Orders of automorphisms of smooth plane curves for the automorphism groups to be cyclic

For a fixed integer d≥4\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$d\ge 4$$\end{document}, the list of groups that appear as automorphism groups of smooth plane curves whose degree is d is unknown, except for d=4\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$d=4$$\end{document} or 5. Harui showed a certain characteristic about structures of automorphism groups of smooth plane curves. Badr and Bars began to study for certain orders of automorphisms and try to obtain exact structures of automorphism groups of smooth plane curves. In this paper, based on the result of T. Harui, we extend Badr–Bars study for different and new cases, mainly for the cases of cyclic groups that appear as automorphism groups.


Introduction
In this paper, we work over C. Let C be a smooth plane curve of degree d ≥ 4, and Aut(C) be the automorphism group of C. The group Aut(C) is a finite subgroup of the group PGL(3, C) = Aut(P 2 ) [17]. For d = 4, Henn [11] and Kuribayashi-Komiya [14] independently give a list of precise automorphism groups that appear in a smooth plane curve of degree 4, see [1] for a survey of both results for quartics. For d = 5, Badr and Bars give a precise list [2]. In [6], Harui provides a list of group structures that may be displayed as automorphism groups of smooth plane curves of degree d ≥ 4. Part of his list are expressed using exact sequences of groups, and exact structures are unknown. Especially when the degree d is fixed, his list does not clearly state group structures which are actually realized as automorphism groups of smooth plane curves of degree d.
We fix d ≥ 4, and let G d be the set consisting of cyclic groups C yc m , dihedral groups D m , the tetrahedral group A 4 , the octahedral group S 4 , and the icosahedral group A 5 where m ≤ d − 1.
Let F d be the Fermat curve defined by X d + Y d + Z d = 0, and K d be the Klein curve defined by Z X d−1 + XY d−1 + Y Z d−1 = 0. The group structures of the automorphism groups F d and K d are known [4,19].
In [3], Badr and Bars study the group structure of Aut(C) when ord(σ ) = k(d − 2), k(d − 1), or kd for k ≥ 2, and show the following theorem. Theorem 1.1 (Please see [3] for details) Let C be a smooth plane curve of degree d ≥ 4, Aut(C) be the automorphism group of C, σ ∈ Aut(C) be an automorphism, and k ≥ 2.
T. Hayashi (B) Faculty of Agriculture, Kindai University, Nakamaticho 3327-204, Nara, Nara 631-8505, Japan E-mail: haya4taro@gmail.com 1. If ord(σ ) = k(d − 1), then Aut(C) is a cyclic group. 2. If ord(σ ) = kd, then Aut(C)) is a subgroup of Aut(F d ) or there is a normal subgroup H of Aut(C) such that the quotient group Aut(C)/H is one of G d . 3. If ord(σ ) = k(d − 2), d > 6, and d = 10, then there is a normal cyclic group H such that the quotient group Aut(C)/H is a dihedral group.
Our first main theorem is the following Theorem 1.2 which is a continuation work of Theorem 1.1 based on [6]. We write the greatest common divisor of positive integers s and t as gcd(s, t). Let C be a smooth plane curve of degree d ≥ 4, and σ be the automorphism group of C such that ord(σ ) ≥ 4. Notice that if gcd(ord(σ ), d − 1) ≥ 2, then ord(σ ) divides (d − 1)d or (d − 1) 2 , if gcd(ord(σ ), d) ≥ 2, then ord(σ ) divides (d − 1)d or (d − 2)d, and if gcd(ord(σ ), d − 2) ≥ 2, then ord(σ ) divides (d − 2)d. Theorem 1.2 Let C be a smooth plane curve of degree d ≥ 4, and Aut(C) be the automorphism group of C. We assume that there is an automorphism σ ∈ Aut(C) such that ord(σ ) ≥ 8.
Notice that if the order ord(σ ) of σ divides (d − 1) 2 but not d − 1, then ord(σ ) is not the product of different prime numbers, i.e. ord(σ ) = p 1 × · · · × p r where p i is a prime number, and p i = p j for 1 ≤ i < j ≤ r . In particular, ord(σ ) = 2, 3, 5, 6, 7. For d ≥ 6, the (1) and (2) of Theorem 1.2 uses smaller orders than the (1) of Theorem 1.1 to give sufficient conditions for Aut(C) to be a cyclic group. Please see Examples 3.4 and 3.6 . When ord(σ ) is not a multiple of d − 2, either case of the (3) of Theorem 1.2 is occurred. Please see Examples 3.9 and 3.10 . For the case ord(σ ) divides (d − 2)(d − 1) + 1, please see Theorem 3.11.
Among smooth plane curves of degree d ≥ 4, characterizations of the Fermat curve F d using the structure or order of the automorphism group is known [6,18]. Our second main theorem is a characterization of the Fermat curve F d not by the order or structure of the automorphism group, but by the orders of two automorphisms and an automorphism obtained by their composition. Theorem 1.3 Let C be a smooth plane curve of degree d ≥ 4, and Aut(C) be the automorphism group of C. The curve C has two automorphism σ and ς such that ord(σ ) = 2d, ord(ς ) = 3, and ord(ς σ 2 ) = 3 if and only if the curve C is projectively equivalent to the Fermat curve F d .
Section 2 is preliminary. We will explain the basic facts of the automorphism groups of smooth plane curves and the Galois point. To show Theorem 1.3, we will use the Galois point as a crucial step. Let C be a smooth plane curve of degree d ≥ 4, and σ ∈ Aut(C). Based on [3], we will explain the relationship between the order of σ and the defining equation of C. To show Theorem 1.2, we will use the relationship. In Sect. 3, we will show Theorems 1.2 and 1.3.

Preliminary
Let C be a smooth plane curve of degree d ≥ 4, and Aut(C) be the automorphism group of the curve C. Recall that the group Aut(C) is a finite subgroup of the group PGL(3, C).
From here, we will explain the automorphism groups of the Klein curve K d and the Fermat curve F d which are the famous and important smooth plane curves [4,19]. First, we explain the automorphism group Aut(F d ) of the Fermat curve F d . Let N F d be a subgroup of the group Aut(F d ) generated by ⎛ ⎝ e d 0 0 Let S 3 be the symmetric group of degree 3.
2. There is a split short exact sequence of groups: Notice that

Corollary 2.3 Let F d be the Fermat curve for d ≥ 4. The order of an automorphism
If the order of q(σ ) is 2, then σ 2 ∈ N F d , and hence the order of σ divides 2d. We assume that the order of q(σ ) is 3. By The characterization of the Fermat curve F d using the order of the automorphism group is known as follows.
Theorem 2.4 [18] Let C be a smooth plane curve of degree d ≥ 4, and Aut(C) be the automorphism group of C. If the order of Aut(C) is 6d 2 , then C is projectively equivalent to the Fermat curve F d .
We  and there exists a split short exact sequence of groups: The characterization of the Klein curve K d using the order of an automorphism is known as follows.
Theorem 2.7 [3,6] Let C be a smooth plane curve of degree d ≥ 4. If C has an automorphsim of order (d − 2)(d − 1) + 1, then C is projectively equivalent to the Klein curve K d .
For a homogeneous polynomial F : Definition 2.9 [6] Let C 0 be a smooth plane curve of degree d ≥ 4. A pair (C, G) of a smooth plane curve C and a subgroup G of the group Aut(C) is said to be a descendant of C 0 if C is defined by a homogeneous polynomial whose core is a defining polynomial of C 0 , and G is a subgroup of the group Aut(C 0 ) as a subgroup of the group PGL(3, C) under a suitable coordinate system. Notice that the group H l is contained in the center of PBD(2, 1). Harui provides a list of group structures that may be displayed as automorphism groups of smooth plane curves as follows.
Theorem 2.11 [6] Let C be a smooth plane curve of degree d ≥ 4, Aut(C) be the automorphism group of C, and G a subgroup of Aut(C). Then one of the following holds.
(a-i) The group G is a cyclic group. (a-ii) The group G fixes a point not lying on the curve C, by replacing the local coordinate system if necessary, G ⊂ PBD(2, 1) and H l ⊂ G for some l such that l divides d, H l is contained in the center of G, and the quotient group G/H l is one of G d . In addition, if G/H l ∼ = D m and H l is non-trivial, then m divides (c) The group G is primitive and conjugate to the icosahedral group A 5 , the Klein group PSL(2, F 7 ), the alternating group A 6 , the Hessian group H 216 of order 216, or its subgroup of order 36 or 72.
The order of an element of the groups A 5 , PSL(2, F 7 ), A 6 , and H 216 is 7 or less. Therefore, if the automorphism group Aut(C) of a smooth plane curve C of degree d ≥ 4 has an element of order 8 or more, the pair (C, Aut(C)) is not one of the case (c) of Theorem 2.11.
From here, based on [3], we will explain the relationship between the order of an automorphism and the defining equation of a smooth plane curve (see also [8]). Let C be a smooth plane curve of degree d ≥ 4, and σ be an automorphism of C. We set Fix(σ ) := {x ∈ C | g(x) = x}. By replacing the local coordinate system, we may assume that σ is defined by a diagonal matrix A :=  No.
The following theorem holds.
Theorem 2.12 [3] By replacing the local coordinate system, the number n(σ ), the polynomial F * , the diagonal matrix A, and the order ord(σ ) are one of Table 1.
Theorem 2.12 determines the orders of automorphisms, that is, the cyclic groups acting on smooth plane curves of degree d ≥ 4. In [10], the structures of abelian groups which are not cyclic groups and act on smooth plane curves of degree d ≥ 4 are determined. Besides, in [9], the structures of groups which act freely on smooth plane curves of degree d ≥ 4 are determined.

Definition 2.13 Let C (d−1)d be a smooth plane curve defined by
The automorphism groups of the curves Theorem 2.14 [3,6] We have the following.
is generated by two automorphism σ and ς := The characterization of the curves C (d−1)d , C (d−1) 2 , and C (d−2)d using the order of an automorphism is known as follows.
Theorem 2.15 [3] Let C be a smooth plane curve of degree d ≥ 4.
1. If C has an automorphsim of order (d − 1)d, then C is projectively equivalent to the curve C (d−1)d .

If C has an automorphsim of order
Let C be a smooth plane curve of degree d ≥ 4, Aut(C) be the automorphism group of C, and σ be an then C is uniquely determined, and hence the structure of the group Aut(C) is also uniquely determined.
From here, we will explain the Galois point. Let P 1 be the projective line. The gonality of a curve C is gon (C) := min{deg ϕ | ϕ : C → P 1 }. If C is a smooth plane curve of degree d ≥ 3, then gon(C) = d − 1, and the gonality of C is given by a projection π p : C → P 1 at a point p ∈ C [17]. [16,20] Let C be an irreducible plane curve of degree d ≥ 4. A point p ∈ P 1 is called a Galois point of C if the projection π p : C → P 1 is a Galois cover. Moreover, if p ∈ C (resp. p ∈ P 2 \C), then the point is said to be an inner (resp. outer) Galois point. [16,20] Let C is a smooth plane curve of degree d ≥ 4. If p ∈ P 2 is an inner (resp. outer) Galois point of C, then the Galois group of the Galois cover π p : C → P 1 is a cyclic group of order d − 1 (resp. d) which is generated by an automorphism σ of C.

Definition 2.18
Let C be an irreducible plane curve of degree d ≥ 4. We denote the number of inner (resp. outer) Galois points of C by δ(C) (resp. δ (C)). [16,20]) Let C is a smooth plane curve of degree d ≥ 4. We have the following.

5.
If δ(C) = 1, then Aut(C) is a cyclic group. 6. If δ (C) = 1, then the pair (C, Aut(C)) is the case (a-ii) of Theorem 2.11. In particular, let G be the Galois group of the Galois cover π p : C → P 1 where p ∈ P 2 is the Galois point of C. The quotient group Theorem 2.20 [3,10] Let C be a smooth plane curve in of degree d ≥ 4, and σ be an automorphism of C.
1. If ord(g) = d − 1, then |Fix(g)| = 2 if and only if C has an inner Galois point p, and the group g is the Galois group of the Galois cover π p : C → P 1 .
for k ≥ 2, then C has an inner Galois point p, and the group σ k is the Galois group of the Galois cover π p : C → P 1 . 3. If ord(g) = d, then |Fix(g)| = 0 if and only if C has an outer Galois point p, and the group g is the Galois group of the Galois cover π p : C → P 1 . 4. If ord(σ ) = kd for k ≥ 2, then C has an outer Galois point p, and the group σ k is the Galois group of the Galois cover π p : C → P 1 .
Other characterizations of smooth plane curves with Galois points are known [12,13].
Recall that for d ≥ 4, G d is a set consisting of cyclic groups C yc m , dihedral groups D m , the tetrahedral group A 4 , the octahedral group S 4 , and the icosahedral group A 5 where m ≤ d − 1. By Theorems 2.19 and 2.20 , we have the following.
Theorem 2.21 [3] Let C be a smooth plane curve of degree d, and Aut(C) be the automorphism group of C. We take an automorphism σ ∈ Aut(C).

Proof of Theorems 1.2, and 1.3
First, we will show the case (1) of Theorem 1.2. We will establish Lemma 3.1 and 3.2 in preparation for that. Recall that PSL(2, F 7 ) is the Klein group, A 6 is the alternating group, and H 216 is the Hessian group of order 216. Lemma 3.1 Let C be a smooth plane curve of degree d ≥ 4, and Aut(C) be the automorphism group of C. If there is an automorphism σ ∈ Aut(C) such that ord(σ ) ≥ 8, and ord(σ ) divides (d − 1) 2 but does not divide d − 1, then the pair (C, Aut(C)) is the case (a-i) or (a-ii) of Theorem 2.11.
Proof Let σ ∈ Aut(C) be an automorphism such that ord(σ ) ≥ 8, and ord(σ ) divides (d − 1) 2 but does not divide d − 1. Since orders of the elements of the groups A 5 , PSL(2, F 7 ), A 6 , and H 216 is 7 or less, we get that the pair (C, Aut(C)) is not the case (c) of Theorem 2.11. Since ord(σ ) ≥ 5, and ord(σ ) divides (d − 1) 2 but does not divide d − 1 Corollary 2.6, we get that the pair (C, Aut(C)) is not the case (b-ii) of Theorem 2.11. Same, since Corollary 2.3, the pair (C, Aut(C)) is not the case (b-i) of Theorem 2.11. Lemma 3.2 Let C be a smooth plane curve of degree d ≥ 4, and Aut(C) be the automorphism group of C. We assume that the pair (C, Aut(C)) is the case (a-ii) of Theorem 2.11. We take an automorphism σ . Let F be the defining equation of C, and A be a matrix defining σ . By replacing the local coordinate system, we get that F * and A are one of Table 1 (PBD(2, 1))B = PBD(2, 1). Therefore, by replacing the local coordinate system, we get that σ is defined by a diagonal matrix, [0 : 0 : 1] / ∈ C, and Aut(C) ⊂ PBD(2, 1). Let F be the defining equation of C. Since the automorphism σ is defined by a diagonal matrix, and Theorem 2.12, by multiplying X ,Y , and Z by a constant or permuting X and Y , we get that the pair (C, σ ) is one of Table 1, i.e. F * and A are one of Table 1. Multiplying X ,Y , and Z by a constant and permuting X and Y do not change the situation where [0 : 0 : 1] / ∈ C and Aut(C) ⊂ PBD(2, 1). Therefore, we get this lemma.
The following Theorem 3.3 corresponds to the (1) of Theorem 1.2. Proof Let σ ∈ Aut(C) be an automorphism such that ord(σ ) ≥ 5, and ord(σ ) divides (d − 1) 2 but does not divide d − 1. By Lemma 3.1, the pair (C, Aut(C)) is the case (a-i) or (a-ii) of Theorem 2.11. We assume that the pair (C, Aut(C)) is the case (a-ii) of Theorem 2.11. From here, we will show that the group Aut(C) is a cyclic group. By the case (a-ii) of Theorem 2.11, there is a normal cyclic subgroup H of Aut(C) such that H is contained in the center of Aut(C), the order of H divides d, and the quotient group Since H is contained in the center of Aut(C), σ, H = σ ⊕ H . Since gcd(d, (d − 1) 2 ) = 1, the group σ, H is a cyclic group. Let ς be a generator of σ, H , and v := |H |. Then ord(ς ) = v × ord(σ ). By Theorem 2.12, we get that v = 1, and hence the subgroup H is trivial. Therefore, the group Aut(C) is one of G d . Then the cyclic group σ is isomorphic to a subgroup of the groups C yc m , D m , A 4 , S 4 , or A 5 where m ≤ d − 1. Since ord(σ ) ≥ 8, and orders of the elements of the groups A 4 , S 4 , and A 5 is 5 or less, we get that the group Aut(C) is not one of the groups A 4 , S 4 , and A 5 .
We assume that the group Aut(C) is a dihedral group D m where m ≤ d − 1. Since ord(σ ) ≥ 8, there is an automorphism τ ∈ Aut(C) of order 2 such that τ σ τ = σ −1 . Since Lemma 3.2, ord(σ ), and Theorem 2.12, by replacing the local coordinate system, we get that [0 : 0 : 1] / ∈ C, and Aut(C) ⊂ PBD(2, 1), and the pair (C, σ ) is No. 6 of Table 1, i.e. C and σ are defined as follows: that the defining equation of the curve C has a Z X d−1 term but no ZY d−1 term. Therefore, the group Aut(C) is a cyclic group.
We give an example of a smooth plane curve which satisfies the condition of the (1) of Theorem 1.2 but not the condition of the (1) of Theorem 1.1.

Example 3.4
Let C be a smooth plane curve of degree d := 3(2k + 1) defined by The curve C has an automorphism σ of order l: By Theorem 3.3, the automorphism group Aut(C) is a cyclic group. From here, we will show that the order of Aut(C) is l. Let ς be a generator of Aut(C), and t := ord(ς ). Since σ and ς are commutative, the automorphism ς is defined by a diagonal matrix. Since d−1 Next, we will show the (2) of Theorem 1.2.

Theorem 3.5 Let C be a smooth plane curve of degree d ≥ 4, and Aut(C) be the automorphism group of C.
We assume that C has an automorphism σ such that ord(σ ) ≥ 8 and gcd(ord(σ ), d) ≥ 2.
Since the order of an element of S 4 , A 4 , and A 5 is at most 5, if gcd(l, d − 1) ≥ 6, then the quotient group Aut(C)/H is a cyclic group, and hence Aut(C) is a cyclic group. Let s be an element of S 4 , and l be the order of s. Then l is 2, 3, or 4. If l = 3 or 4, then there is a subgroup G l of S 4 such that s ∈ G l and G l is the dihedral group of degree l. Therefore, if gcd(l, d − 1) ≥ 3, then the quotient group Aut(C)/H is not S 4 . In the same way, we get that if gcd(l, d − 1) ≥ 3, then the quotient group Aut(C)/H is not A 5 . In addition, since the order of an element of the group A 4 is 2 or 3, if gcd(l, d − 1) ≥ 4, then the group Aut(C) is a cyclic group.
We give an example of a smooth plane curve which satisfies the condition of the (2) of Theorem 1.2 but not the condition of the (1) of Theorem 1.1.

Example 3.6
Let C be a smooth plane curve of degree d := 6(3k + 1) defined by . The curve C has an automorphism σ of order l: By Theorem 3.5, the automorphism group Aut(C) is a cyclic group. From here, we will show that the order of Aut(C) is l. Let ς be a generator of Aut(C), and t := ord(ς ). Since σ and ς are commutative, the automorphism ς is defined by a diagonal matrix. Since 3 d−1 2 divides t, by Theorem 2.12, we get that  Proof Let σ ∈ Aut(C) be an automorphism such that gcd(ord(σ ), d − 2) ≥ 12. Since ord(σ ) ≥ 8, the pair (C, Aut(C)) is not the case (c) of Theorem 2.11. Since Lemmas 2.3 and 2.6 , the pair (C, Aut(C)) is neither the case (b-i) nor the case (b-ii) of Theorem 2.11. Therefore, the pair (C, Aut(C)) is the case (a-i) or (a-ii) of Theorem 2.11. By Theorem 2.12, if gcd(ord(σ ), d − 2) ≥ 12, then ord(σ ) divides (d − 2)d. The following Theorem 3.  Proof Let σ ∈ Aut(C) be an automorphism such that gcd(ord(σ ), d − 2) ≥ 12. By Lemma 3.7, the pair (C, Aut(C)) is the case (a-i) or (a-ii) of Theorem 2.11. We assume that the pair (C, Aut(C)) is the case (a-ii) of Theorem 2.11. Since Lemma 3.2, ord(σ ) and Theorem 2.12, by replacing the local coordinate system, [0 : 0 : 1] / ∈ C, and Aut(C) ⊂ PBD(2, 1), and the pair (C, σ ) is No. 7 of Table 1 As be seen from the following two Examples 3.9 and 3.10 , either case of Theorem 3.8 is occurred. Therefore, if ord(σ ) divides (d − 2)d, then the structure of Aut(C) is may not determined by ord(σ ). Example 3.9 Let C 1 be a smooth plane curve of degree d := 8k + 6 defined by . The curve C 1 has an automorphism From here, we will show that the automorphism group Aut(C 1 ) is a cyclic group of order l. We assume that Aut(C 1 ) is not a cyclic group. Since gcd(ord(σ ), d − 2) ≥ 12, by Theorem 3.8, there is a normal cyclic group H such that H is contained in the center of Aut(C 1 ), the order of H divides d, and the quotient group ∈ H , and Aut(C 1 )/H is D m , there is an automorphism ς such that ς / ∈ H , ς 2 ∈ H , and ςσ ςσ ∈ H . Recall that since the pair (C 1 , Aut(C 1 )) is the case (a-ii) of Theorem 2.11, there is one invariant point p ∈ P 2 \C 1 under the action of Aut(C 1 ). From the defining matrix of the automorphism σ , the above-mentioned p is 1 : 0]}, we get that a 11 = a 22 = 0. This contradicts that ς is an automorphism of C 1 from the form of the defining equation of C 1 . Therefore, the group Aut(C 1 ) is a cyclic group.
Next, we assume that Aut(C) is not a cyclic group. Since the group Aut(C) has an element whose order l ≥ 8, the pair (C, Aut(C)) is not the case (c) of Theorem 2.11. By Corollary 2.3, the pair (C, Aut(C)) is not the case (b-i) of Theorem 2.11. We assume that the pair (C, Aut(C)) is the case (a-ii) of Theorem 2.11. Since Lemma 3.2, by replacing the local coordinate system if necessary, we may assume that [0 : 0 : 1] / ∈ C, and the automorphism σ is defined by a diagonal matrix. Since ord(σ ) divides (d − 2)(d − 1) + 1, and ord(σ ) > 3, by Theorem 2.12, the pair (C, σ ) is No. 10 of Table 1, and hence n(σ ) = 3. This contradicts that [0 : 0 : 1] / ∈ C. Therefore, the pair (C, Aut(C)) is not the case (a-ii) of Theorem 2.11. As a result, the pair (C, Aut(C)) is the case (b-ii) of Theorem 2.11, i.e. the pair (C, Aut(C)) is a descendant of the Klein curve K d .
By the following two Examples 3.12 and 3.13 , if ord(σ ) divides (d − 2)(d − 1) + 1, then the structure of Aut(C) is may not determined by ord(σ ). Example 3.12 Let C 3 be a smooth plane curve of degree d := 7k + 6 defined by From here, we show that the automorphism group Aut(C 3 ) is a cyclic group of order l. We assume that Aut(C 3 ) is not a cyclic group. Since ord(σ ) divides (d − 2)(d − 1) + 1, ord(σ ) ≥ 8, and Theorem 3.11, there is an automorphism ς such that ord(ς ) = 3, and ς ∩ σ = {id C 3 }. Let τ ∈ Aut(C 3 ) such that ord(τ ) is the largest order among the automorphisms of Aut(C 3 ). Since ord(σ ) = 3, and Theorem 2.5, ς i = σ for some i. Then ord(σ ) divides ord(τ ). By Theorem 2.12, the pair (C, h) is No. 10 of Table 1 This contradicts that the form of the defining equation of C 3 . Therefore, Aut(C 3 ) is a cyclic group. Next, we show that the order of the group Aut( Let τ ∈ Aut(C 3 ) be a generator. Since Aut(C 3 ) is a cyclic group, τ i = for some i. In particular, ord(σ ) divides ord (τ ). By Theorem 2.12, the pair (C 3 , τ ) is No. 10 of Table 1. Then Fix(τ ) is a set of three points. Since τ i = σ for some i, we get that Fix  This implies that m divides 2 7 ((d − 2)(d − 1) + 1). Therefore, Aut(C 3 ) is a cyclic group of order l = 1 7 ((d − 2)(d − 1) + 1). Example 3.13 Let C 4 be a smooth plane curve of degree d := 7k + 6 defined by Then Aut(C 4 ) is not a cyclic group, and the quotient group Aut(C 4 )/ g is C yc 3 .
Lastly, we will characterize the Fermat curve F d using an automorphism of order 2d and an automorphism of order 3 for a smooth plane curve of degree d. Theorem 3.14 Let C be a smooth plane curve whose degree d ≥ 4, and Aut(C) be the automorphism group of C. If C has two automorphism σ and ς such that ord(σ ) = 2d, ord(ς ) = 3, and ord(ς σ 2 ) = 3, then C is projectively equivalent to the Fermat curve F d .
Proof Let σ ∈ Aut(C) be an automorphism such that ord(σ ) = 2d. By Theorem 2.20, C has an outer Galois point. Let δ (C) be the number of outer Galois points of C. By Theorem 2.19, δ (C) = 1 or 3, and if δ (C) = 3, then C is projectively equivalent to the Fermat curve F d . We assume that δ (C) = 1. By Theorem 2.19, the pair (C, Aut(C)) is the case (a-ii). Since the order of σ and Lemma 3.2, by replacing the local coordinate system, Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.