Redei [1] gave the definition of a minimal non-abelian group

It is an important and difficult subject in group theory to study the effect of the structure of group

In this thesis, we study the action of Aut(

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

[9]. Let Ω = {_{Ω} denote a symmetric group on Ω. A homomorphism _{Ω} is called an action of ^{x}^{a}^{b} = ^{ab}, a, b

[10, 11]. For ^{G}^{a} |a

[12,13,14]. If

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

[17,18,19,20]. ^{τ} is a subgroup of G and K is isomorphic to K^{τ}^{−1} ∈

[1].

[21, 22].

[23]. ^{i}y^{j}, y → x^{r}y^{s}. Then, G^{τ}, y^{τ}

[23]. ^{pn}^{pm}^{y}^{1+pn−1} >,

_{1}, _{2}, _{3}, _{4} >,

_{1}, _{2}, _{3}, _{4}, _{5}, _{6} >,

_{1}, _{2}, _{3}, _{4} _{5} >,

_{5} : ^{s}, y → y, s^{pn−1}, ^{n}.

In this article, we study the finite

Since ^{τ}

Next, we can prove that ^{τ}^{τ}^{−1} ∈ Aut(^{τ−1}. Because ^{τ−1} is a subgroup of ^{τ−1} = ^{−1}. Thus, ^{τ}

If _{1} and _{2} are subgroups of _{1} ≠ _{2}, then

So, _{Ω}. Thus, _{Ω}.

In addition, ∀^{τσ}^{τ}^{σ} , _{Ω}. This completes the proof of the theorem.

We notice that ^{τ}^{τ}

For an abelian

_{1} _{2} _{m} > and G is not a cyclic group with order p^{mn}, where m_{1}, _{2},⋯ , _{m} have the same order p^{n}.

When _{1}, _{2},⋯ , _{m}_{1} is ^{s}_{2} is ^{t}

Let
_{1} is a maximal subgroup, with the type invariant (^{s}^{−1}, ^{t}, o_{3}),..., _{m}

Let
_{2} is also a maximal subgroup, with the type invariant (^{s}, p^{t}^{−1}, _{3}), ... _{m}_{1}, _{2},..., _{m}^{n}

Let us prove necessity. All the maximal subgroups of _{i}, a_{1}, _{2},..., _{m−1} is a set of minimum generating system with the same order ^{n}_{1} and _{2} be any maximal subgroups of

We need to point out that _{j}, b_{1}, _{2},..., _{m−1} is also a set of minimum generating system, _{j}, _{1}, _{2},..., _{m−1} with order ^{n}, i_{1} and _{2} are all (^{n}^{−1}, ^{n},..., p^{n}_{1} and _{2} are isomorphic. Hence, we obtain that all maximal subgroups of

_{1} _{2} _{m} >, g_{1}, _{2},..., _{m}^{n}_{1} and _{2} are any maximal subgroups of _{i}, a_{1}, _{2},..., _{m−1} is a set of minimum generating system with the same order ^{n}

Let _{1} _{i}, g_{2} _{1}, _{3} _{2},..., _{m}_{m−1}. It is easy to check that

Hence,

For minimal non-abelian ^{2} with ^{2}) > 1. For metacyclic groups, Φ(^{p}^{p}

First, Φ(^{2} > and _{1} =< _{2} =< _{3} =_{1}, _{2}, _{3}}. Since these three groups are cyclic groups of order four, they are isomorphic.

Next, let us consider the orbit of _{1}, where
^{3} ^{2}^{2}^{3}^{3}^{3}, ^{2} ^{2}, 1

We can prove ^{2}, ^{3} ∈ Aut(^{3} is an identity transformation of

_{2}

(1) First, Φ(^{2}, ^{2} > and

Let us prove that the necessary and sufficient condition for maximal subgroups isomorphism is _{1} and _{2} are isomorphic, we claim ^{2} = 1. So, _{1} =< _{2} =< ^{2}, _{1} and _{2} are not isomorphic. This is a contradiction.

Since _{1} and _{2} are isomorphic, the type invariant (2^{m−1}, 2^{n}) of _{1} is the same as (2^{n−1}, 2^{m}) of _{2}. So, _{1}, _{2} and _{3} are all (2^{m−1}, 2^{m}). Then, _{1}, _{2} and _{3} are isomorphic. Then, _{1}, _{2} and _{3} are pairwise isomorphic if and only if

(2) Next, let us consider the orbit of Aut(_{1}, where

According to Lemma 4, let ^{r}b^{s}, b → a^{u}b^{v}, r, s, u, v^{τ}^{τ}

When _{1}, _{2}, _{3}, _{4} >. We obtain that

In fact, by computation

When _{1}, _{2}, _{3}, _{4}, _{5}, _{6} >. We obtain that

In fact, by computation,

We obtain that
^{b}^{1+2n−1}, i.e. ^{−1}^{2n−1}, then, ^{2n−1} and ^{−2n−1}. Hence, by computation,

Then,

Therefore, the length of the orbit containing _{1} is one and the length of the orbit containing _{2} is two. Thus, Aut(

_{2}(

First, Φ(^{2}, ^{2},

Then, Ω = {_{1}, _{2}, _{3}}.

Let us prove that the necessary and sufficient condition for maximal subgroups isomorphism is _{1} has type invariants (2^{n}, 2^{m−1}, 2) and the abelian two-group _{2} has type invariants (2^{m}, 2^{n−1}, 2). Since _{1} and _{2} are isomorphic, we have _{1}, _{2} and _{3} are the same. So, they are isomorphic.

Thus, _{1}, _{2} and _{3} are pairwise isomorphic if and only if

Next, let us consider the orbit of Aut(_{1}, where

Letting _{1} be an identical transformation,
_{2} : _{2} ∈ Aut(_{3} : _{3} ∈ Aut(

_{p}

_{p} (n, m) with n

First, ^{p}, b^{p}_{1}, _{2}, _{1},..., _{p−1}}.

We can prove that _{1}, _{2}, _{1},..., _{p−1} are pairwise isomorphic if and only if

Let ^{r}b^{s}, b → a^{u}b^{v}, r, s, u, v^{τ}^{τ}_{1} : ^{p}^{+1}, _{2} : ^{p}a, b → b, τ_{3} : _{4} : ^{1+p}, _{5} : ^{s}^{0}, _{0} = ^{pn−1}, ^{n}

Then,

_{p} (n, m,

First, Φ(^{p}, b^{p}

^{p}, c >, M_{i}^{s}b^{p}, a^{s}b, b^{p}, c >, s_{0},⋯ , _{p−1}}.

We can prove that _{0},⋯ , _{p−1} are pairwise isomorphic if and only if ^{s}b, b → a^{u}b^{v}, u^{σ}_{s}, s

We can also let ^{σ}_{s}, s^{τ}

In this paper, for Aut(

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