let $M=f_{\gamma _1(G)}$ be the matrix of its action on the quotient by $\gamma _1(G)$. Since $f(g_3)=g_3^{ab-cd}$, then M has determinant different from 1. We can have the following cases:

• (i) M is diagonalizable with eigenvalues $\lambda \le \mu$ and $\lambda \mu \ne 1$. So there exist x, y, z, t such that $h=h(x,y,z,t,0,0)$ satisfies:

  $$\begin{aligned} h f h^{-1}=f(\lambda , 0,0,\mu ,\delta ,\rho ) \end{aligned}$$

  for some $\delta ,\rho$. Conjugating $f(\lambda , 0,0,\mu ,\delta ,\rho )$ by $w=w(1,0,0,1,x,y)$ with $x=\frac{\delta }{\lambda (\mu -1)}$ and $y=\frac{\rho }{\mu (\lambda -1)}$ we obtain $f_{\lambda ,\mu }$. Finally, note that $f_{\lambda ,\mu }$ and $f_{\mu ,\lambda }$ are conjugate by $w=w(0,1,1,0,0,0)$.

• (ii) M has one eigenvalue $\lambda \ne 1$ of multiplicity 2 and it is not diagonalizable. So there exist x, y, z, t such that $h=h(x,y,z,t,0,0)$ satisfies:

  $$\begin{aligned} h f h^{-1}=f(\lambda , 0,1,\lambda ,\delta ,\rho ) \end{aligned}$$

  for some $\delta ,\rho$. Conjugating $f(\lambda ,0,1,\lambda ,\delta ,\rho )$ by $w=w(1,0,0,1,x,y)$ with $x=\frac{\delta }{\lambda (\lambda -1)}$ and $y=\frac{\rho -x}{\lambda (\lambda -1)}$ we obtain $h_{\lambda }$.

• (iii) M has no eigenvalues. So there exist x, y, z, t such that $h=h(x,y,z,t,0,0)$ satisfies

  $$\begin{aligned} h f h^{-1}=f(0, 1,-a,-b,\delta ,\rho ) \end{aligned}$$

  for some $\delta ,\rho$ and $a\ne 1$ such that $q(x)=x^{2}+bx+a$ is an irreducible polynomial. Conjugating $f(0, 1,-a,-b,\delta ,\rho )$ by $w=w(1,0,0,1,x,y)$ with $x=\frac{(a+b)\delta +\rho }{q(a)}$ (since p is irreducible x is well defined) and $y=ax-\delta$ we obtain $f_q$.

The automorphisms in the list are not conjugate since the induced automorphisms on $G/\gamma _1(G)$ are not. In particular, there are $\frac{(p-1)(p-3)}{2}$ different $f_{\lambda ,\mu }$, $p-3$ different $h_\lambda$ and $\frac{(p-1)^2}{2}$ irreducible polynomial with constant term different from 1 (since there are $\frac{p-1}{2}$ irreducible polynomials of degree two and constant term equal to 1 (Yucas 2006, Theorem 3.5). So the number of conjugacy classes is

$$\begin{aligned} \frac{(p-1)(p-3)}{2}+\frac{(p-1)^2}{2}+p-3=(p-1)(p-2)+p-3=p^2-2p-1. \end{aligned}$$

$\square$

Proposition 7.2

A set of representatives of conjugacy classes of automorphisms f such that $Fix(f)=Z(G)$ and $Fix(f_{\gamma _1(G)})=1$ is:

• (i) $f_{\lambda } = f(\lambda , 0,0,\lambda ^{-1},0,0)$, where $\lambda ,\lambda ^{-1}, \in \{2,\ldots , p-1\}$ and $\lambda < \lambda ^{-1}$.

• (ii) $h_{-1} = f(-1, 1,0,-1,0,0)$.

• (iii) $f_{q,1} = f(0, 1,-1,-b,0,0)$, where $q=x^2+bx+1$ is an irreducible polynomial.

The number of such conjugacy classes is p.

Proof

The proof is analogous to the proof of Proposition 7.1, taking into account that we are now dealing with automorphisms with $Fix(f)=Z(G)$, and so $f_{\gamma _1(G)}$ has determinant 1.

There are $\frac{p-1}{2}$ different $f_\lambda$ and $\frac{p-1}{2}$ irreducible polynomials of degree two and constant term equal to 1 (Yucas 2006, Theorem 3.5). Hence there are p such conjugacy classes. $\square$

1.2 Conjugacy classes of automorphisms of the relevant groups of order $p^4$ for $p>3$

1.2.1 Case $G=G_7$

In this section we will use the power-commutator presentation of the group $G_7$ as

$$\begin{aligned} G_7=\langle g_1, g_2, g_3, g_4, \, | \, [g_2,g_1]=g_3, \, [g_3,g_1]=g_4\rangle . \end{aligned}$$

We will denote such group just by G and by M the subgroup generated by $g_2,g_3$ and $g_4$. Note that $\gamma _1(G)$ is generated by $g_3$ and $g_4$. Using an inductive argument it is easy to see the following formulas we are freely using in the computations with no further reference:

$$\begin{aligned} g_1^{-a}g_2g_1^a= & {} g_2 g_3^a g_4^{\frac{a(a-1)}{2}}, \end{aligned}$$

(18)

$$\begin{aligned} g_1^{-a}g_3 g_1^a= & {} g_3 g_4^a,\end{aligned}$$

(19)

$$\begin{aligned} (g_1^a g_2^b g_3^c)^n= & {} g_1^{an} g_2^{nb} g_3^{nc+ab\left( {\begin{array}{c}n\\ 2\end{array}}\right) }g_4^{ac\left( {\begin{array}{c}n\\ 2\end{array}}\right) +b\rho (a,n)}, \end{aligned}$$

(20)

where $n\in {\mathbb {Z}}$ $a,b,c\in \{ 0,\ldots , p-1\}$ and $\rho (a,n)=\frac{a}{2}\left( {\begin{array}{c}n\\ 2\end{array}}\right) (\frac{a(2n-1)}{3}-1)$. Following (Girnat 2018) and since M is an elementary abelian subgroup of rank 3, the automorphisms of G are given by the following collection of data:

$$\begin{aligned} h= \left\{ \begin{array}{l} H=h|_M=\begin{bmatrix} x &{} 0 &{} 0\\ y &{} tx&{} 0\\ z &{} ty+xt\frac{t-1}{2} &{} t^2x \end{bmatrix} , \qquad g_1\mapsto g_1^t g_2^{u} g_3^{\alpha } g_4^{\beta }\end{array}\right. , \end{aligned}$$

where $u,y,z,\alpha ,\beta \in \{0,\ldots , p-1\}$ and $x,t\in \{1,\ldots ,p-1\}$. A direct computation tells us that $|{\text {Aut}}\left( {G}\right) |=(p-1)^2p^5$. We denote h by $h(t,u,\alpha ,\beta ,x,y,z)$, and it is straightforward to verify that h satifies (A) if and only if $t x=1$ and $t\ne 1$.

Lemma 7.3

Let $f=f(a,b,\alpha ,\beta ,a^{-1},c,d)\in {\text {Aut}}\left( {G}\right)$ and let $f_{M}$ be the action of f on the quotient by M. Then $f_{M}=a$ is invariant under conjugation.

Proof

Let f and g be conjugated. Then $f_{M}$ and $g_{M}$ are conjugate in ${\text {Aut}}\left( {G/M}\right)$. Since $G/M\cong {\mathbb {Z}}_p$, then $f_M=g_M$. $\square$

Proposition 7.4

A set of representatives of conjugacy classes of automorphisms satisfying (A) with $a\ne \pm 1, p-1$ is given by

• (i) $f_a=f_a(a,0,0,0,a^{-1},0,0)$.

• (ii) $g_a=g_a(a,1,0,0,a^{-1},0,0)$.

In particular, there are $2(p-3)$ such conjugacy classes.

Proof

Let $h=h(t,u,\alpha ,\beta ,x,y,z)\in C_{{\text {Aut}}\left( {G}\right) }(f_a)$ or $h=h(t,u,\alpha ,\beta ,x,y,z)\in C_{{\text {Aut}}\left( {G}\right) }(f_a)$, then $H=h|_M$ centralizes $F=f_a|_M=g_a|_M$ and so $y=z=0$.

If $h=h(t,u,\alpha ,\beta ,x,0,0)\in C_{{\text {Aut}}\left( {G}\right) }(f_a)$, then $h_{\gamma _1(G)}$ centralizes ${f_a}_{\gamma _1(G)}$ and so $u=0$. Then $h=h(t,0,\alpha ,\beta ,x,0,0)\in C_{{\text {Aut}}\left( {G}\right) }(f_a)$ if and only if $\alpha =0$. These conditions on the coefficients of h are also sufficient for h to centralize $f_a$. Accordingly $|C_{{\text {Aut}}\left( {G}\right) }(f_a)|=p(p-1)^2$, since t, x and $\beta$ are free.

On the other hand, $h=h(t,u,\alpha ,\beta ,x,0,0)\in C_{{\text {Aut}}\left( {G}\right) }(g_a)$, if and only if $t,u,\alpha$ and x are solutions of the following system of linear equations over ${\mathbb {Z}}_p$:

$$\begin{aligned} \left\{ \begin{array}{l} u(a^{2}-1)=a(t-x) \\ \alpha (a-1)=a\left( {\begin{array}{c}t\\ 2\end{array}}\right) -tu\left( {\begin{array}{c}a\\ 2\end{array}}\right) \\ \alpha (a-1)(1-ta)=2u\rho (t,a)-2\rho (a,t) \end{array}\right. . \end{aligned}$$

A straightforward calculation shows that if $tx = 1$ then $u, \alpha$ are uniquely determined by x. Therefore $|C_{{\text {Aut}}\left( {G}\right) }(g_a)|=p(p-1)$.

The automorphisms $f_a$ and $g_a$ are not conjugate since the sizes of their centralizers are different. The sizes of conjugacy classes of $f_a$ and $g_a$ are respectively $p^4$ and $(p-1)p^4$ and the number of automorphisms of G satisfying (A) with fixed a is $p^5=p^4+(p-1)p^4$. So $f_a$ and $g_a$ are the representatives of their conjugacy classes. $\square$

We now discuss the case $a=-1$.

Lemma 7.5

Let $f=f(-1,b,\alpha ,\beta ,-1,c,d)\in {\text {Aut}}\left( {G}\right)$ satisfying condition (A). Then:

• (i) f is conjugate to

  $$\begin{aligned} f_0= \left\{ \begin{array}{l} F_0=\begin{bmatrix} -1 &{} 0 &{} 0\\ 0 &{} 1&{} 0\\ \frac{c(c+1)+2d}{2} &{} -1 &{} -1 \end{bmatrix} ,\qquad g_1\mapsto g_1^{-1} g_2^{b} g_3^{\gamma } g_4^{\delta } \end{array}.\right. \end{aligned}$$

  for some $\gamma ,\delta ,\in \{0,\ldots ,p-1\}$.

• (ii) If $g(-1,b,\alpha ,\beta ,-1,0,s)$ and $h(-1,b^\prime ,\alpha ^\prime ,\beta ^\prime ,-1,0,s^\prime )$ are conjugate, then $s=s^{\prime }r^2$ for some $r\ne 0$.

Proof

(i) Conjugating f by $h=h(1,0,0,0,1,\frac{c}{2},0)$ you obtain $f_0$.

(ii) The second statement follows by direct computations (it is enough to consider the restriction of h and g to M). $\square$

According to Lemma 7.5, every automorphism is conjugate to one of the standard form $g(-1,b,\alpha ,\beta ,-1,0,s)$. It follows from item (ii) that automorphisms with $s=0$ and $s^\prime \ne 0$ are not conjugate. The following Lemma provides a list of representatives of conjugacy classes in the case $s=0$.

Proposition 7.6

A set of representatives of the conjugacy classes of elements of the form $f=f(-1,b,\alpha ,\beta ,-1,c,d)\in {\text {Aut}}\left( {G}\right)$ satisfying (A) with $c(c+1)+2d=0$ is:

• (i) $\phi =\phi (-1,0,0,0,-1,0,0)$.

• (ii) $\varepsilon =\varepsilon (-1,0,0,1,-1,0,0)$.

• (iii) $\sigma =\sigma (-1,1,0,0,-1,0,0)$.

Proof

The automorphisms $\phi$ and $\varepsilon$ are not conjugate since they have different orders and $\sigma$ is not conjugate to $\phi$ nor to $\varepsilon$ since ${\sigma }_{\gamma _1(G)}$ is not conjugate to ${\phi }_{\gamma _1(G)}$ nor to ${\varepsilon }_{\gamma _1(G)}$. According to Lemma 7.5(i), we can assume that $f=f(-1,b,\alpha ,\beta ,-1,0,0)\in {\text {Aut}}\left( {G}\right)$.

It is easy to verify that $f^2=1$ if and only if $b=0$ and $\alpha +\beta =0$ and in particular $\phi$ has order 2. Let $f=f(-1,0,\alpha ,-\alpha ,0,0)$ be an automorphism of order 2. The automorphism $h=h(1,0,-\frac{\alpha }{2},0,1,0,0)$ fulfils $\phi =h^{-1}fh$.

Assume that $f_{\gamma (G)}= -I$ and that $f^2\ne 1$. Then $\varepsilon =h^{-1} f h$ for $h=(1,0,-\frac{\alpha }{2},0,\alpha +\beta ,0,0)\in {\text {Aut}}\left( {G}\right)$ ($\alpha +\beta \ne 0$).

Assume that $f_{\gamma (G)}\ne -I$ and let $h=h(1,0,-\frac{\alpha }{2},0,b,0,\beta )$. Then $\sigma =h^{-1} f h$. $\square$

Now we consider the conjugation classes of automorphisms of the form $f=f(-1,b,\alpha ,\beta ,-1,0,s)$ with $s\ne 0$.

Proposition 7.7

Let w be a quadratic non-residue modulo p. A set of representatives of conjugacy classes of elements of the form $f=f(-1,b,\alpha ,\beta ,-1,c,d)\in {\text {Aut}}\left( {G}\right)$ satisfying (A) with $c(c+1)+2d\ne 0$ is:

• (i) $f_1=f_1(-1,0,0,0,-1,0,1)$.

• (ii) $f_w=f_w(-1,0,0,0,-1,0,w)$.

• (iii) $h_{1}=h_{1}(-1,1,0,0,-1,0,1)$.

• (iv) $h_{w}=h_{w}(-1,1,0,0,-1,0,w)$.

Proof

The automorphisms $f_s$ and $h_s$, for $s=1,w$ are not conjugate since $\left( {f_s}\right) _{\gamma _1(G)}$ and $\left( {h_s}\right) _{\gamma _1(G)}$ are not conjugate. Moreover, $f_1$ and $f_w$ (resp. $h_1$ and $h_{w}$) are not conjugate, since ${f_{1}}|_{M}$ and ${f_{w}}|_{M}$ (resp. ${h_{1}}|_{M}$ and ${h_{w}}|_{M}$) are not conjugate by Lemma 7.5(ii).

Let $f\in {\text {Aut}}\left( {G}\right)$. We can assume that $f=f(-1,b,\alpha ,\beta ,-1,0,k)$ according to Lemma 7.5(i). If $k=t^2s$ for some $t\ne 0$ and $s\in \{1,w\}$ conjugating f by $h=h(t,0,0,0,0,0)$ we can assume that $f=f(-1,b^\prime ,\alpha ,\beta ,-1,0,s)$.

If $f_{\gamma _1(G)}\ne -I$ then conjugating f by $h=h(1,u,\gamma ,0,0,z)$ where $u=2b^\prime -1$, $\gamma =\frac{b^\prime -\alpha }{2}$ and $z=\beta -\gamma +us$ we get $h_s$.

If $f_{\gamma _1(G)}= -I$ then conjugating f by $h=h(1,u,\gamma ,0,0,0)$ where $u=\frac{\alpha }{s}$ and $\gamma =\frac{\alpha -u}{2}$ we get $f_s$. $\square$

1.2.2 Case $G=G_8$

In this section we will use the power-commutator presentation of the group $G_8$ as

$$\begin{aligned} G_8=\langle g_1, g_2, g_3, g_4, \, | \, [g_2,g_1]=g_3, \, [g_3,g_1]=g_1^p=g_4\rangle . \end{aligned}$$

We will denote such group just by G and by M the subgroup generated by $g_2,g_3$ and $g_4$. Note that $\gamma _1(G)$ is generated by $g_3$ and $g_4$. In particular note that (18), (19) and (20) holds for G. According to Girnat (2018) and since M is an elementary abelian subgroup of rank 3, the automorphisms of G are given by:

$$\begin{aligned} h= \left\{ \begin{array}{l} H=h|_M=\begin{bmatrix} t^{-1} &{} 0 &{} 0\\ y &{} 1&{} 0\\ z &{} ty+\frac{t-1}{2} &{} t \end{bmatrix} ,\qquad g_1\mapsto g_1^t g_2^{u} g_3^{\alpha } g_4^{\beta } \end{array}.\right. \end{aligned}$$

where $u,y,z,\alpha ,\beta \in \{0,\ldots , p-1\}$ and $t\in \{1,\ldots ,p-1\}$. We have that $|{\text {Aut}}\left( {G}\right) |=(p-1)p^5$. We denote h by $h(t,u,\alpha ,\beta ,y,z)$, and it is straightforward to verify that h satifies (A) if and only if $t \ne 1$. Note that for every fixed a there exists $p^5$ automorphisms satisfying (A).

As for $G_7$ we have the following (see Lemma 7.5).

Lemma 7.8

Let $f=f(a,b,\alpha ,\beta ,c,d)\in {\text {Aut}}\left( {G}\right)$ and let $f_{M}$ be the action of f on the quotient by M. Then $f_{M}=a$ is invariant under conjugation.

Proposition 7.9

Let $a\ne 1,p-1$. A set of representatives of conjugacy classes of automorphisms of G satisfying (A) is:

$$\begin{aligned} f_{a,b}= \left\{ \begin{array}{l} F_a=\begin{bmatrix} a^{-1} &{} 0 &{} 0\\ 0 &{} 1&{} 0\\ 0 &{} \frac{a-1}{2} &{} a \end{bmatrix} , \quad g_1\mapsto g_1^{a}g_4^b \end{array}, \right. \end{aligned}$$

for $b\in \{0,\ldots p-1\}$. In particular, there are $p(p-3)$ such conjugacy classes.

Proof

We compute the centralizer of $f_{a,b}$. We have that $h=h(t,u,\alpha ,\beta ,y,z)\in C_{a,b}=C_{{\text {Aut}}\left( {G}\right) }(f_{a,b})$ if and only if $\alpha =u=y=z=0$. Hence $|C_{a,b}|=p(p-1)$. It is easy to see that if $f_{a,b}$ and $f_{a,c}$ are conjugate, then $b=c$. Since for every a there are $p^5$ automorphisms of G satisfying (A) and

$$\begin{aligned} \sum _{b\in \{ 0,\ldots ,p-1\}}[{\text {Aut}}\left( {G}\right) :C_{a,b}]=p\frac{(p-1)p^5}{p(p-1)}=p^5, \end{aligned}$$

we are done. $\square$

Now we discuss the case $a=p-1$.

Lemma 7.10

Let $f=f(p-1,b,\alpha ,\beta ,c,d)\in {\text {Aut}}\left( {G}\right)$ satisfying condition (A). Then

• (i) f is conjugate to $f_0=f_0(p-1,b,\gamma ,\delta ,0, \frac{c(c+1)+2d}{2})$, i.e.

  $$\begin{aligned} f_0= \left\{ \begin{array}{l} F_0=\begin{bmatrix} -1 &{} 0 &{} 0\\ 0 &{} 1&{} 0\\ \frac{c(c+1)+2d}{2} &{} -1 &{} -1 \end{bmatrix} , \qquad g_1\mapsto g_1^{-1} g_2^{b} g_3^{\gamma } g_4^{\delta } \end{array},\right. \end{aligned}$$

  for some $\gamma ,\delta \in \{0,\ldots ,p-1\}$.

• (ii) If $g(p-1,b,\alpha ,\beta ,0,s)$ and $h(p-1,b^\prime ,\alpha ^\prime ,\beta ^\prime ,0,s^\prime )$ are conjugate, then $s^{\prime }=st^2$ and $b^\prime =\frac{b}{t^2}$ for some $t\ne 0$. Therefore $bs=b^\prime s^\prime$.

Proof

(i) Conjugating $f=f(p-1,b,\alpha ,\beta ,c,d)$ by $h=h(1,0,0,0,\frac{c}{2},0)$ we obtain $f_0$.

(ii) Let $g(p-1,b,\alpha ,\beta ,0,s)$ and $h(p-1,b^\prime ,\alpha ^\prime ,\beta ^\prime ,0,s^\prime )$ be conjugate by some $w=w(t,u,\gamma ,\delta ,x,y,z)$. Since $g_{\gamma _1(G)}$ and $h_{\gamma _1(G)}$ are conjugate by $w_{\gamma _1(G)}$ then $s^\prime =s t^2$ and $b^\prime =\frac{b}{t^2}$. $\square$

Moreover, if f and g are conjugate then also $f_{\gamma _1(G)}$ and $g_{\gamma _1(G)}$ are conjugate. If $f_{\gamma _1(G)}=-I$ then also $g_{\gamma _1(G)}=-I$. Let us show the representatives of classes of automorphisms with this property and for which (A) holds.

Lemma 7.11

Let $a=p-1$ and w be a quadratic non-residue modulo p. A set of representatives of conjugacy classes of automorphisms of G satisfying (A) and such that $f_{\gamma _1(G)}= -I$ and $\frac{c(c+1)+2d}{2}\ne 0$ is:

$$\begin{aligned} f_{s}= \left\{ \begin{array}{l} F_{s}=\begin{bmatrix} -1 &{} 0 &{} 0\\ 0 &{} 1&{} 0\\ s &{} -1 &{} -1 \end{bmatrix} , \qquad g_1\mapsto g_1^{p-1} , \end{array}\right. \end{aligned}$$

for $s\in \{1,w\}$.

Proof

The automorphism $h(t,u,\alpha ,\beta ,y,z)$ centralizes $f_s$ if and only if $t=\pm 1$, $y=0$, $u=\pm sz$, $\alpha =\frac{kz}{2}$. Then the centralizer of $f_s$ has size $2p^2$. Therefore

$$\begin{aligned} 2\frac{p^5(p-1)}{2p^2}=p^3(p-1). \end{aligned}$$

Let $f(p-1,b,\alpha ,\beta ,-1,c,-\frac{c(c+1)}{2})$ be an automorphism satisfying (A). Then $f_{\gamma _1(G)}=-I$ if and only if $b=0$. The number of such automorphisms is exactly $p^3(p-1)$. According to Lemma 7.10$f_1$ and $f_w$ are not conjugate, therefore they are a set of representatives of such automorphisms. $\square$

Proposition 7.12

A set of representatives of conjugacy classes of automorphisms of G satisfying (A) and such that $f_{\gamma _1(G)}= -I$ and $\frac{c(c+1)+2d}{2}= 0$ is:

$$\begin{aligned} f_{\beta }= \left\{ \begin{array}{l} F=\begin{bmatrix} -1 &{} 0 &{} 0\\ 0 &{} 1&{} 0\\ 0 &{} -1 &{} -1 \end{bmatrix} , \qquad g_1\mapsto g_1^{p-1}g_4^\beta , \end{array}\right. \end{aligned}$$

for $\beta \in \{0,\ldots ,p-1\}$. In particular, there are p such conjugacy classes.

Proof

Let $C_\beta =C_{{\text {Aut}}\left( {G}\right) }(f_\beta )$. Then $h=h(t,u,\alpha ,\gamma ,y,z)\in C_\beta$ if and only if $y=0$ and $\alpha =\frac{tu}{2}$. Hence $|C_\beta |=(p-1)p^3$ and then $[{\text {Aut}}\left( {G}\right) :C_\beta ]=p^2$. Moreover if $f_\beta$ and $f_\delta$ are conjugate then $\beta =\delta$. Since there are $p^3$ automorphisms of G satisfying the hypothesis then $\{f_\beta \,\mid \,\beta \in \{0,\ldots ,p-1\}\}$ is a set of representatives. $\square$

Let now compute the representatives of classes of automorphisms f such that $f_{\gamma _1(G)}\ne -I$.

Proposition 7.13

Let $a=p-1$ and w be a quadratic non-residue modulo p. A set of representatives of conjugacy classes of automorphisms of G satisfying (A) and such that $f_{\gamma _1(G)}\ne -I$ is:

$$\begin{aligned} f_{k,s}= \left\{ \begin{array}{l} F_{k,s}=\begin{bmatrix} -1 &{} 0 &{} 0\\ 0 &{} 1&{} 0\\ k &{} -1 &{} -1 \end{bmatrix} , \qquad g_1\mapsto g_1^{p-1} g_2^{s}, \end{array}\right. \end{aligned}$$

for $k\in \{0,\ldots p-1\}, s\in \{1,w\}$. In particular, there are 2p such conjugacy classes.

Proof

Clearly the $f_{k,s}$ are not conjugate by Lemma 7.10 (ii). Let $h=h(t,u,\gamma ,\delta ,y,z)\in {\text {Aut}}\left( {G}\right)$. Then $h\in C_{k,s}=C_{{\text {Aut}}\left( {G}\right) }(f_{k,s})$ if and only if $t=1, y=0, u=3\alpha$ and $z=\frac{3k\alpha -s}{s}$ or $t= -1,y=0 ,u= -3\alpha +s$ and $z=\frac{ku}{s}$. Therefore $|C_{k,s}|=2p^2$.

According to Lemma 7.10(i), the automorphisms conjugate to $f_{k,s}$ are the automorphisms $f=f(p-1,b,\alpha ,\beta ,p-1,c,d)$ for which:

$$\begin{aligned} b \frac{c(c+1)+2d}{2}=ks. \end{aligned}$$

(21)

So for every pair k, s there are $p^3(p-1)$ many choices of $b,\alpha ,\beta ,c,d$ for which (21) holds. Therefore,

$$\begin{aligned} \sum _{k \in \{0,\ldots p-1 \}} \sum _{s\in \{1,w\}} [{\text {Aut}}\left( {G}\right) :C_{k,s}]=p\left( \frac{p^3(p-1)}{2}+\frac{p^3(p-1)}{2}\right) =p^4(p-1), \end{aligned}$$

and so $\{f_{k,s}\,\mid \,k\in \{0,\ldots ,p-1\}\, , s\in \{1,w\}\}$ is a set of representatives. $\square$

1.2.3 Case $G=G_9, G_{10}$

In this section we will use the power-commutator presentation of the group $G_i$, for $i=9,10$ as

$$\begin{aligned} G_i=\langle g_1, g_2, g_3, g_4, \, | \, [g_2,g_1]=g_3, \, [g_3,g_1]=g_2^p=g_4^w\rangle , \end{aligned}$$

where w is either 1 or a primitive root of unity. According to (Girnat 2018, Section 5.7, 5.8) the automorphisms of $G_i$ have the same syntactic description and since $g_3,g_4$ generate an elementary abelian subgroup of rank 2 they can be represented as

$$\begin{aligned} h_{j}= \left\{ \begin{array}{l} g_1\mapsto g_1^{(-1)^ j} g_3^{\alpha } g_4^{\beta }, \qquad H=\begin{bmatrix} (-1)^ j t &{} 0\\ (-1)^ j u &{} t \end{bmatrix}\\ g_2\mapsto g_2^{t} g_3^{u} g_4^{\gamma } , \end{array}.\right. \end{aligned}$$

for $j=0,1$ and denoted by $h_{j}(\alpha ,\beta ,t,u,\gamma )$ where $u,\alpha ,\beta ,\gamma \in \{0,\ldots , p-1\}$ and $t\in \{1,\ldots ,p-1\}$. Then $|{\text {Aut}}\left( {G}\right) |=2(p-1)p^4$. It is straightforward to verify that $h_{j}$ satisfies (A) if and only $j=1$ and

$$\begin{aligned} h_{1}= \left\{ \begin{array}{l} g_1\mapsto g_1^{- 1} g_3^{\alpha } g_4^{\beta },\qquad H=\begin{bmatrix} 1 &{} 0\\ - u &{} -1 \end{bmatrix} ,\\ g_2\mapsto g_2^{-1} g_3^{u} g_4^{\gamma } \end{array}.\right. \end{aligned}$$

and so there are $p^4$ many of them.

Lemma 7.14

If $g(\alpha ,\beta ,-1,0,\gamma )$ and $h(\epsilon ,\rho ,-1,0,\delta )$ are conjugate, then $\gamma =\delta$.

Proof

Let $g=g(\alpha ,\beta ,-1,0,\gamma )$ and $h=h(\epsilon ,\rho ,-1,0,\delta )$ be conjugate. By looking at the image of $g_2$ of g and h, we have $\delta =\gamma$. $\square$

Proposition 7.15

A set of representatives of conjugacy classes of automorphisms of $G_i$, $i=9,10$, satisfying (A) is:

$$\begin{aligned} f_{k, \gamma }= \left\{ \begin{array}{l} g_1\mapsto g_1^{- 1} g_3^{k}, \qquad F=\begin{bmatrix} 1 &{} 0\\ 0 &{} -1 \end{bmatrix} , \\ g_2\mapsto g_2^{-1} g_4^{\gamma } \end{array}.\right. \end{aligned}$$

for $\gamma \in \{0,\ldots p-1\}$ and $k=0,1$. In particular there are 2p such conjugacy classes.

Proof

Let $h_{j}=h_{j}(\alpha ,\beta ,x,y,z)$ and $C_{k,\gamma }= C_{{\text {Aut}}\left( {G_i}\right) }(f_{k, \gamma })$. Then

$$\begin{aligned} h_{0}\in C_{k,\gamma }\Leftrightarrow & {} y=0 ,\quad \alpha =0,\quad k(1-x)=0\\ h_{1}\in C_{k,\gamma }\Leftrightarrow & {} y=0 ,\quad \alpha =k, \quad k(x+1)=0. \end{aligned}$$

So we have that

$$\begin{aligned} |C_{k, \gamma }|= \left\{ \begin{array}{l} 2p^2(p-1), \quad \text { if } k =0\\ 2p^2,\qquad \qquad \text { otherwise.} \end{array}.\right. \end{aligned}$$

In particular $f_{k,\gamma }$ and $f_{0,\gamma }$ are not conjugate since their centralizers have different sizes. There are $p^4$ automorphisms satisfying (A). By virtue of Lemma 7.14(ii), $f_{\gamma ,k}$ and $f_{\gamma ,l}$ are not conjugate if $k\ne l$. Therefore,

$$\begin{aligned} \sum _{\gamma \in \{0,\ldots p-1 \}} \sum _{k=0}^1 [{\text {Aut}}\left( {G_i}\right) :C_{k,\gamma }]=p (p^2+p^2(p-1))=p^4, \end{aligned}$$

and the statement follows. $\square$