On the Ranks of the Alternating Group $A_{n}$

Abstract

Let G be a finite group and X be a conjugacy class of G. The rank of X in G,  denoted by rank(G : X),  is defined to be the minimal number of elements of X generating G. In this paper we establish some general results on the ranks of certain conjugacy classes of elements for simple alternating group $A_{n}$. We apply these general results together with the structure constants method to determine the ranks of all the non-trivial classes of $A_{8}$ and $A_{9}$.

1 Introduction

Generation of finite groups by suitable subsets is of great interest and has many applications to groups and their representations. For example, the computations of the genus of simple groups can be reduced to the generations of the relevant simple groups (see Woldar [1] for details). Also Di Martino et al. [2] established a useful connection between generation of groups by conjugates and the existence of elements representable by almost cyclic matrices. Their motivation was to study irreducible projective representations of sporadic simple groups. Recently, more attention was paid to the generation of finite groups by conjugate elements. In his Ph.D. thesis [3], Ward considered the generation of a simple group by conjugate involutions satisfying certain conditions.

We are interested in the generation of finite simple groups by the minimal number of elements from a given conjugacy class of the group. This motivates the following definition.

Definition 1

Let G be a finite simple group and X be a conjugacy class of G. The rank of X in G,  denoted by rank(G : X), is defined to be the minimal number of elements of X generating G.

One of the applications of ranks of conjugacy classes of a finite group is that they are involved in the computations of the covering number of the finite simple group (see Zisser [4]).

In [5,6,7], the second author computed the ranks of involuntary classes of the Fischer sporadic simple group $Fi_{22}.$ He found that $rank(Fi_{22}{:}2B) = rank(Fi_{22}{:}2C)$$= 3,$ while $rank(Fi_{22}{:}2A) \in \{5,6\}.$ The work of Hall and Soicher [8] implies that $rank(Fi_{22}{:}2A) = 6.$ Then in a considerable number of publications (for example, but not limited to, see [9,10,11,12,13,14] or [7]) Moori, Ali and Ibrahim explored the ranks for various sporadic simple groups. In this article we prove some general results on the ranks of certain conjugacy classes of elements for the simple alternating group $A_{n}.$ Then we apply these general results together with the structure constants method to determine the ranks for all non-trivial conjugacy classes of $A_{8}$ and $A_{9}$. In this paper we follow the notation of [15].

2 Preliminaries

Let G be a finite group and $C_{1}, C_{2}, \ldots , C_{k}$ be $k\ge 3$ (not necessarily distinct) conjugacy classes of G with $g_{1}, g_{2}, \ldots , g_{k}$ being representatives for these classes, respectively.

For a fixed representative $g_{k} \in C_{k}$ and for $g_{i} \in C_{i},\ 1\le i \le k-1,$ denote by $\Delta _{G} = \Delta _{G}(C_{1}, C_{2}, \ldots , C_{k})$ the number of distinct $(k-1)$-tuples $(g_{1}, g_{2}, \ldots , g_{k-1})$ such that $g_{1}g_{2}\ldots g_{k-1} = g_{k}.$ This number is known as class algebra constant or structure constant. With $\mathrm {Irr}(G) = \{\chi _{1}, \chi _{2}, \ldots , \chi _{r}\}$ being the set of complex irreducible characters of G,  the number $\Delta _{G}$ is easily calculated from the character table of G through the formula

$$\begin{aligned} \Delta _{G}(C_{1}, C_{2}, \ldots , C_{k}) = \frac{\prod \nolimits _{i=1}^{k-1}|C_{i}| }{|G|} \sum _{i=1}^{r} \frac{\chi _{i}(g_{1}) \chi _{i}(g_{2}) \ldots \chi _{i}(g_{k-1}) \overline{\chi _{i}(g_{k})} }{(\chi _{i}(1_{G}))^{k-2}}. \end{aligned}$$

(1)

Also for a fixed $g_{k} \in C_{k}$ we denote by $\Delta ^{*}_{G}(C_{1}, C_{2}, \ldots , C_{k})$ the number of distinct $(k-1)$-tuples $(g_{1}, g_{2}, \ldots , g_{k-1})\in C_{1} \times C_{2} \times \cdots \times C_{k-1}$ satisfying

$$\begin{aligned} g_{1}g_{2}\ldots g_{k-1} = g_{k}\quad \text {and}\quad \langle g_{1}, g_{2}, \ldots , g_{k-1} \rangle = G. \end{aligned}$$

(2)

Definition 2

If $\Delta ^{*}_{G}(C_{1}, C_{2}, \ldots , C_{k}) > 0,$ the group G is said to be $(C_{1}, C_{2}, \ldots , C_{k})$-generated.

Furthermore, if $H \le G$ is any subgroup containing a fixed element $g_{k}\in C_{k},$ we let $\Sigma _{H}(C_{1}, C_{2}, \ldots , C_{k})$ be the total number of distinct $(k-1)$-tuples $(g_{1}, g_{2},\ldots , g_{k-1})$ such that $g_{1}g_{2}\ldots g_{k-1} = g_{k}$ and $\langle g_{1}, g_{2},\ldots , g_{k-1} \rangle \le H.$ The value of $\Sigma _{H}(C_{1}, C_{2}, \ldots , C_{k})$ can be obtained as a sum of the structure constants $\Delta _{H}(c_{1}, c_{2}, \ldots , c_{k})$ of H-conjugacy classes $c_{1}, c_{2}, \ldots , c_{k}$ such that $c_{i} \subseteq H\bigcap C_{i}.$

Theorem 2.1

Let G be a finite group and H be a subgroup of G containing a fixed element g such that $\gcd (o(g), [N_{G}(H){:}H] ) = 1.$ Then the number h(g, H) of conjugates of H containing g is $\chi _{H}(g),$ where $\chi _{H}(g)$ is the permutation character of G with action on the conjugates of H. In particular,

$$\begin{aligned} h(g,H) = \displaystyle \sum _{i=1}^{m} \frac{|C_{G}(g)|}{|C_{N_{G}(H)}(x_{i})|}, \end{aligned}$$

where $x_{1}, x_{2}, \ldots , x_{m}$ are representatives of the $N_{G}(H)$-conjugacy classes fused to the G-class of g.

Proof

See, for example, Ganief and Moori [16,17,18]. $\square$

The above number h(g, H) is useful in giving a lower bound for $\Delta ^{*}_{G}(C_{1}, C_{2}, \ldots , C_{k}),$ namely $\Delta ^{*}_{G}(C_{1}, C_{2}, \ldots , C_{k}) \ge \Theta _{G}(C_{1}, C_{2}, \ldots , C_{k}),$ where

$$\begin{aligned} \Theta _{G}(C_{1}, C_{2}, \ldots , C_{k}) = \Delta _{G}(C_{1}, C_{2}, \ldots , C_{k}) - \displaystyle \sum h(g_{k}, H)\Sigma _{H}(C_{1}, C_{2}, \ldots , C_{k}), \end{aligned}$$

(3)

$g_{k}$ is a representative of the class $C_{k}$, and the sum is taken over all the representatives H of G-conjugacy classes of maximal subgroups containing elements of all the classes $C_{1}, C_{2}, \ldots , C_{k}.$

Since we have all the maximal subgroups of the sporadic simple groups (except for $G = \mathbb {M}$ the Monster group), it is possible to build a small subroutine in GAP [19] or Magma [20] to compute the values of $\Theta _{G} = \Theta _{G}(C_{1}, C_{2}, \ldots , C_{k})$ for any collection of conjugacy classes of a sporadic simple group.

If $\Theta _{G} > 0$ then certainly G is $(C_{1}, C_{2}, \ldots , C_{k})$-generated. In the case $C_{1} = C_{2} = \cdots = C_{k-1} = C$ then G can be generated by $k-1$ elements suitably chosen from C and hence $rank(G{:}C) \le k-1$.

We now quote some results for establishing generation and non-generation of finite simple groups. These results are also important in determining the ranks of the finite simple groups.

Lemma 2.2

(e.g., see Ali and Moori [14] or Conder et al. [21]) Let G be a finite simple group such that G is (lX, mY, nZ)-generated. Then G is $(\underbrace{lX, lX, \ldots , lX}_{m-times}, (nZ)^{m})$-generated.

Proof

Since G is (lX, mY, nZ)-generated group, it follows that there exist $x\in lX$ and $y \in mY$ such that $xy\in nZ$ and $\langle x,y \rangle = G.$ Let $N:= \langle x,x^{y}, x^{y^{2}}, \ldots , x^{y^{m-1}} \rangle .$ Then $N\unlhd G.$ Since G is simple group and N is non-trivial subgroup we obtain that $N=G.$ Furthermore, we have

$$\begin{aligned} xx^{y}x^{y^{2}}x^{y^{m-1}}= & {} x(yxy^{-1})(y^{2}xy^{-2}) \cdots (y^{m-1}xy^{1-m}) \\= & {} (xy)^{m}\in (nZ)^{m}. \end{aligned}$$

Since $x^{y^{i}}\in lX$ for all i,  the result follows. $\square$

Corollary 2.3

(e.g., see Ali and Moori [14]) Let G be a finite simple group such that G is (lX, mY, nZ)-generated. Then $rank(G{:}lX) \le m.$

Proof

Follows immediately by Lemma 2.2. $\square$

Lemma 2.4

(e.g., see Ali and Moori [14]) Let G be a finite simple (2X, mY, nZ)-generated group. Then G is $(mY, mY, (nZ)^{2})$-generated.

Proof

Since G is (2X, mY, nZ)-generated group, it is also (mY, 2X, tK)-generated group. The result follows immediately by Lemma 2.2. $\square$

Corollary 2.5

If G is a finite simple (2X, mY, nZ)-generated group, then $rank(G{:}mY) = 2.$

Proof

By Lemma 2.4 and Corollary 2.3 we have $rank(G{:}mY) \le 2.$ But a non-abelian simple group cannot be generated by one element. Thus, $rank(G{:}mY) = 2.$$\square$

The following two results are in some cases useful in establishing non-generation for finite groups.

Lemma 2.6

(e.g., see Ali and Moori [14] or Conder et al. [21]) Let G be a finite centerless group. If $\Delta ^{*}_{G}(C_{1}, C_{2}, \ldots , C_{k})< |C_{G}(g_{k})|,\ g_{k} \in C_{k},$ then $\Delta ^{*}_{G}(C_{1}, C_{2}, \ldots , C_{k}) = 0$ and therefore G is not $(C_{1}, C_{2}, \ldots , C_{k})$-generated.

Proof

We prove the contrapositive of the statement, that is if $\Delta ^{*}_{G}(C_{1}, C_{2}, \ldots , C_{k}) > 0$ then $\Delta ^{*}_{G}(C_{1}, C_{2}, \ldots , C_{k})\ge |C_{G}(g_{k})|,$ for a fixed $g_{k} \in C_{k}.$ So let us assume that $\Delta ^{*}_{G}(C_{1}, C_{2}, \ldots , C_{k})> 0.$ Thus, there exists at least one $(k-1)$-tuple $(g_{1}, g_{2},\ldots , g_{k-1}) \in C_{1} \times C_{2} \times \cdots \times C_{k-1}$ satisfying Eq. (2). Let $x\in C_{G}(g_{k}).$ Then we obtain

$$\begin{aligned} x(g_{1}g_{2}\ldots g_{k-1})x^{-1} = (xg_{1}x^{-1})(xg_{2}x^{-1})\cdots (xg_{k-1}x^{-1}) = (xg_{k}x^{-1}) = g_{k}. \end{aligned}$$

Thus, the $(k-1)$-tuple $(xg_{1}x^{-1},xg_{2}x^{-1},\ldots , xg_{k-1}x^{-1})$ will generate G. Moreover, if $x_{1}$ and $x_{2}$ are distinct elements of $C_{G}(g_{k}),$ then the $(k-1)$-tuples $(x_{1}g_{1}x_{1}^{-1},x_{1}g_{2}x_{1}^{-1},\ldots , x_{1}g_{k-1}x_{1}^{-1})$ and $(x_{2}g_{1}x_{2}^{-1},x_{2}g_{2}x_{2}^{-1},\ldots , x_{2}g_{k-1}x_{2}^{-1})$ are also distinct since G is centerless. Thus, we have at least $|C_{G}(g_{k})|\; (k-1)$-tuples $(g_{1}, g_{2},\ldots , g_{k-1})$ generating G. Hence, $\Delta ^{*}_{G}(C_{1}, C_{2}, \ldots , C_{k}) \ge |C_{G}(g_{k})|.$$\square$

The following result is due to Ree [22].

Theorem 2.7

Let G be a transitive permutation group generated by permutations $g_{1},g_{2}, \ldots , g_{s}$ acting on a set of n elements such that $g_{1}g_{2} \ldots g_{s} = 1_{G}.$ If the generator $g_{i}$ has exactly $c_{i}$ cycles for $1\le i \le s,$ then $\sum \nolimits _{i=1}^{s}c_{i} \le (s-2)n + 2.$

Proof

See, for example, Ali and Moori [14].

The following result is due to Conder et al. [21] and Scott [23].

Theorem 2.8

(Scott’s theorem) Let $g_{1},g_{2}, \ldots , g_{s}$ be elements generating a group G with $g_{1}g_{2} \ldots g_{s} = 1_{G}$ and ${\mathbb {V}}$ be an irreducible module for G with $\dim {\mathbb {V}} = n \ge 2.$ Let $C_{{\mathbb {V}}}(g_{i})$ denote the fixed point space of $\langle g_{i} \rangle$ on ${\mathbb {V}}$, and let $d_{i}$ be the codimension of $C_{{\mathbb {V}}}(g_{i})$ in ${\mathbb {V}}.$ Then $\sum \nolimits _{i=1}^{s}d_{i} \ge 2n.$

With $\chi$ being the ordinary irreducible character afforded by the irreducible module ${\mathbb {V}}$ and $\mathbf {1}_{\langle g_{i}\rangle }$ being the trivial character of the cyclic group $\langle g_{i}\rangle ,$ the codimension $d_{i}$ of $C_{{\mathbb {V}}}(g_{i})$ in ${\mathbb {V}}$ can be computed using the following formula [16]:

$$\begin{aligned} d_{i}= & {} \dim ({\mathbb {V}}) - \dim (C_{{\mathbb {V}}}(g_{i})) = \dim ({\mathbb {V}}) - \langle \chi {\downarrow }^{G}_{\langle g_{i}\rangle }, {\mathbf {1}}_{\langle g_{i}\rangle } \rangle \nonumber \\= & {} \chi (1_{G}) - \frac{1}{|\langle g_{i}\rangle |}\displaystyle \sum _{j=0}^{o(g_{i})-1} \chi (g_{i}^{j}). \end{aligned}$$

(4)

3 Some General Results on Ranks of $A_{n}$

In this section we give some general results on the ranks for certain conjugacy classes of elements of the simple group $A_{n}.$

The alternating group $A_{n},\ n\ge 5$ has $\lfloor \frac{n}{3} \rfloor$ conjugacy classes of elements of order 3. The cycle structures of these classes are $3^{m}1^{n-3m},\ 1 \le m \le \lfloor \frac{n}{3} \rfloor .$ For $m= 1,$ let 3A denote the class of elements of $A_{n}$ of cycle structure (a, b, c). In this section we determine the rank of this class in $A_{n}.$ We will use $(A_{n})_{[k_{1},k_{2},\ldots , k_{r}]}$ to denote the subgroup of $A_{n}$ fixing the points $k_{1},k_{2},\ldots , k_{r}$, and if it fixes a single point $k_{i},$ we use $(A_{n})_{k_{i}}.$

Lemma 3.1

$rank(A_{5}{:}3A) = 2.$

Proof

We claim that $A_{5} = \langle (1,2,3),(1,4,5)\rangle .$ We have $(1,2,3)(1,4,5) = (1,4,5,2,3),$ which has order 5. This implies that $15{\displaystyle \mid }|\langle (1,2,3),(1,4,5)\rangle |.$ By looking at the maximal subgroups of $A_{5}$ (see the ${\mathbb {A}}{\mathbb {T}}{\mathbb {L}}{\mathbb {A}}{\mathbb {S}}$ [24] for example) we can see that there is no maximal subgroup of $A_{5}$ with order divisible by 15. It follows that $\langle (1,2,3),(1,4,5)\rangle = A_{5}$ and hence $rank(A_{5}{:}3A) = 2.$$\square$

Lemma 3.2

$rank(A_{n}{:}3A) \ne 2,\ \forall \ n \ge 6.$

Proof

Suppose that $x, y \in 3A$ of $A_{n},\ n \ge 6$, and let $x = (a,b,c)$ and $y = (d,e,f).$ If $xy = yx$ then $\langle x,y\rangle \cong {\mathbb {Z}}_{3}\times {\mathbb {Z}}_{3}.$ If $xy \ne yx,$ then x and y are not disjoint cycles and have some common points, i.e., $\{a,b,c\} \bigcap \{d,e,f\} \ne \phi .$ Thus, the number of moved points by $\langle x,y\rangle$ is at most 5 and it follows that $\langle x,y\rangle \le A_{5}.$ Hence, $rank(A_{n}{:}3A) \ne 2$ for $n \ge 6.$$\square$

Lemma 3.3

$rank(A_{6}{:}3A) = 3.$

Proof

We show that $A_{6} = \langle (1,2,3),(1,4,5), (1,5,6)\rangle .$ Let $H = \langle (1,2,3),(1,4,5)\rangle .$ Then $H \cong A_{5}$ and $H = (A_{6})_{6},$ which is a maximal subgroup of $A_{6}.$ Since $(1,5,6) \not \in H,$ we have $\langle H, (1,5,6)\rangle = A_{6}.$ Since $rank(A_{6}{:}3A) \ne 2$ by Lemma 3.2, it follows that $rank(A_{6}{:}3A) = 3.$$\square$

Now we state and prove an important theorem on the rank of the class 3A of $A_{n},\ n\ge 5.$

Theorem 3.4

For the alternating group $A_{n},\ n\ge 5$, we have

$$\begin{aligned} rank(A_{n}{:}3A) = \left\{ \begin{array}{ll} \frac{n-1}{2} &{} \text {if}\ n\ \text {is odd}, \\ \frac{n}{2} &{} \text {if}\ n\ \text {is even}. \end{array} \right. \end{aligned}$$

Proof

We use the mathematical induction on n. The result is true for $n=5$ and $n = 6$ by Lemmas 3.1 and 3.3, respectively. We will show that

$$\begin{aligned} A_{n} = \left\{ \begin{array}{ll} \langle (1,2,3),(1,4,5), (1,6,7), \ldots , (1,n-3,n-2),(1,n-1,n) \rangle &{} \text {if}\ n\ \text {is odd}, \\ \langle (1,2,3),(1,4,5), (1,6,7), \ldots , (1,n-2,n-1),(1,n-1,n) \rangle &{} \text {if}\ n\ \text {is even}. \end{array} \right. \end{aligned}$$

(5)

Suppose that the result is true for n odd; then, we will show that the result will be true for $n+1$ and $n+2.$ So assume that Eq. (5) is true for n odd. Note that if $H = \langle (1,2,3),(1,4,5), (1,6,7), \ldots , (1,n-1,n)\rangle ,$ then Eq. (5) implies that $A_{n} \cong (A_{n+1})_{n+1} = H.$ Since $(1,n,n+1) \in A_{n+1}{\setminus }A_{n}$ and H is a maximal subgroup of $A_{n+1},$ we have

$$\begin{aligned} K = \langle H, (1,n,n+1)\rangle = A_{n+1}. \end{aligned}$$

(6)

Since $n+1$ is even, we have proven the result for the even case. Now since K is a maximal subgroup of $A_{n+2}$ ($K = (A_{n+2})_{n+2}$), then $\langle K, (1,n+1,n+2)\rangle = A_{n+2}$ as $(1,n+1,n+2) \in A_{n+2}{\setminus }A_{n+1}.$ Thus,

$$\begin{aligned} \langle H, (1,n,n+1), (1,n+1,n+2)\rangle = A_{n+2}. \end{aligned}$$

(7)

We now show that we do not need the element $(1,n,n+1)$ in Eq. (7) to generate $A_{n+2};$ that is, $(1,n,n+1)$ is redundant. Let $\alpha = (1,n-1,n) \in H$ and $\beta = (1,n+1,n+2).$ Then $\alpha ^{\beta } = (n-1, n,n+1) := \gamma$ and $\gamma ^{\alpha } = (1,n+1,n) = (1,n,n+1)^{-1}.$ This shows that $\langle H, (1,n,n+1), (1,n+1,n+2)\rangle$ can actually be reduced to $\langle H, (1,n+1,n+2)\rangle .$ That is $\langle H, (1,n+1,n+2)\rangle = A_{n+2},$ i.e.,

$$\begin{aligned} \langle (1,2,3),(1,4,5), (1,6,7), \ldots , (1,n-1,n), (1,n+1,n+2)\rangle = A_{n+2}, \end{aligned}$$

completing the proof. $\square$

Next we turn to the generation of $A_{n}$ by the classes of n-cycles for n odd and $(n-1)$-cycles for n even.

Note 1

• (i). It is well known that for n odd, the class of the n-cycles of the symmetric group $S_{n}$ splits into two classes (same size) in $A_{n}.$ For n even, we have no n-cycles in $A_{n}$, but the class of the $(n-1)$-cycles splits into two classes (same size) in $A_{n}.$ The reason for both cases is obvious since if $\alpha$ is an n-cycle (for n odd) or an $(n-1)$-cycle (for n even), then $C_{S_n}(\alpha ) = C_{A_n}(\alpha )=\langle \alpha \rangle .$

• (ii). Assume that $\alpha$ is an n-cycle (for n odd) or an $(n-1)$-cycle (for n even) in $A_n$. If $h\in S_n{\setminus } A_n$, then $\alpha$ and $\alpha ^h$ are not conjugate in $A_n$. If $\alpha \sim \alpha ^h$ in $A_n$, then we have $g\in A_n$ such that $\alpha = (\alpha ^h)^g$ which implies that $hg\in C_{S_n}(\alpha ) = C_{A_n}(\alpha ),$ a contradiction since hg is an odd permutation and cannot be in $A_n$.

The following lemma is useful in the determination of the ranks of the classes of n- or $(n-1)$-cycles of $A_{n}.$

Lemma 3.5

Let G be a primitive subgroup of $S_{n}.$ If G contains a 3-cycle, then $G \ge A_{n}.$

Proof

See Cameron [25]. $\square$

Now we give an important result showing the ranks for the classes of n-cycles and $(n-1)$-cycles of $A_{n}.$

Theorem 3.6

For $n\ge 5,$ we have $\underbrace{rank(A_{n}{:}nX)}_{n\ \text {is odd}} = 2 = \underbrace{rank(A_{n}{:}(n-1)X)}_{n\ \text {is even}},\; X\in \{A,B\}.$

Proof

We consider the case when n is odd and the even case follows easily. For $n \ge 5$ odd, let $\alpha _{1} = (1,2,3,\ldots ,n)$ and $\alpha _{2} = (1,2,3,\ldots ,n, n-1)$ be representatives of the two classes nA and nB of the n-cycle classes of $A_{n}$, respectively. (Note that $\alpha _{1}$ and $\alpha _{2}$ are conjugate in $S_{n},$ but not in $A_{n}$ as $\alpha _{2} = \alpha _{1}^{(n-1,n)}$, see Note 1 (ii).) Also let $\beta _{1} = (1,4,5,6,7,\ldots , n,2,3)$ and $\beta _{2} = (1,4,5,6,7,\ldots , n-2, n, n-1, 2,3).$ We handle the class nA, and the result for the other class nB follows similarly. Let $H:= \langle \alpha _{1}, \beta _{1}\rangle .$ Clearly, $H \le S_{n}$ and in fact, since H contains even permutations only, it follows that $H \le A_{n}.$ We are aiming to show that the equality holds; that is, $H = A_{n}.$ To establish the converse inequality ($H \ge A_{n}$) we need to show that H is primitive in $S_{n}$ and contains a 3-cycle element. Since $\alpha _{1},\ \beta _{1} \in H,$ we have

$$\begin{aligned} H \ni \alpha _{1}^{-1}\beta _{1}= & {} (1,n,n-1,n-2,\ldots ,4,3,2)(1,4,5,6,7,\ldots , n,2,3) \\= & {} (1,3,n),\ \text {a}\ 3\text {-cycle element.} \end{aligned}$$

Now since H contains n-cycle elements (at least $\alpha _{1}$ and $\beta _{1}$), it follows by O’Nan–Scott theorem (see, for example, Theorem 2.4 of Wilson [26]) that H cannot be of type (i) or (ii) of maximal subgroups of $S_{n}$ (as subgroups of these two types cannot have n-cycle elements). Hence, H is a primitive subgroup of $S_{n}$, and since H contains a 3-cycle element ($\alpha _{1}^{-1}\beta _{1} = (1,3,n)$), it follows by Lemma 3.5 that $H \ge A_{n}.$ This together with the information $H \le A_{n}$ implies that $H = A_{n}.$

So far, we proved that the group H,  generated by the two n-cycles $\alpha _{1}$ and $\beta _{1},$ is just $A_{n}.$ Next we prove that $\alpha _{1}$ and $\beta _{1}$ are conjugate in H. But this follows as $(1,2,3) \in 3A \subset A_{n} = H$, and it is easy to see that $\beta _{1} = \alpha _{1}^{(1,2,3)},$ i.e., $\beta _{1} \in nA.$ Thus, $rank(A_{n}{:}nA) = 2.$ The result holds similarly for the class nB of $A_{n},\ n$ odd, by letting $H:= \langle \alpha _{2}, \beta _{2}\rangle ,$ where in a similar manner it will be shown that $H = A_{n}$ and also one can see that $\beta _{2} = \alpha _{2}^{(1,2,3)},$ i.e., $\beta _{2} \in nB.$ Thus, $rank(A_{n}{:}nB) = 2.$

The even case follows consequently from the odd case, because if n is even then the treatment of the two classes $(n-1)A$ and $(n-1)B$ of the $(n-1)$-cycles of $A_{n}$ reduce to the odd case. Hence, the result. $\square$

4 Ranks of the Classes of $A_{8}$ and $A_{9}$

In this section we apply the general results discussed in previous sections, namely Sects. 2 and 3, to the groups $A_{8}$ and $A_{9}.$ We determine the ranks for all conjugacy classes.

4.1 Ranks of $A_{8}$

The group $A_{8}$ is a simple group of order $20160 = 2^{6}\times 3^{2} \times 5 \times 7.$ By the ${\mathbb {A}}{\mathbb {T}}{\mathbb {L}}{\mathbb {A}}{\mathbb {S}}$ the group $A_{8}$ has exactly 14 conjugacy classes of its elements and 6 conjugacy classes of its maximal subgroups. Representatives of these classes of maximal subgroups can be taken as follows:

$$\begin{aligned} \begin{array}{llll} H_{1} = A_{7} &{} &{} &{} H_{2} = 2^{3}{:}L_{3}(2) \\ H_{3} = 2^{3}{:}L_{3}(2) &{} &{} &{} H_{4} = S_{6} \\ H_{5} = 2^{4}{:}(S_{3} \times S_{3}) &{} &{} &{} H_{6} = (A_{5}\times 3){:}2. \\ \end{array} \end{aligned}$$

We firstly list in Table 1 the values of $h(g,H_{i})$ for all the non-identity classes and maximal subgroups $H_{i},\ 1\le i \le 6,$ of $A_{8}.$

Table 1 Values $h(g,H_{i}),\ 1\le i \le 6$ for non-identity classes and maximal subgroups of $A_{8}$

We start our investigation on the ranks of the non-trivial classes of $A_{8}$ by looking at the two classes of involutions 2A and 2B. It is well known that two involutions generate a dihedral group. Thus, the lower bound of the rank of an involuntary class in a finite group $G \ne D_{2n}$ (the dihedral group of order 2n) is 3.

In this subsection we let $G = A_{8}.$

Lemma 4.1

$rank(G{:}2Z) \ne 3,$ for $Z \in \{A,B\}.$

Proof

We show that the group G is not (2Z, 2Z, 2Z, nX)-generated group for $Z\in \{A,B\}$ and for any non-trivial conjugacy class nX of G. We start with the case $Z = A.$ The direct computations yield $\Delta _{G}(2A,2A,2A, nX) = 0$ for $nX \in T_{1}:= \{3A,5A,6A,15A,15B\}.$ Thus, G is not (2A, 2A, 2A, nX)-generated group for any class nX in $T_{1}.$ The group $A_{8}$ has a 14-dimensional complex irreducible module ${\mathbb {V}}.$ For any conjugacy class nX,  let $d_{nX} = \dim ({\mathbb {V}}/C_{{\mathbb {V}}}(nX))$ denote the codimension of the fixed space (in ${\mathbb {V}}$) of a representative of nX. Using Eq. (4) together with the power maps associated with the character table of $A_{8}$ given in the ${\mathbb {A}}{\mathbb {T}}{\mathbb {L}}{\mathbb {A}}{\mathbb {S}},$ we were able to compute all the values of $d_{nX}$ for all non-trivial classes nX of G,  and we list these values in Table 2.

Table 2 $d_{nX} = \dim ({\mathbb {V}}/C_{{\mathbb {V}}}(nX)),\ nX$ is a non-trivial class of G and $\dim ({\mathbb {V}}) = 14$

Now if $A_{8}$ is (2A, 2A, 2A, nX)-generated group, then by Scott’s theorem (see Theorem 2.8) we must have $d_{2A} + d_{2A} + d_{2A} + d_{nX} \ge 2 \times 14.$ However, it is clear from Table 2 that $3 \times d_{2A} + d_{nX} < 28,$ for each $nX \in T_{2}:= \{2A, 2B, 3B, 4A, 4B, 6B, 7A, 7B\}$ and therefore $A_{8}$ is not (2A, 2A, 2A, nX)-generated group, for any $nX \in T_{2}.$ Since G is not (2A, 2A, 2A, nX)-generated group, for any $nX \in T_{1}\bigcup T_{2},$ it follows that $rank(G{:}2A) \ne 3.$

We now turn to the other case $Z = B$ to show that G is not (2B, 2B, 2B, nX)-generated group, for all non-trivial conjugacy classes nX of G. We use similar arguments to the above case. We know that the group $A_{8}$ has a 7-dimensional complex irreducible module ${\mathbb {V}}.$ Let $d_{nX}$ be the codimension of the fixed space in ${\mathbb {V}},\ \dim ({\mathbb {V}}) = 7$ of a representative of nX. Similarly, we list in Table 3 the values of $d_{nX}$ for all non-trivial classes nX of G.

Table 3 $d_{nX} = \dim ({\mathbb {V}}/C_{{\mathbb {V}}}(nX)),\ nX$ is a non-trivial class of G and $\dim ({\mathbb {V}}) = 7$

Now if $A_{8}$ is (2B, 2B, 2B, nX)-generated group, then we must have $d_{2B} + d_{2B} + d_{2B} + d_{nX} \ge 2 \times 7.$ However, it is clear from Table 3 that $3 \times d_{2B} + d_{nX} < 14,$ for all non-trivial classes nX of $A_{8}$ and therefore $A_{8}$ is not (2B, 2B, 2B, nX)-generated group, for any class nX of $A_{8}.$ This establishes the non-generation of $A_{8}$ by three conjugate involutions from class 2B. Thus, $rank(G{:}2B) \ne 3,$ completing the proof. $\square$

Note 2

Observe that the non-generation of $A_{8}$ by three conjugate involutions from class 2A can be established without the need of computing the structure constants $\Delta _{G}(2A,2A,2A, nX),\ nX \in T_{1},$ as it is clear from Table 2 that $3 \times d_{2A} + d_{nX} < 28$ for all the non-trivial classes nX of $A_{8}.$

Lemma 4.2

The group $A_{8}$ is (2A, 4B, 15A)- and (2B, 4A, 15B)-generated group.

Proof

Let $a_{1} := (1,2)(3,4)(5,6)(7,8) \in 2A,\$$a_{2} := (5,6)(7,8) \in 2B,\$$b_{1} := (2,3)(4,6,7,5) \in 4B$ and $b_{2} := (1,2,6,8)(3,5,4,7) \in 4A.$ Then $\langle a_{1},b_{1}\rangle = \langle a_{2},b_{2}\rangle = A_{8}$ with $a_{1}b_{1} = (1,3,6,4,2)(5,7,8) \in 15A$ and $a_{2}b_{2} =(1,2,6,4,7)(3,5,8) \in 15B.$ Thus, $A_{8}$ is (2A, 4B, 15A)- and (2B, 4A, 15B)-generated group. $\square$

Proposition 4.3

$rank(G{:}2Z) = 4,$ for $Z\in \{A,B\}.$

Proof

Since by Lemma 4.2, $A_{8}$ is (2A, 4B, 15A)- and (2B, 4A, 15B)-generated group, it follows by applications of Lemma 2.2 that $A_{8}$ is $(2A,2A,2A,2A,(15A)^{4})$- and $(2B,2B,2B,2B,(15B)^{4})$-generated group. Thus, $rank(A_{8}{:}2Z) \le 4,$ for $Z\in \{A,B\}.$ Since $rank(A_{8}{:}2Z) \ne 3$ by Lemma 4.1, it follows that $rank(A_{8}{:}2Z) = 4,$ for $Z\in \{A,B\}.$$\square$

Remark 1

The generation of $A_{8}$ by four suitable involutions from class 2A or 2B can be established using the structure constant method. For example, the direct computations show that $\Delta _{G}(2Z,2Z,2Z,2Z,15A) = 3375\ (74250)$ for $Z = A\ \text {and}\ B$, respectively. From Table 1 we see that $h(g,H_{i}) = 0$ for $g \in 15A$ and $i \in \{1,2,3,4,5\},$ while $h(g,H_{6}) = 1$ for $g \in 15A.$ The computations show that $\Sigma _{H_{6}}(2Z,2Z,2Z,2Z,15A) = 0\ (10125)$ for $Z = A\ \text {and}\ B$, respectively. It follows that

$$\begin{aligned} \Theta _{G}(2Z,2Z,2Z,2Z,15A)= & {} \Delta _{G}(2Z,2Z,2Z,2Z,15A)\\&- \displaystyle \sum _{i=1}^{6} h(g_{i}, H_{i})\Sigma _{H_{i}}(2Z,2Z,2Z,2Z,15A) \\= & {} 3375 - 0 = 3375\; (74250 - 10125 = 64125) \\&\text {for}\ Z = A\ \text {and}\ B\ \text {, respectively}, \end{aligned}$$

establishing the generation of G by the tuple (2Z, 2Z, 2Z, 2Z, 15A). Hence, the result.

Proposition 4.4

$rank(G{:}3A) = 4.$

Proof

Direct application of Theorem 3.4. $\square$

Remark 2

We can prove Proposition 4.4 without using Theorem 3.4. The rank of class 3A in G can still be established using the structure constant method together with the results of Sect. 2. The direct computations together with applications of Lemma 2.6 and Theorem 2.8 reveal that G is neither (3A, 3A, nX)- nor (3A, 3A, 3A, nX)-generated group for any non-trivial class nX of G. Thus, $rank(G{:}3A) \not \in \{2,3\}.$ It is easy to show that G is (3A, 4A, 15A)-generated group. Now it follows by applications of Lemma 2.2 that G is $(3A,3A,3A,3A,(15A)^{4})$-generated group. Thus, $rank(G{:}3A) \le 4.$ Since $rank(G{:}3A) \not \in \{2,3\}$, we deduce that $rank(G{:}3A) = 4.$

Proposition 4.5

$rank(G{:}7X) = 2,$ for $X\in \{A,B\}.$

Proof

Direct application of Theorem 3.6. $\square$

The above result can be established using the structure constant method. In the next proposition, we give the ranks of all the remaining non-trivial classes of $A_{8}$ including the classes 7A and 7B.

Proposition 4.6

Let $T := \{3B, 4A, 4B, 5A, 6A, 6B, 7A, 7B, 15A, 15B\}.$ Then $rank(G{:}nX) = 2$ for any $nX\in T.$

Proof

The aim here is to show that G is an (nX, nX, 15A)-generated group for any $nX \in T{\setminus } \{15A\}.$ For all the classes $nX \in T{\setminus } \{15A\}$ we give in Table 4 some information about $\Delta _{G}(nX, nX, 15A),$$h(15A,H_{6}),$$\Sigma _{H_{6}}(nX, nX, 15A)$ and $\Theta _{G}(nX, nX, 15A),$ where by $h(15A,H_{6})$ we mean the number of conjugate subgroups of $H_{6}$ that contain a fixed element of 15A.

Table 4 Some information on the classes $nX \in T$

The last column of Table 4 establishes the generation of G by the triple (nX, nX, 15A) for all $nX \in T{\setminus } \{15A\}.$ Thus, $rank(G{:}nX) = 2$ for all $nX \in T{\setminus } \{15A\}.$

Now it is possible to show that G is (15A, 15A, nX) for any non-trivial class nX of G. For example, we have $\Delta _{G}(15A, 15A, 7Y) = 91,$$h(g,H_{1}) = h(g,H_{2}) = h(g,H_{3}) = 1$ and $\Sigma _{H_{i}}(15A, 15A, 7Y) = 0$ for all $1 \le i \le 6$ and $Y\in \{A,B\}.$ It follows that $\Theta _{G}(15A, 15A, 7Y) = 91 - 0 = 91,$ showing the generation of G by the triple (15A, 15A, 7Y). Thus, $rank(G{:}15A) = 2.$ Hence, the result. $\square$

Remark 3

For most of the classes $nX \in T$ of Proposition 4.6, the result can also be proved using Lemma 2.5 together with the facts that the group $A_{8}$ is (2B, 4A, 15B)-, (2A, 4B, 15A)-, (2A, 5A, 15A)-, (2A, 6A, 15B)-, (2A, 6B, 15A)-, (2A, 7A, 15A)-, (2A, 7B, 15B)-, (2A, 15A, 15B)- and (2A, 15B, 15A)-generated group.

Table 5 Values $h(g,H_{i}),\ 1\le i \le 8$, for non-identity classes and maximal subgroups of $A_{9}$

Now we gather the results on ranks of the non-trivial classes of $A_{8}.$

Theorem 4.7

Let G be the alternating group $A_{8}.$ Then

• 1. $rank(G{:}2A) = rank(G{:}2B) = rank(G{:}3A) = 4.$

• 2. $rank(G{:}nX) = 2$ for all $nX \not \in \{1A, 2A, 2B, 3A\}.$

Proof

The result follows by Propositions 4.3, 4.4 and 4.6. $\square$

4.2 Ranks of $A_{9}$

The group $A_{9}$ is a simple group of order $181440 = 2^{6}\times 3^{4} \times 5 \times 7.$ By the ${\mathbb {A}}{\mathbb {T}}{\mathbb {L}}{\mathbb {A}}{\mathbb {S}}$ the group $A_{9}$ has exactly 18 conjugacy classes of its elements and 8 conjugacy classes of its maximal subgroups. Representatives of these classes of maximal subgroups can be taken as follows:

$$\begin{aligned} \begin{array}{llll} H_{1} = A_{8} &{} &{} &{} H_{2} = S_{7} \\ H_{3} = (A_{6}\times 3){:}2 &{} &{} &{} H_{4} = L_{2}(8){:}3 \\ H_{5} = L_{2}(8){:}3 &{} &{} &{} H_{6} = (A_{5}\times A_{4}){:}2 \\ H_{7} = 3^{3}{:}S_{4} &{} &{} &{} H_{8} = 3^{2}{:}2A_{4}. \end{array} \end{aligned}$$

We firstly list in Table 5 the values of $h(g,H_{i})$ for all the non-identity classes and maximal subgroups $H_{i},\ 1\le i \le 8,$ of $A_{9}.$

We start our investigation on the ranks of the non-trivial classes of $A_{9}$ by looking at the two classes of involutions 2A and 2B.

From now on let $G = A_{9}.$

Lemma 4.8

$rank(G{:}2A) \ne 3.$

Proof

We show that the group G is not (2A, 2A, 2A, nX)-generated group for any non-trivial conjugacy class nX of G. The group $A_{9}$ has an 8-dimensional complex irreducible module ${\mathbb {V}}.$ For any conjugacy class nX,  let $d_{nX} = \dim ({\mathbb {V}}/C_{{\mathbb {V}}}(nX))$ denote the codimension of the fixed space (in ${\mathbb {V}}$) of a representative of nX. Using Eq. (4) together with the power maps associated with the character table of $A_{9}$ given in the ${\mathbb {A}}{\mathbb {T}}{\mathbb {L}}{\mathbb {A}}{\mathbb {S}},$ we were able to compute all the values of $d_{nX}$ for all non-trivial classes nX of G,  and we list these values in Table 6.

Table 6 $d_{nX} = \dim ({\mathbb {V}}/C_{{\mathbb {V}}}(nX)),\ nX$ is a non-trivial class of G and $\dim ({\mathbb {V}}) = 8$

Now if $A_{9}$ is (2A, 2A, 2A, nX)-generated group, then by Scott’s theorem (see Theorem 2.8) we must have $d_{2A} + d_{2A} + d_{2A} + d_{nX} \ge 2 \times 8.$ However, it is clear from Table 6 that $3 \times d_{2A} + d_{nX} < 16,$ for all the non-trivial classes nX of G. Therefore, $A_{9}$ is not (2A, 2A, 2A, nX)-generated group for any non-trivial class nX of G and it follows that $rank(G{:}2A) \ne 3.$

Lemma 4.9

The group $A_{9}$ is (2A, 4B, 9B)-generated group.

Proof

Let $a := (6,7)(8,9) \in 2A$ and $b := (1,3,6,9)(2,5,4,7) \in 4B.$ Then $\langle a,b\rangle = A_{9}$ with $ab = (1,3,6,2,5,4,7,9,8) \in 9B.$ Thus $A_{9}$ is (2A, 4B, 9B)-generated group. $\square$

Proposition 4.10

$rank(G{:}2A) = 4.$

Proof

Since by Lemma 4.9, $A_{9}$ is (2A, 4B, 9B)-generated group, it follows by applications of Lemma 2.2 that $A_{9}$ is $(2A,2A,2A,2A,(9B)^{4})$-generated group. Thus, $rank(A_{9}{:}2A) \le 4.$ Since $rank(A_{9}{:}2A) \ne 3$ by Lemma 4.8, it follows that $rank(A_{9}{:}2A) = 4.$$\square$

Remark 4

The generation of $A_{9}$ by four suitable involutions from class 2A can be established using the structure constant method. For example, the direct computations show that $\Delta _{G}(2A,2A,2A,2A,9A) = 59049.$ From Table 5 we see that $h(g,H_{i}) = 0$ for $g \in 9A$ and $i \in \{1,2,\ldots ,8\}{\setminus } \{4,7\},$ while $h(g,H_{4}) = 3$ and $h(g,H_{7}) = 1$ for $g \in 9A.$ However, the computations show that $\Sigma _{H_{4}}(2A,2A,2A,2A,9A) = \Sigma _{H_{7}}(2A,2A,2A,2A,9A) = 0.$ It follows that

$$\begin{aligned} \Theta _{G}(2A,2A,2A,2A,9A)= & {} \Delta _{G}(2A,2A,2A,2A,9A)\\&- \displaystyle \sum _{i=1}^{8} h(g_{i}, H_{i})\Sigma _{H_{i}}(2A,2A,2A,2A,9A) \\= & {} 59049 - 0 = 59049, \end{aligned}$$

establishing the generation of G by the tuple (2A, 2A, 2A, 2A, 9A). Hence, the result.

Lemma 4.11

The group $A_{9}$ is (2B, 3C, 15A)-generated group.

Proof

Let $a := (2,4)(3,6)(5,7)(8,9) \in 2B$ and $b := (1,8,2)(3,5,7)(4,6,9) \in 3C.$ Then $\langle a,b\rangle = A_{9}$ with $ab = (1,8,4)(2,6,5,3,9) \in 15A.$ Thus, $A_{9}$ is (2B, 3C, 15A)-generated group. $\square$

Proposition 4.12

$rank(G{:}2B) = 3.$

Proof

Since by Lemma 4.11, $A_{9}$ is (2B, 3C, 15A)-generated group, and it follows by applications of Lemma 2.2 that $A_{9}$ is $(2B,2B,2B,(15A)^{4})$-generated group. Thus, $rank(A_{9}{:}2B) \le 3.$ Since two involutions generate dihedral group, it follows that $rank(A_{9}{:}2B) \ne 2$ and hence we have $rank(A_{9}{:}2B) = 3.$$\square$

Proposition 4.13

$rank(G{:}3A) = 4.$

Proof

Direct application of Theorem 3.4. $\square$

Remark 5

Similarly to Remark 2, generation of $G = A_{9}$ by four suitable elements from class 3A can be established using the structure constant method together with the results of Sect. 2. The direct computations together with applications of Lemma 2.6 and Theorem 2.8 reveal that G is neither (3A, 3A, nX)- nor (3A, 3A, 3A, nX)-generated group for any non-trivial class nX of G. Thus, $rank(G{:}3A) \not \in \{2,3\}.$ Now the direct computations yield $\Delta _{G}(3A,3A,3A, 3A, 9A)$$= 729.$ From Table 5 we see that $h(g,H_{i}) = 0$ for $g \in 9A$ and $i \in \{1,2,\ldots ,8\}{\setminus } \{4,7\},$ while $h(g,H_{4}) = 3$ and $h(g,H_{7}) = 1$ for $g \in 9A.$ However, the computations show that $\Sigma _{H_{4}}(3A, 3A,3A, 3A, 15A) = \Sigma _{H_{7}}(3A, 3A,3A, 3A, 15A) = 0$ and it follows that

$$\begin{aligned} \Theta _{G}(3A,3A,3A, 3A, 9A)= & {} \Delta _{G}(3A,3A,3A,3A, 9A)\\&- \displaystyle \sum _{i=1}^{8} h(g_{i}, H_{i})\Sigma _{H_{i}}(3A,3A,3A,3A, 9A) \\= & {} 729 - 0 = 729, \end{aligned}$$

establishing the generation of $A_{9}$ by the tuple (3A, 3A, 3A, 3A, 9A). Thus, $rank(G{:}3A) \le 4.$ Since $rank(G{:}3A) \not \in \{2,3\}$, we deduce that $rank(G{:}3A) = 4.$

Proposition 4.14

$rank(G{:}9X) = 2,$ for $X\in \{A,B\}.$

Proof

Direct application of Theorem 3.6. $\square$

The above result can be established using the structure constant method. In the next proposition, we give the ranks of all the remaining non-trivial classes of $A_{9}$ including the classes 9A and 9B.

Proposition 4.15

Let $T := \{3B, 3C, 4A, 4B, 5A, 6A, 6B, 7A, 9A, 9B, 10A, 12A, 15A, 15B\}.$ Then $rank(G{:}nX) = 2$ for any $nX\in T.$

Proof

The aim here is to show that G is an (nX, nX, 9A)-generated group for any $nX \in T{\setminus } \{9A\}.$ For all the classes $nX \in T{\setminus } \{9A\}$ we give in Table 7 some information about $\Delta _{G} = \Delta _{G}(nX, nX, 9A),$$h(9A,H_{4}),\ h(9A,H_{7}),$$\Sigma _{H_{4}} = \Sigma _{H_{4}}(nX, nX, 9A),\ \Sigma _{H_{7}} = \Sigma _{H_{7}}(nX, nX, 9A)$ and $\Theta _{G} = \Theta _{G}(nX, nX, 9A),$ where by $h(9A,H_{4}),$$h(9A,H_{7})$ we mean the number of conjugate subgroups of $H_{4}$ (resp. $H_{7}$) that contain a fixed element of 9A.

Table 7 Some information on the classes $nX \in T$

The last column of Table 7 establishes the generation of G by the triple (nX, nX, 9A) for all $nX \in T{\setminus } \{9A\}.$ Thus, $rank(G{:}nX) = 2$ for all $nX \in T{\setminus } \{9A\}.$

Now it is possible to show that G is (9A, 9A, nX) for any non-trivial class nX of G. For example, we have $\Delta _{G}(9A, 9A, 15A) = 2280,$$h(g,H_{1}) = h(g,H_{3}) = h(g,H_{6}) = 1$ and $\Sigma _{H_{i}}(9A, 9A, 15A) = 0$ for all $1 \le i \le 8.$ It follows that $\Theta _{G}(9A, 9A, 15A) = 2280 - 0 = 2280,$ showing the generation of G by the triple (9A, 9A, 15A). Thus, $rank(G{:}9A) = 2.$ Hence, the result. $\square$

Remark 6

For most of the classes $nX \in T$ of Proposition 4.15, the result can also be proved using Corollary 2.5 together with the facts that the group $A_{9}$ is (2B, 3C, 15A)-, (2B, 4A, 15A)-, (2A, 4B, 9B)-, (2B, 5A, 9B)-, (2B, 6A, 9A)-, (2B, 6B, 15B)-, (2B, 7A, 5A)-, (2A, 9A, 7A)-, (2A, 9B, 7A)-, (2B, 10A, 7A)-, (2B, 15A, 5A)- and (2B, 15B, 5A)-generated group. We also note that $A_{9}$ is not (2X, 3B, nY)-generated group for $X \in \{A,B\}$ and for any non-trivial conjugacy class nY of G.

We now gather the results on the ranks of all non-trivial classes of $A_{9}$.

Theorem 4.16

Let G be the alternating group $A_{9}.$ Then

• 1. $rank(G{:}nA) = 4$ for $n \in \{2,3\}.$

• 2. $rank(G{:}2B) = 3.$

• 3. $rank(G{:}nX) = 2$ for all $nX \not \in \{1A, 2A, 2B, 3A\}.$

Proof

The result follows by Propositions 4.10, 4.12, 4.13 and 4.15. $\square$