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# Action of Aut(G) on the set of maximal subgroups of p-groups

###### Recibido: 25 Jan 2022
Detalles de la revista
Formato
Revista
eISSN
2444-8656
Primera edición
01 Jan 2016
Calendario de la edición
2 veces al año
Idiomas
Inglés
Introduction

Redei [1] gave the definition of a minimal non-abelian group G, whose subgroups are abelian except G itself, and categorised them into three categories: quaternion group, metacyclic groups and non-metacyclic groups. Hermann [2] gave the composition of groups, whose all maximal subgroups are pairwise isomorphic, with remainder classes numbering 2. Mann [3] gave the relationship between the number of remainder classes and nilpotent classes of G, whose maximal subgroups are isomorphic to each other. The complete classification of maximal automorphism p-groups with nilpotent class 3 can be seen in a previous paper [4].

It is an important and difficult subject in group theory to study the effect of the structure of group G on Aut(G). For some abelian groups, their automorphism groups are non-abelian. Therefore, the nature of the group does not necessarily hold for its automorphism group. It is not easy to find the composition of Aut(G), even if it is not easy to find the order of Aut(G). There are some results on the order of Aut(G) in other previous works [5,6,7,8].

In this thesis, we study the action of Aut(G) on the set of all maximal subgroups of some p-groups.

Preliminaries

In this section, we give some known definitions and lemmas, which will be frequently used.

Definition 1

[9]. Let Ω = {α, β, γ,⋯} and SΩ denote a symmetric group on Ω. A homomorphism φ from group G to SΩ is called an action of G on Ω. In other words, each element xG corresponds to a one-to-one transformation φ(x) of Ω, where φ(x) : ααx, satisfying (αa)b = αab, a, bG, α ∈ Ω, α ∈ Ω. If Ker φ = G, then we say that G acts on Ω trivially.

Definition 2

[10, 11]. For α ∈ Ω, αG = {αa |aG} is called the orbit of G containing α.

Definition 3

[12,13,14]. If G has only one orbit on Ω, i.e. Ω itself, we say that G acts on Ω transitively.

Definition 4

[12, 15, 16]. The number of elements in an orbit is called the length of the orbit.

Lemma 1

[17,18,19,20]. Suppose that K is a subgroup of group G, τAut(G). Then, (1) Kτ is a subgroup of G and K is isomorphic to Kτ; (2) τ−1Aut(G).

Lemma 2

[1]. Let G be a minimal non-abelian p-group. Then, G is one of the following groups:

Quaternion group $​.$ < a,{\kern 1pt} b|{a^4}{\kern 1pt} = {\kern 1pt} 1,{\kern 1pt} {b^2}{\kern 1pt} = {\kern 1pt} {a^2},{\kern 1pt} {a^b}{\kern 1pt} = {\kern 1pt} {a^{ - 1}} > .

Metacyclic group $Mp(n, m) = , with n≥2, m≥1$ {M_p}(n,{\kern 1pt} m){\kern 1pt} = {\kern 1pt} < a,{\kern 1pt} b|{a^{{p^n}}}{\kern 1pt} = {\kern 1pt} {b^{{p^m}}}{\kern 1pt} = {\kern 1pt} 1,{\kern 1pt} {a^b}{\kern 1pt} = {\kern 1pt} {a^{1 + {p^{n - 1}}}} > ,\;with\;n \ge 2,{\kern 1pt} m \ge 1

Non-metacyclic group $Mp(n, m, 1) = , where m+n≥3 if p=2.$ {M_p}(n,{\kern 1pt} m,{\kern 1pt} 1){\kern 1pt} = {\kern 1pt} < a,{\kern 1pt} b|{a^{{p^n}}} = {b^{{p^m}}} = {c^p} = 1,{\kern 1pt} [a,{\kern 1pt} b] = c,{\kern 1pt} [a,{\kern 1pt} c] = [b,{\kern 1pt} c] = 1 > ,\;where\;m + n \ge 3\;if\;p = 2.

Lemma 3

[21, 22]. A finite p-group G is minimal non-abelian if and only if Φ(G) = Z(G) and d(G) = 2.

Lemma 4

[23]. Suppose that G is a finite p-group, G =< x, y >, τAut(G), where τ : x → xiyj, y → xrys. Then, G =< xτ, yτ >⇔ isjr ≡ kp + l, where kZ, l = 1, 2,⋯ , p − 1.

Lemma 5

[23]. Let G =< x, y|xpn = ypm = 1, xy = x1+pn−1 >, with n ≥ 2, m ≥ 1. Then,

if p = 2, n = m = 2, we have Aut(G) =< τ1, τ2, τ3, τ4 >, where $τ1:x→x2n−1, y→y;τ2:x→y2x, y→y;τ3:x→x, y→yx;τ4:x→x, y→y2m−1.$ {\tau _1}:x \to {x^{{2^n} - 1}},{\kern 1pt} y \to y;{\tau _2}:x \to {y^2}x,{\kern 1pt} y \to y;{\tau _3}:x \to x,{\kern 1pt} y \to yx;{\tau _4}:x \to x,{\kern 1pt} y \to {y^{{2^m} - 1}}.

if p = 2, m ≥ 3, 3 ≤ n ≤ m, we have Aut(G) =< τ1, τ2, τ3, τ4, τ5, τ6 >, where $τ1:x→x5, y→y; τ2:x→x2n−1, y→y; τ3:x→y2m−n+1x, y→y;τ4:x→x, y→yx; τ5:x→x, y→y5; τ6:x→x, y→y2m−1$ \matrix{ {{\tau _1}:x \to {x^5},{\kern 1pt} y \to y;{\kern 1pt} {\tau _2}:x \to {x^{{2^n} - 1}},{\kern 1pt} y \to y;{\kern 1pt} {\tau _3}:x \to {y^{{2^{m - n + 1}}}}x,{\kern 1pt} y \to y;} \hfill \cr {{\tau _4}:x \to x,{\kern 1pt} y \to yx;{\kern 1pt} {\tau _5}:x \to x,{\kern 1pt} y \to {y^5};{\kern 1pt} {\tau _6}:x \to x,{\kern 1pt} y \to {y^{{2^m} - 1}}} \hfill \cr }

if p > 2, mn, we have Aut(G) =< τ1, τ2, τ3, τ4 > ã < τ5 >, where $τ1:x→xp+1, y→y; τ2:x→ypx, y→y; τ3:x→x, y→yx; τ4:x→x, y→y1+p;$ {\tau _1}:x \to {x^{p + 1}},{\kern 1pt} y \to y;{\kern 1pt} {\tau _2}:x \to {y^p}x,{\kern 1pt} y \to y;{\kern 1pt} {\tau _3}:x \to x,{\kern 1pt} y \to yx;{\kern 1pt} {\tau _4}:x \to x,{\kern 1pt} y \to {y^{1 + p}};

τ5 : x → xs, y → y, s = tpn−1, t is the primitive root of modulo pn.

Results and discussion

In this article, we study the finite p-group G; Ω = {K | K is any maximal subgroup of G}, and we use P to represent the action of Aut(G) on Ω.

Action
Theorem 3.1.1

Suppose that G is a p-group, τAut(G), assume $P(τ):K→Kτ, K∈Ω.$ P(\tau ):K \to {K^\tau },{\kern 1pt} K \in \Omega . Then, P is an action of Aut(G) on Ω.

Proof

Since K is maximal, we obtain that Kτ is a subgroup of G, under isomorphism τ.

Next, we can prove that Kτ is maximal. Assume that Kτ is a proper set of M and M is a subgroup of G. By Lemma 1, τ−1 ∈ Aut(G), we obtain that K is a proper set of Mτ−1. Because Mτ−1 is a subgroup of G and K is maximal, we have Mτ−1 = G. Then, M = G, under isomorphism τ−1. Thus, Kτ is a maximal in G. So, P(τ) is a map from Ω to itself.

If K1 and K2 are subgroups of G, where K1K2, then $K1τ≠K2τ$ K_1^\tau \ne K_2^\tau , under isomorphism τ.

So, P(τ) is an injection from Ω to itself. Moreover, P(τ) is a surjection from Ω to itself because Ω is a finite set. So, P(τ) is a one-to-one transformation of Ω. That is, P(τ) ∈ SΩ. Thus, P is a map from Aut(G) to SΩ.

In addition, ∀K ∈ Ω, Kτσ = (Kτ)σ , τ, σ ∈ Aut(G). So, P is a homomorphism from Aut(G) to SΩ. This completes the proof of the theorem.

We notice that K is isomorphic to Kτ, under isomorphism τ, and both K and Kτ are maximal in G. Next, we consider the conditions for maximal subgroups to be isomorphic, before we study the actions of Inn(G) and Aut(G) on Ω.

For an abelian p-group, we know that G is not a complete group and Inn(G) acts on the set Ω trivially. Next, we consider the action of Aut(G) on Ω.

Research on abelian p-groups
Theorem 3.2.1

Suppose G =< g1 > × < g2 > ×...× < gm > and G is not a cyclic group with order pmn, where m ≥ 2 and n ≥ 1. Then, the maximal subgroups of G are isomorphic if and only if g1, g2,⋯ , gm have the same order pn.

Proof

When n = 1, it is obviously true. Let n > 1, because G has the form $G=××…×; hence, ΦG=××…×.$ G = < {g_1} > \times < {g_2} > \times \ldots \times < {g_m} > ;\;{\rm{hence,}}\;\Phi G = < g_1^p > \times < g_2^p > \times \ldots \times < g_m^p > . Let us prove adequacy. We claim that g1, g2,⋯ , gm have the same order. Otherwise, we assume that the order of g1 is ps and the order of g2 is pt, where st.

Let $K1=×××…×$ {K_1} = < g_1^p > \times < {g_2} > \times < {g_3} > \times \ldots \times < {g_m} > . We notice that K1 is a maximal subgroup, with the type invariant (ps−1, pt, o(g3),..., o(gm)).

Let $K2=×××…×$ {K_2} = < {g_1} > \times < g_2^p > \times < {g_3} > \times \ldots \times < {g_m} > . We find that K2 is also a maximal subgroup, with the type invariant (ps, pt−1, o(g3), ... o(gm)). Their type invariants are different, they are not isomorphic. This is a contradiction. So, the order of g1, g2,..., gm is pn.

Let us prove necessity. All the maximal subgroups of G have the next form $×××…×$ < g_i^p > \times < {a_1} > \times < {a_2} > \times \ldots \times < {a_{m - 1}} > , and gi, a1, a2,..., am−1 is a set of minimum generating system with the same order pn. Let K1 and K2 be any maximal subgroups of G. Without losing generality, let $K1=×××…×,K2=×××…×.$ \matrix{ {{K_1} = < g_i^p > \times < {a_1} > \times < {a_2} > \times \ldots \times < {a_{m - 1}} > ,} \hfill \cr {{K_2} = < g_j^p > \times < {b_1} > \times < {b_2} > \times \ldots \times < {b_{m - 1}} > .} \hfill \cr }

We need to point out that gj, b1, b2,..., bm−1 is also a set of minimum generating system, gj, b1, b2,..., bm−1 with order pn, ij, 1 ≤ i, j ≤ m. Then, the type invariants of K1 and K2 are all (pn−1, pn,..., pn) So, K1 and K2 are isomorphic. Hence, we obtain that all maximal subgroups of G are isomorphic to each other. This completes the proof of the theorem.

Theorem 3.2.2

Assume that G is an abelian p-group and G is not cyclic. Then, P is transitive.

Proof

G has the decomposition form: G =< g1 > × < g2 > × ... × < gm >, g1, g2,..., gm with the same order pn. Assume that K1 and K2 are any maximal subgroups of G. Without losing generality, let $K1=×××…×$ {K_1} = < g_1^p > \times < {g_2} > \times < {g_3} > \times \ldots \times < {g_m} > ; $K2=×××…×$ {K_2} = < g_i^p > \times < {a_1} > \times < {a_2} > \times \ldots \times < {a_{m - 1}} > and gi, a1, a2,..., am−1 is a set of minimum generating system with the same order pn.

Let τ : g1 → gi, g2 → a1, g3 → a2,..., gmam−1. It is easy to check that τ ∈ Aut(G) and $K1τ=K2$ K_1^\tau = {K_2} . Then, all maximal subgroups of G are pairwise isomorphic.

Hence, $K1Aut(G)={K1τ|τ∈Aut(G)}=Ω$ K_1^{{\rm{Aut}}(G)} = \left\{ {K_1^\tau |\tau \in {\rm{Aut}}(G)} \right\} = \Omega , i.e. the action P of Aut(G) on Ω is transitive. This completes the proof of the theorem.

For minimal non-abelian p-groups G, Z(G) = Φ(G). For the quaternion group, Φ(G) contains element a2 with o(a2) > 1. For metacyclic groups, Φ(G) contains element ap With o(ap) > 1. For non-metacyclic groups, Φ(G) contains element c with order p. Overall, Z(G) ≠ {1}. Then, all minimal non-abelian p-groups are not complete groups. In addition, all elements in the set Ω are normal, the length of each orbit of Inn(G) is one, and the action of Inn(G) acts on the set Ω is trivial. Next, we consider the action of Aut (G) on Ω.

Research on minimal non-abelian 2-groups
Theorem 3.3.1

For quaternion group G, we have Aut(G) acting on the set Ω transitively, i.e. P is transitive.

Proof

First, Φ(G) =< a2 > and d(G) = 2. Then, G has three maximal subgroups, as follows: K1 =< a >, K2 =< b >, K3 =<ab>. Then, Ω = {K1, K2, K3}. Since these three groups are cyclic groups of order four, they are isomorphic.

Next, let us consider the orbit of G containing K1, where $K1Aut(G)={K1τ|τ∈Aut(G)}$ K_1^{{\rm{Aut}}(G{\rm{)}}} = \left\{ {K_1^\tau |\tau \in {\rm{Aut}}(G)} \right\} . Let τ : a → b, b → ab, ab → a, a3 → a2b, a2b → a3b, a3b → a3, a2 → a2, 1 1.

We can prove τ ∈ Aut(G) easily. Then, τ2, τ3 ∈ Aut(G), τ3 is an identity transformation of G. By computation, we have $K1τ=K2$ K_1^\tau = {K_2} , $K1τ2=K3$ {\kern 1pt} K_1^{{\tau ^2}} = {K_3} , $K1τ3=K1$ K_1^{{\tau ^3}} = {K_1} . So, $K1Aut(G)=Ω$ K_1^{{\rm{Aut}}(G{\rm{)}}} = \Omega . Then, P is transitive. This completes the proof of the theorem.

Theorem 3.3.2

For group M2 (n, m) in Lemma 2, if n = m ≥ 2, Aut(G) has two orbits and P is non-transitive.

Proof

(1) First, Φ(G) = < a2, b2 > and d(G) = 2. Then, G has three maximal subgroups, as follows: $K1=<Φ(G), a>==;K2=<Φ(G), b>==;K3=<Φ(G), ab>==​.$ \matrix{ {{K_1} = < \Phi (G),{\kern 1pt} a > = < {a^2},{\kern 1pt} {b^2},{\kern 1pt} a > = < {b^2},{\kern 1pt} a > ;} \hfill \cr {{K_2} = < \Phi (G),{\kern 1pt} b > = < {a^2},{\kern 1pt} {b^2},{\kern 1pt} b > = < {a^2},{\kern 1pt} b > ;} \hfill \cr {{K_3} = < \Phi (G),{\kern 1pt} ab > = < {a^2},{\kern 1pt} {b^2},{\kern 1pt} ab > = < {a^2},{\kern 1pt} ab > .} \hfill \cr }

Let us prove that the necessary and sufficient condition for maximal subgroups isomorphism is n = m. Since K1 and K2 are isomorphic, we claim m > 1. Conversely, if m = 1, there is b2 = 1. So, K1 =< a > and K2 =< a2, b > is non-cyclic. Then, K1 and K2 are not isomorphic. This is a contradiction.

Since K1 and K2 are isomorphic, the type invariant (2m−1, 2n) of K1 is the same as (2n−1, 2m) of K2. So, n = m. Conversely, if n = m, the type invariants of K1, K2 and K3 are all (2m−1, 2m). Then, K1, K2 and K3 are isomorphic. Then, K1, K2 and K3 are pairwise isomorphic if and only if n = m ≥ 2.

(2) Next, let us consider the orbit of Aut(G) containing K1, where $K1Aut(G)={K1τ|τ∈Aut(G)}​.$ K_1^{{\rm{Aut}}(G{\rm{)}}} = \left\{ {K_1^\tau |\tau \in {\rm{Aut}}(G)} \right\}.

According to Lemma 4, let τ : a → arbs, b → aubv, r, s, u, vN,τ ∈ Aut(G). Because an isomorphism changes a generator into a generator, aτ and bτ have the same definition relationship as a and b where rvus ≡ 1 (modulo 2). By Lemma 5, we study two cases as follows: n = m = 2 and n = m ≥ 2.

When n = m = 2, Aut(G) =< τ1, τ2, τ3, τ4 >. We obtain that $K1τ1=, where τ1:a→a3, b→b,K1τ2=, where τ2:a→b2a, b→b,K1τ3=, where τ3:a→a, b→ba,K1τ4=, where τ4:a→a, b→b3.$ \matrix{ {K_1^{{\tau _1}} = < {a^3},{\kern 1pt} {b^2} > ,\;{\rm{where}}\;{\tau _1}:a \to {a^3},{\kern 1pt} b \to b,} \hfill \cr {K_1^{{\tau _2}} = < {b^2},{\kern 1pt} {b^2}a > ,\;{\rm{where}}\;{\tau _2}:a \to {b^2}a,{\kern 1pt} b \to b,} \hfill \cr {K_1^{{\tau _3}} = < a,{\kern 1pt} {{(ba)}^2} > ,\;{\rm{where}}\;{\tau _3}:a \to a,{\kern 1pt} b \to ba,} \hfill \cr {K_1^{{\tau _4}} = < a,{\kern 1pt} {b^6} > ,\;{\rm{where}}\;{\tau _4}:a \to a,{\kern 1pt} b \to {b^3}.} \hfill \cr }

In fact, by computation $K1τi=K1$ K_1^{{\tau _i}} = {K_1} , i = 1,2,3,4. Then, $K1τ=K1$ K_1^\tau = {K_1} , ∀τ ∈ Aut(G). Thus, $K1Aut(G)=K1$ K_1^{{\rm{Aut}}(G)} = {K_1} . We obtain that $K2τ1=$ K_2^{{\tau _1}} = < {a^6},{\kern 1pt} b > , $K2τ2=<(b2a)2, b>$ {\kern 1pt} K_2^{{\tau _2}} = < ({b^2}a{)^2},{\kern 1pt} b > , $K2τ3=$ K_2^{{\tau _3}} = < {a^2},{\kern 1pt} ba > , $K2τ4=$ K_2^{{\tau _4}} = < {a^2},{\kern 1pt} {b^3} > . By computation, $K2τ3=K3$ K_2^{{\tau _3}} = {K_3} , $K2τi=K2$ {\kern 1pt} K_2^{{\tau _i}} = {K_2} , i = 1,2,4. Then, $K2Aut(G)={K2,K3}$ K_2^{{\rm{Aut}}(G)} = \left\{ {{K_2},{K_3}} \right\} .

When n = m > 2, Aut(G) =< τ1, τ2, τ3, τ4, τ5, τ6 >. We obtain that $K1τ1=, where τ1:a→a5, b→b,K1τ2=, where τ2:a→a−1, b→b,K1τ3=, where τ3:a→b2a, b→b,K1τ4=, where τ4:a→a, b→ba,K1τ5=, where τ5:a→a, b→b5,K1τ6=, where τ6:a→a, b→b−1$ \matrix{ {K_1^{{\tau _1}} = < {a^5},{\kern 1pt} {b^2} > ,\;{\rm{where}}\;{\tau _1}:a \to {a^5},{\kern 1pt} b \to b,} \hfill \cr {K_1^{{\tau _2}} = < {a^{ - 1}},{\kern 1pt} {b^2} > ,\;{\rm{where}}\;{\tau _2}:a \to {a^{ - 1}},{\kern 1pt} b \to b,} \hfill \cr {K_1^{{\tau _3}} = < {b^2},{\kern 1pt} {b^2}a > ,\;{\rm{where}}\;{\tau _3}:a \to {b^2}a,{\kern 1pt} b \to b,} \hfill \cr {K_1^{{\tau _4}} = < a,{\kern 1pt} {{(ba)}^2} > ,\;{\rm{where}}\;{\tau _4}:a \to a,{\kern 1pt} b \to ba,} \hfill \cr {K_1^{{\tau _5}} = < a,{\kern 1pt} {b^{10}} > ,\;{\rm{where}}\;{\tau _5}:a \to a,{\kern 1pt} b \to {b^5},} \hfill \cr {K_1^{{\tau _6}} = < a,{\kern 1pt} {b^{ - 2}} > ,\;{\rm{where}}\;{\tau _6}:a \to a,{\kern 1pt} b \to {b^{ - 1}}} \hfill \cr }

In fact, by computation, $K1τi=K1$ K_1^{{\tau _i}} = {K_1} , i = 1,2,3,4,5,6. Then, $K1τ=K1$ K_1^\tau = {K_1} , ∀τ ∈ Aut(G). Thus, $K1Aut(G)=K1$ K_1^{{\rm{Aut}}(G)} = {K_1} .

We obtain that $K2τ1=$ K_2^{{\tau _1}} = < {a^{10}},{\kern 1pt} b > , $K2τ2=$ {\kern 1pt} K_2^{{\tau _2}} = < {a^{ - 2}},{\kern 1pt} b > , $K2τ3=<(b2a)2, b>$ {\kern 1pt} K_2^{{\tau _3}} = < ({b^2}a{)^2},{\kern 1pt} b > , $K2τ4=$ {\kern 1pt} K_2^{{\tau _4}} = < {a^2},{\kern 1pt} ba > , $K2τ5=$ {\kern 1pt} K_2^{{\tau _5}} = < {a^2},{\kern 1pt} {b^5} > , $K2τ6=$ K_2^{{\tau _6}} = < {a^2},{\kern 1pt} {b^{ - 1}} > . By computation, $K2τi=K2$ K_2^{{\tau _i}} = {K_2} , i = 1,2,3,5,6. Since ab = a1+2n−1, i.e. b−1ab = aa2n−1, then, ab = baa2n−1 and ba = aba−2n−1. Hence, by computation, $K2τ4 = = = = K3$ K_2^{{\tau _4}}{\kern 1pt} = {\kern 1pt} < {a^2},{\kern 1pt} ba > {\kern 1pt} = {\kern 1pt} < {a^2},{\kern 1pt} ab{a^{ - {2^{n - 1}}}} > {\kern 1pt} = {\kern 1pt} < {a^2},{\kern 1pt} ab > {\kern 1pt} = {\kern 1pt} {K_3} .

Then, $K2Aut(G)={K2, K3}$ K_2^{{\rm{Aut}}(G)} = \{ {K_2},{\kern 1pt} {K_3}\} .

Therefore, the length of the orbit containing K1 is one and the length of the orbit containing K2 is two. Thus, Aut(G) does not act on the set Ω transitively. This completes the proof of the theorem.

Theorem 3.3.3

For group M2(n, m, 1), when n = m ≥ 2, P is transitive.

Proof

First, Φ(G) =< a2, b2, c > and d(G) = 2. Then, G has three subgroups that are maximal, as follows: $K1=<Φ(G), a>=, K2 = <Φ(G), b>=andK3=<Φ(G), ab>==​​.$ \matrix{ {{K_1} = < \Phi (G),{\kern 1pt} a > = < a,{\kern 1pt} {b^2},{\kern 1pt} c > ,{\kern 1pt} {K_2}{\kern 1pt} = {\kern 1pt} < \Phi (G),{\kern 1pt} b > = < b,{\kern 1pt} {a^2},{\kern 1pt} c > {\rm{and}}} \hfill \cr {{K_3} = < \Phi (G),{\kern 1pt} ab > = < {a^2},{\kern 1pt} {b^2},{\kern 1pt} c,{\kern 1pt} ab > = < ab,{\kern 1pt} {b^2},{\kern 1pt} c > .} \hfill \cr }

Then, Ω = {K1, K2, K3}.

Let us prove that the necessary and sufficient condition for maximal subgroups isomorphism is n = m. Note that the abelian two-group K1 has type invariants (2n, 2m−1, 2) and the abelian two-group K2 has type invariants (2m, 2n−1, 2). Since K1 and K2 are isomorphic, we have n = m. Conversely, if n = m, the type invariants of K1, K2 and K3 are the same. So, they are isomorphic.

Thus, K1, K2 and K3 are pairwise isomorphic if and only if n = m ≥ 2.

Next, let us consider the orbit of Aut(G) containing K1, where $K1Aut(G)={K1σ | σ∈Aut(G)}$ K_1^{{\rm{Aut}}(G)} = \left\{ {K_1^\sigma {\kern 1pt} |{\kern 1pt} \sigma \in {\rm{Aut}}(G)} \right\}

Letting σ1 be an identical transformation, $K1σ1=K1$ K_1^{{\sigma _1}} = {K_1} . Letting σ2 : a → ab, b → b, by an easy computation, σ2 ∈ Aut(G) and $K1σ2=K3$ K_1^{{\sigma _2}} = {K_3} . Letting σ3 : a → b, b → a, we have σ3 ∈ Aut(G) and $K1σ3=K2$ K_1^{{\sigma _3}} = {K_2} . Then, $K1Aut(G)=Ω$ K_1^{{\rm{Aut}}(G)} = \Omega . Therefore, P is transitive. This completes the proof of the theorem.

Research on group Mp (n, m) with p>2
Theorem 3.4.1

For group Mp (n, m) with n = m ≥ 2, P is non-transitive.

Proof

First, d(G) = 2 and Φ(G) =< ap, bp >. Then, G has p + 1 subgroups that are maximal, as follows: $K1=<Φ(G), a>=,K2=<Φ(G), b>=,Ms=<Φ(G), abs>s=1,2,…,p−1.$ \matrix{ {{K_1} = < \Phi (G),{\kern 1pt} a > = < {b^p},{\kern 1pt} a > ,} \hfill \cr {{K_2} = < \Phi (G),{\kern 1pt} b > = < {a^p},{\kern 1pt} b > ,} \hfill \cr {{M_s} = < \Phi (G),{\kern 1pt} a{b^s} > s = 1,2, \ldots ,p - 1.} \hfill \cr } Then, Ω = {K1, K2, M1,..., Mp−1}.

We can prove that K1, K2, M1,..., Mp−1 are pairwise isomorphic if and only if n = m ≥ 2.

Let τ : a → arbs, b → aubv, r, s, u, vN, τ ∈ Aut(G). Because an isomorphism changes a generator into a generator, aτ and bτ have the same definition relationship as a and b, where prvus. By Lemma 5, we only consider that τ1 : a → ap+1, b → b, τ2 : a → bpa, b → b, τ3 : a → a, b → ba, τ4 : a → a, b → b1+p, τ5 : a → as0, b → b, s0 = tpn−1, t is the primitive root of modulo pn.

Then, $K1τ1=$ K_1^{{\tau _1}} = < {a^{1 + p}},{\kern 1pt} b > , $K1τ2=$ {\kern 1pt} K_1^{{\tau _2}} = < {b^p}a,{\kern 1pt} {b^p} > , $K1τ3=$ K_1^{{\tau _3}} = < a,{\kern 1pt} {(ba)^p} > , $K1τ4=$ K_1^{{\tau _4}} = < a,{\kern 1pt} {b^{p(1 + p)}} > , $K1τ5=$ K_1^{{\tau _5}} = < {a^{{s_0}}},{\kern 1pt} {b^p} > . In fact, by computation, $K1τi=K1$ K_1^{{\tau _i}} = {K_1} , i = 1,⋯ , 5. So, $K1τ=K1$ K_1^\tau = {K_1} , ∀τ ∈ Aut(G). Thus, $K1Aut(G)=K1$ K_1^{{\rm{Aut}}(G)} = {K_1} . So, Aut(G) does not act on the set Ω transitively. This completes the proof of the theorem.

Theorem 3.4.2

For group Mp (n, m, 1), when m = n ≥ 2, P is transitive.

Proof

First, Φ(G) =< c, ap, bp > and d(G) = 2. So, G has p + 1 maximal subgroups, as follows:

K =< Φ(G), a >=< a, bp, c >, Mi =< Φ(G), asb >=< ap, asb, bp, c >, s = 0,1,⋯ , p − 1. Then, we get Ω = {K, M0,⋯ , Mp−1}.

We can prove that K, M0,⋯ , Mp−1 are pairwise isomorphic if and only if n = m ≥ 2. Let σ ∈ Aut(G), σ : a → asb, b → aubv, u = sv + kp, u, v, kN, s = 1,⋯ , p − 1. By computation, Kσ = Ms, s = 1,⋯ , p − 1.

We can also let τ ∈ Aut(G), τ : a → b, b → a. By computation, Kσ = Ms, s = 0. Letting τ be an identical transformation, Kτ = K. Then, $K1Aut(G)=Ω$ K_1^{{\rm{Aut}}(G)} = \Omega . Therefore, P is transitive. This completes the proof of the theorem.

Conclusion

In this paper, for Aut(G) of finite abelian p-groups and minimal non-abelian p-groups, we study their actions on Ω. We get some results about them. When G is a non-cyclic abelian p-group, we obtain that the action P of Aut(G) on Ω is transitive. For a quaternion group, Aut(G) has only one orbit and P is transitive; when G is a non-metacyclic group, which is minimal non-abelian, the result is the same as for the quaternion group. When G is a metacyclic group, which is minimal non-abelian, P is non-transitive.

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