Braces whose additive group has a cyclic maximal subgroup

Abstract

The problem of constructing all the non-degenerate involutive set theoretic solutions of the Yang-Baxter equation recently has been reduced to the problem of describing all the right braces. In particular, the classification of all finite right braces is fundamental in describing all such solutions of the Yang-Baxter equation. Let H be a right brace of order p^n, (H,+) \cong{\mathbb{Z}}_p\times {\mathbb{Z}}_{p^{n-1}}, where n⩾4 and p is odd prime. In this paper we prove \mathrm{S}\mathrm{o}\mathrm{c}\left(H\right)\ne 1 and classify all right braces H such that |\mathrm{S}\mathrm{o}\mathrm{c}\left(H\right)|=p^{n-1}.

Keywords

Brace / finite p-group / Yang-Baxter equation

1 Introduction

In 2007, \mathrm{R}\mathrm{u}\mathrm{m}\mathrm{p} introduced a new algebraic structure called brace in [5]. In 2014, Cedó et al. introduced right brace and left brace in [3]. Let H be a nonempty set with two operations “+” and “\circ”. If (H,+) is a abelian group, (H,\circ ) is a group and (a+b)\circ c=a\circ c+b\circ c-c for every a,b,c\in H, then (H,+,\circ ) is called a right brace. Left braces are defined analogously, but replacing the last property by a\circ (b+c)=a\circ b+a\circ c-a for every a,b,c\in H. A left or right brace is simply called a brace. Braces in [6] are all right braces. Let (H,+,\circ ) be a right brace. Let's define a new operation “\star” such that a\star b=b\circ a for every a,b\in H. Then (H,+,\star ) is a left brace. In this paper, braces are all right braces.

In 2016, Cedó et al. in [2] gave a construction method of all the non-degenerate involutive set-theoretic solutions of Yang-Baxter equations, which depends on the structure of braces. Thus, a classification of braces is necessary. In 2015, Bachiller gave a classification of braces of order p^2 and p^3 in [1]. There are 27 pairwise non-isomorphic braces of order 2^3 and 6p+19 pairwise non-isomorphic braces of order p^3 when p is odd. In 2018, Dietzel classified braces of order p^{2}q such that q>p+1 in [4], where p,q are two primes. Rump classified cyclic braces which are prime power order in [6] and not prime power order in [7]. In this paper, we study braces whose additive group has a cyclic maximal subgroup.

Recall the definitions of the structure map of H and \mathrm{S}\mathrm{o}\mathrm{c}(H). Let H be a brace, b\in H and b^{\mu }:H\to H be a map such that a\mapsto a\circ b-b. Then b^{\mu }\in \mathrm{A}\mathrm{u}\mathrm{t}(H,+) and \mu is a homomorphism from (H,\circ ) to \mathrm{A}\mathrm{u}\mathrm{t}(H,+). Then \mu is called the structure map of H and \mathrm{S}\mathrm{o}\mathrm{c}(H)=\{b\in H\mid b^{\mu }=id\}.

Let H be a brace of order p^n, (H,+)\cong {\mathbb{Z}}_p\times {\mathbb{Z}}_{p^{n-1}}, where n⩾4 and p is an odd prime. We have |\mathrm{S}\mathrm{o}\mathrm{c}(H)|\in \{1,p,p^2,\cdots ,p^n\}. Obviously, if |\mathrm{S}\mathrm{o}\mathrm{c}(H)|= p^n, then a\circ b=a+b for a,b\in H. Thus, we only need to study H such that |\mathrm{S}\mathrm{o}\mathrm{c}(H)|\ne p^n. In this paper, we prove that \mathrm{S}\mathrm{o}\mathrm{c}(H)\ne 1 in Theorem 3.1 and classify all braces H such that |\mathrm{S}\mathrm{o}\mathrm{c}(H)|=p^{n-1} in Theorem 3.2.

2 Preliminaries

We use notations and symbols about braces as in [5] and about groups as in [8]. The following Theorem 2.1 gives a construction method of braces.

Theorem 2.1 [1, Theorem 2.1]　 Let (H,+) be an abelian group, (B,+,\circ ) be a brace. Let \sigma :(B,\circ )\to \mathrm{A}\mathrm{u}\mathrm{t}(H,+) be an injective morphism, and h:(H,+)\to (B,+) be a surjective morphism. Suppose that x^{b^{\sigma }h}=x^{h{b}^{{\mu }^{\prime }}} for all b\in B and x\in H, where {\mu }^{\prime } is the structure map of B. Then, the multiplication over H given by

x\circ y=x^{y^{h\sigma }}+y,\mathrm{\forall }x,y\in H,

defines a structure of brace on H such that h is a morphism of braces, \mathrm{S}\mathrm{o}\mathrm{c}(H)=\mathrm{K}\mathrm{e}\mathrm{r}(h) and H/\mathrm{S}\mathrm{o}\mathrm{c}(H)\cong B as braces.

Two of these structures, determined by h,\sigma and h^{\prime },{\sigma }^{\prime }, respectively, are isomorphic if and only if there exists an F\in \mathrm{A}\mathrm{u}\mathrm{t}(H,+) such that

F^{-1}{x}^{h\sigma }F=(x^F{)}^{h^{\prime }{\sigma }^{\prime }},

for all x\in H.

Conversely, suppose that G is a brace. Then, the map \sigma :(G/\mathrm{S}\mathrm{o}\mathrm{c}(G),\circ )\to \mathrm{A}\mathrm{u}\mathrm{t}(G,+) induced by \mu :(G,\circ )\to \mathrm{A}\mathrm{u}\mathrm{t}(G,+), and the natural map h:G\to G/\mathrm{S}\mathrm{o}\mathrm{c}(G) satisfy the above properties.

Proposition 2.1 [1, Remark 2.2]　 There are two special uses of the isomorphism condition F^{-1}{x}^{h\sigma }F=(x^F{)}^{h^{\prime }{\sigma }^{\prime }}.

(1) Let (H,+,{\circ }_1) and (H,+,{\circ }_2) be two braces determined by h_1,\sigma and h_2,\sigma, respectively. If there is F\in \mathrm{A}\mathrm{u}\mathrm{t}(H,+) such that F{b}^{\sigma }=b^{\sigma }F for all b\in H/\mathrm{S}\mathrm{o}\mathrm{c}(H), and h_2=F{h}_1, then (H,+,{\circ }_1) and (H,+,{\circ }_2) are isomorphic.

(2) Let (H,+,{\circ }_1) be a brace determined by h_1 and {\sigma }_1. If there is F\in \mathrm{A}\mathrm{u}\mathrm{t}(H,+) such that F^{-1}{b}^{{\sigma }_1}F=b^{{\sigma }_2} for all b\in H/\mathrm{S}\mathrm{o}\mathrm{c}(H), then there is h_2 such that the brace (H,+,{\circ }_2) determined by h_2,{\sigma }_2 is isomorphic to (H,+,{\circ }_1).

Lemma 2.1 [1, Proposition 2.4]　 Let (G,+,\circ ) be a brace of order p. Then

(G,+)\cong (G,\circ )\cong {\mathbb{Z}}_p,x\circ y=x+yforallx,y\in G.

3 The classification of braces in title with \mathbf{|}\mathbf{S}\mathbf{o}\mathbf{c}\mathbf{(}\mathit{H}\mathbf{)}|\mathbf{=}{\mathit{p}}^{\mathit{n}\mathbf{-}\mathbf{1}}

Let a\in \mathbb{Z}, p be a prime, n\in \mathbb{N} and n⩾4. Assume

\overline{a}=\{a+kp\mid k\in \mathbb{Z}\},\hat{a}=\{a+k{p}^{n-1}\mid k\in \mathbb{Z}\}.

Obviously \overline{a}\in {\mathbb{Z}}_p,\hat{a}\in {\mathbb{Z}}_{p^{n-1}}. In the following text, we always hold (H,+)\cong {\mathbb{Z}}_p\times {\mathbb{Z}}_{p^{n-1}}, H=\{(\overline{a_1},\widehat{a_2})|a_1,a_2\in \mathbb{Z}\} and (\overline{a_1},\widehat{a_2})+(\overline{b_1},\widehat{b_2})=(\overline{a_1+b_1},\widehat{a_2+b_2}).

We only need to determine the operation \circ of H for determining (H,+,\circ ). Firstly, we study \mathrm{A}\mathrm{u}\mathrm{t}(H,+). Let \alpha \in \mathrm{A}\mathrm{u}\mathrm{t}(H,+), (\overline{1},\hat{0}{)}^{\alpha }=(\overline{f_{11}},\widehat{f_{12}}) and (\overline{0},\hat{1}{)}^{\alpha }=(\overline{f_{21}},\widehat{f_{22}}). Obviously, (H,+)=⟨(\overline{1},\hat{0}), (\overline{0},\hat{1})⟩, o((\overline{1},\hat{0}))=p and o((\overline{0},\hat{1}))=p^{n-1}. Then (H,+)=⟨(\overline{1},\hat{0}{)}^{\alpha }, (\overline{0},\hat{1}{)}^{\alpha }⟩, o((\overline{1},\hat{0}{)}^{\alpha })=p and o((\overline{0},\hat{1}{)}^{\alpha })=p^{n-1}. Thus

p^{n-2}\mid f_{12},p\nmid f_{11}{f}_{22}.

Then we can assume (\overline{1},\hat{0}{)}^{\alpha }=(\overline{f_{11}},\widehat{p^{n-2}{f}_{12}}). Thus, \mathrm{\forall }(\overline{x_1},\widehat{x_2})\in (H,+),

\begin{array}{rl}(\overline{x_1},\widehat{x_2}{)}^{\alpha }& =x_1(\overline{1},\hat{0}{)}^{\alpha }+x_2(\overline{0},\hat{1}{)}^{\alpha }\\ & =x_1(\overline{f_{11}},\widehat{p^{n-2}{f}_{12}})+x_2(\overline{f_{21}},\widehat{f_{22}})\\ & =(\overline{f_{11}{x}_1+f_{21}{x}_2},\widehat{p^{n-2}{f}_{12}{x}_1+f_{22}{x}_2}).\end{array}

We stipulate (\overline{x_1},\widehat{x_2})\left(\begin{array}{cc}\overline{f_{11}}& \widehat{p^{n-2}{f}_{12}}\\ \overline{f_{21}}& \widehat{f_{22}}\end{array}\right)=(\overline{f_{11}{x}_1+f_{21}{x}_2},\widehat{p^{n-2}{f}_{12}{x}_1+f_{22}{x}_2}). Then (\overline{x_1},\widehat{x_2}{)}^{\alpha }=(\overline{x_1},\widehat{x_2})\left(\begin{array}{cc}\overline{f_{11}}& \widehat{p^{n-2}{f}_{12}}\\ \overline{f_{21}}& \widehat{f_{22}}\end{array}\right) and p\nmid f_{11}{f}_{22}. The following Lemma 3.1 gives more properties of \mathrm{A}\mathrm{u}\mathrm{t}(H,+).

Lemma 3.1　 Let p be a prime, (H,+)\cong {\mathbb{Z}}_p\times {\mathbb{Z}}_{p^{n-1}}, n⩾4. Assume

K=\{\alpha :H\to H|(\overline{x_1},\widehat{x_2}{)}^{\alpha }=(\overline{x_1},\widehat{x_2})A,(\overline{x_1},\widehat{x_2})\in H,A\in I\},

L=\{\alpha :H\to H|(\overline{x_1},\widehat{x_2}{)}^{\alpha }=(\overline{x_1},\widehat{x_2})A,(\overline{x_1},\widehat{x_2})\in H,A\in J\},

where I=\left\{\left(\begin{array}{cc}\overline{f_{11}}& \widehat{p^{n-2}{f}_{12}}\\ \overline{f_{21}}& \widehat{f_{22}}\end{array}\right)|f_{11},f_{12},f_{21},f_{22}\in \mathbb{Z},p\nmid f_{11}{f}_{22}\right\},

J=\left\{\left(\begin{array}{cc}\overline{1}& \widehat{p^{n-2}r}\\ \overline{s}& \widehat{1+pt}\end{array}\right)|r,s,t\in \mathbb{Z}\right\}.

We define the operation * in I as follows:

\left(\begin{array}{cc}\overline{f_{11}}& \widehat{p^{n-2}{f}_{12}}\\ \overline{f_{21}}& \widehat{f_{22}}\end{array}\right)\ast \left(\begin{array}{cc}\overline{f_{11}^{\prime }}& \widehat{p^{n-2}{f}_{12}^{\prime }}\\ \overline{f_{21}^{\prime }}& \widehat{f_{22}^{\prime }}\end{array}\right)=\left(\begin{array}{cc}\overline{f_{11}{f}_{11}^{\prime }}& \widehat{p^{n-2}(f_{11}{f}_{12}^{\prime }+f_{12}{f}_{22}^{\prime })}\\ \overline{f_{22}{f}_{22}^{\prime }}& \widehat{p^{n-2}{f}_{21}{f}_{12}^{\prime }+f_{22}{f}_{22}^{\prime }}\end{array}\right).

Then

(1) \mathrm{A}\mathrm{u}\mathrm{t}(H,+)=K;

(2) K\cong (I,\ast );

(3) J is a \mathrm{S}\mathrm{y}\mathrm{l}\mathrm{o}\mathrm{w}p-subgroup of I;

(4) L\cong J and L is a \mathrm{S}\mathrm{y}\mathrm{l}\mathrm{o}\mathrm{w}p-subgroup of K;

(5) J\cong {\mathrm{M}}_p(1,1,1)\ast {\mathbb{Z}}_{p^{n-2}} when p>2, J\cong {\mathrm{D}}_8\ast {\mathbb{Z}}_{2^{n-3}}\times {\mathbb{Z}}_2 when p=2.

Proof　 (1) Obviously, \mathrm{A}\mathrm{u}\mathrm{t}(H,+)\subseteq K. Since

|\mathrm{A}\mathrm{u}\mathrm{t}(H,+)|=(p^n-p^{n-1})(p^2-p)=p^n(p-1{)}^{2}and|K|⩽|I|=p^n(p-1{)}^2, |\mathrm{A}\mathrm{u}\mathrm{t}(H,+)|⩾|K|. Thus \mathrm{A}\mathrm{u}\mathrm{t}(H,+)=K.

(2) Let \phi :K\to I be a map such that \phi :\alpha \mapsto A. Obviously, \phi is an injection. Since |K|=|I|, \phi is a bijection. We claim \phi is a homomorphism. It only needs to prove ((\overline{x_1},\widehat{x_2}{)}^{{\alpha }_1}{)}^{{\alpha }_2}=(\overline{x_1},\widehat{x_2}{)}^{{\alpha }_1{\alpha }_2} for all {\alpha }_1,{\alpha }_2\in \mathrm{A}\mathrm{u}\mathrm{t}(H,+) and (\overline{x_1},\widehat{x_2})\in H. It needs to prove ((\overline{x_1},\widehat{x_2})A_1)A_2=(\overline{x_1},\widehat{x_2})(A_1\ast A_2) for all A_1,A_2\in K. It can be easily verified.

(3) It follows from (2) that (I,\ast ) is a group. It is easy to see that A\ast B\in J for all A,B\in J. Thus J\le I. Since |I|=p^n(p-1{)}^2 and |J|=p^n, J is a \mathrm{S}\mathrm{y}\mathrm{l}\mathrm{o}\mathrm{w}p-subgroup of I.

(4) Let \phi be the same one in proof of (2). Then \phi {|}_L is an isomorphism of L into J. Thus L\le K. Using the orders of L and K, we get that L is a \mathrm{S}\mathrm{y}\mathrm{l}\mathrm{o}\mathrm{w}p-subgroup of K.

(5) If p>2, then J=⟨\left(\begin{array}{cc}\overline{1}& \hat{0}\\ \overline{1}& \hat{1}\end{array}\right),\left(\begin{array}{cc}\overline{1}& \widehat{p^{n-2}}\\ \overline{0}& \hat{1}\end{array}\right)⟩\ast ⟨\left(\begin{array}{cc}\overline{1}& \hat{0}\\ \overline{0}& \widehat{1+p}\end{array}\right)⟩\cong {\mathrm{M}}_p(1,1,1)\ast {\mathbb{Z}}_{p^{n-2}}; If p=2, then J=⟨\left(\begin{array}{cc}\overline{1}& \hat{0}\\ \overline{1}& \hat{1}\end{array}\right),\left(\begin{array}{cc}\overline{1}& \widehat{2^{n-2}}\\ \overline{0}& \hat{1}\end{array}\right)⟩\ast ⟨\left(\begin{array}{cc}\overline{1}& \hat{0}\\ \overline{0}& \widehat{1+2}\end{array}\right)⟩\times ⟨\left(\begin{array}{cc}\overline{1}& \hat{0}\\ \overline{0}& \widehat{-1}\end{array}\right)⟩\cong {\mathrm{D}}_8\ast {\mathbb{Z}}_{2^{n-3}}\times {\mathbb{Z}}_2.□

Remark 3.1　For the convenience of writing, we use (x_1,x_2) to denote (\overline{x_1},\widehat{x_2}) and \left(\begin{array}{cc}a& b\\ c& d\end{array}\right) to denote \left(\begin{array}{cc}\overline{a}& \hat{b}\\ \overline{c}& \hat{d}\end{array}\right). If \alpha \in \mathrm{A}\mathrm{u}\mathrm{t}(H,+)=K and (x_1,x_2{)}^{\alpha }=(x_1,x_2)A, then A is called the matrix of \alpha. As long as there is no confusion, we also use \alpha =A, \mathrm{A}\mathrm{u}\mathrm{t}(H,+)=I and L=J in the following text.

Theorem 3.1　 Let H be a brace and (H,+)\cong {\mathbb{Z}}_p\times {\mathbb{Z}}_{p^{n-1}}, where n⩾4 and p is an odd prime. Then |\mathrm{S}\mathrm{o}\mathrm{c}(H)|\ne 1.

Proof　Assume that |\mathrm{S}\mathrm{o}\mathrm{c}(H)|=1 and \mu is the structure map of H. Then \mu is a homomorphism from (H,\circ ) to \mathrm{A}\mathrm{u}\mathrm{t}(H,+) and \mathrm{K}\mathrm{e}\mathrm{r}\mu =\mathrm{S}\mathrm{o}\mathrm{c}(H). Since |\mathrm{S}\mathrm{o}\mathrm{c}(H)|=1, (H,\circ )\cong (H,\circ {)}^{\mu }\le \mathrm{A}\mathrm{u}\mathrm{t}(H,+). It follows from the order of (H,\circ ) that (H,\circ {)}^{\mu }\in {\mathrm{S}\mathrm{y}\mathrm{l}}_p(K). By Lemma 3.1, (H,\circ {)}^{\mu }\cong J\cong {\mathrm{M}}_p(1,1,1)\ast {\mathrm{C}}_{p^{n-2}}. Then {\mathrm{\Omega }}_1((H,\circ {)}^{\mu })\cong {\mathrm{M}}_p(1,1,1). Thus ({\mathrm{\Omega }}_1((H,\circ {)}^{\mu }){)}^{{\mu }^{-1}}\cong {\mathrm{M}}_p(1,1,1). Without loss of generality, we assume that (H,\circ {)}^{\mu }\le L. By Lemma 3.1(5), we have

{\mathrm{\Omega }}_1(J)=\left\{\left(\begin{array}{cc}1& p^{n-2}r\\ s& 1+p^{n-2}t\end{array}\right)|r,s,t\in \mathbb{Z}\right\}\cong {\mathrm{M}}_p(1,1,1).

Then using Lemma 3.1(4), we can assume (x_1,x_2{)}^{\mu }=(x_1,x_2)A for all (x_1,x_2)\in ({\mathrm{\Omega }}_1((H,\circ {)}^{\mu }){)}^{{\mu }^{-1}}, where A=\left(\begin{array}{cc}1& p^{n-2}r\\ s& 1+p^{n-2}t\end{array}\right). Then A^k=\left(\begin{array}{cc}1& k{p}^{n-2}r\\ ks& 1+k{p}^{n-2}t+p^{n-2}\left(\begin{array}{c}k\\ 2\end{array}\right)sr\end{array}\right). Since ({\mathrm{\Omega }}_1((H,\circ {)}^{\mu }){)}^{{\mu }^{-1}}\cong {\mathrm{M}}_p(1,1,1), (x_1,x_2{)}^p=(0,0). Obviously, (x_1,x_2)\circ (x_1,x_2)=(x_1,x_2{)}^{(x_1,x_2{)}^{\mu }}+(x_1,x_2). Then

\begin{array}{rl}(0,0)=(x_1,x_2{)}^p& =(x_1,x_2)(A^{p-1}+A^{p-2}+\dots +A^2+A+E)\\ & =(x_1,x_2)\left(\begin{array}{cc}0& {\displaystyle {\sum }_{i=1}^{p-1}ir{p}^{n-2}}\\ {\displaystyle {\sum }_{i=1}^{p-1}is}& p+{\displaystyle {\sum }_{i=1}^{p-1}it{p}^{n-2}+{\displaystyle {\sum }_{i=2}^{p-1}sr{p}^{n-2}}}\end{array}\right)\\ & =\left(x_1,x_2\right)\left(\begin{array}{cc}0& 0\\ 0& p+{\displaystyle {\sum }_{i=2}^{p-1}\left(\begin{array}{c}i\\ 2\end{array}\right)sr{p}^{n-2}}\end{array}\right)\\ & =\left(0,\left(p+\sum _{i=2}^{p-1}\left(\begin{array}{c}i\\ 2\end{array}\right)sr{p}^{n-2}\right)x_2\right).\end{array}

Thus x_2\equiv0(\mathrm{mod}{p}^{n-2}). Therefore, |({\mathrm{\Omega }}_1((H,\circ {)}^{\mu }){)}^{{\mu }^{-1}}|=p^2. This is in contradiction with ({\mathrm{\Omega }}_1((H,\circ {)}^{\mu }){)}^{{\mu }^{-1}}\cong {\mathrm{M}}_p(1,1,1).□

Theorem 3.2　 Let H be a brace and (H,+)\cong {\mathbb{Z}}_p\times {\mathbb{Z}}_{p^{n-1}}, where n⩾4 and p is an odd prime. If |\mathrm{S}\mathrm{o}\mathrm{c}(H)|=p^{n-1}, then (H,+,\circ ) is one of the following pairwise nonisomorphic braces:

(1) (x_1,x_2)\circ (y_1,y_2)=(x_1+y_1,x_2+y_2+p^{n-2}{x}_2{y}_2), (H,\circ )\cong {\mathbb{Z}}_p\times {\mathbb{Z}}_{p^{n-1}};

(2) (x_1,x_2)\circ (y_1,y_2)=(x_1+y_1,x_2+y_2+p^{n-2}{x}_2{y}_1), (H,\circ )\cong {\mathrm{M}}_p(n-1,1);

(3) (x_1,x_2)\circ (y_1,y_2)=(x_1+y_1,x_2+y_2+p^{n-2}{x}_1{y}_2), (H,\circ )\cong {\mathrm{M}}_p(n-1,1);

(4) (x_1,x_2)\circ (y_1,y_2)=(x_1+y_1,x_2+y_2+p^{n-2}{x}_1{y}_1), (H,\circ )\cong {\mathbb{Z}}_p\times {\mathbb{Z}}_{p^{n-1}};

(5) (x_1,x_2)\circ (y_1,y_2)=(x_1+y_1,x_2+y_2+p^{n-2}{x}_1(y_1+y_2)), (H,\circ )\cong {\mathrm{M}}_p(n-1,1);

(6) (x_1,x_2)\circ (y_1,y_2)=(x_1+y_1+x_2{y}_2,x_2+y_2), (H,\circ )\cong {\mathbb{Z}}_p\times {\mathbb{Z}}_{p^{n-1}};

(7) (x_1,x_2)\circ (y_1,y_2)=(x_1+y_1+x_2{y}_2,x_2+y_2+p^{n-2}{x}_1{y}_2+p^{n-2}{x}_2(\genfrac{}{}{0ex}{}{y_2}{2})), (H,\circ )\cong {\mathrm{M}}_p(n-1,1);

(8) (x_1,x_2)\circ (y_1,y_2)=(x_1+y_1+x_2{y}_2,x_2+y_2+p^{n-2}\epsilon x_1{y}_2+p^{n-2}\epsilon x_2(\genfrac{}{}{0ex}{}{y_2}{2})), where \epsilon is a fixed quadratic non-residue modulo p; (H,\circ )\cong {\mathrm{M}}_p(n-1,1).

Proof　Obviously, |H/\mathrm{S}\mathrm{o}\mathrm{c}(H)|=p. Let H/\mathrm{S}\mathrm{o}\mathrm{c}(H)=\{b\mid b\in {\mathbb{Z}}_p\} and {\mu }^{\prime } be the structure map of H/\mathrm{S}\mathrm{o}\mathrm{c}(H). By Lemma 2.1, b^{{\mu }^{\prime }}=id for all b\in H/\mathrm{S}\mathrm{o}\mathrm{c}(H). We apply Theorem 2.1 to find all the brace structures over H.

Assume that h:(H,+)\to (H/\mathrm{S}\mathrm{o}\mathrm{c}(H),+) is a surjective morphism such that (x_1,x_2{)}^h=(x_1,x_2)\left(\begin{array}{c}\alpha \\ \beta \end{array}\right), where \alpha ,\beta \in {\mathbb{Z}}_p and \alpha ,\beta are not both 0.

Assume that \sigma :(H/\mathrm{S}\mathrm{o}\mathrm{c}(H),\circ )\to \mathrm{A}\mathrm{u}\mathrm{t}(H,+) is an injective morphism. Since |H/\mathrm{S}\mathrm{o}\mathrm{c}(H)|=p, |(H/\mathrm{S}\mathrm{o}\mathrm{c}(H),\circ {)}^{\sigma }|=p. Thus (H/\mathrm{S}\mathrm{o}\mathrm{c}(H),\circ {)}^{\sigma } is contained in a Sylow p-subgroup of \mathrm{A}\mathrm{u}\mathrm{t}(H,+). We know that any two Sylow p-subgroups of \mathrm{A}\mathrm{u}\mathrm{t}(H,+) are conjugate. Then we may take (H/\mathrm{S}\mathrm{o}\mathrm{c}(H),\circ {)}^{\sigma }\le L by Proposition 2.1 (2) and Lemma 3.1. Therefore, we obtain (H/\mathrm{S}\mathrm{o}\mathrm{c}(H),\circ {)}^{\sigma }\le J= \left\{\left(\begin{array}{cc}1& p^{n-2}r\\ s& 1+pt\end{array}\right)|r,s\in {\mathbb{Z}}_p,t\in {\mathbb{Z}}_{p^{n-2}}\right\} by Remark 3.1 and Lemma 3.1. Obviously, (H/\mathrm{S}\mathrm{o}\mathrm{c}(H),\circ )=⟨1⟩ and o(1)=p in (H/\mathrm{S}\mathrm{o}\mathrm{c}(H),\circ ). By Lemma 3.1(5), we may take 1^{\sigma }=\left(\begin{array}{cc}1& p^{n-2}r\\ s& 1+p^{n-2}t\end{array}\right), where r,s,t\in {\mathbb{Z}}_p and they are not all 0. By Lemma 2.1, we get b=1^b for every b\in H/\mathrm{S}\mathrm{o}\mathrm{c}(H). Then b^{\sigma }=(1^{\sigma }{)}^b.

By Theorem 2.1, h and \sigma need to satisfy b^{\sigma }h=h{b}^{{\mu }^{\prime }} for every b\in H/\mathrm{S}\mathrm{o}\mathrm{c}(H). Then h and \sigma need to satisfy (x_1,x_2{)}^{b^{\sigma }h}=(x_1,x_2{)}^{h{b}^{{\mu }^{\prime }}} for all (x_1,x_2)\in H. Thus x_2\alpha bs\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}p). By the arbitrariness of x_2 and b, \alpha s\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}p). Thus, we obtain

1^{\sigma }=\left(\begin{array}{cc}1& p^{n-2}r\\ s& 1+p^{n-2}t\end{array}\right),h:(x_1,x_2)\mapsto (x_1,x_2)\left(\begin{array}{c}\alpha \\ \beta \end{array}\right),

where r,s,t,\alpha ,\beta \in {\mathbb{Z}}_p, \alpha s=0, \alpha and \beta are not both 0; and r,s and t are not all 0. Notice that if \alpha, \beta, r,s and t satisfy above condition, then h,\sigma determined by them satisfy b^{\sigma }h=h{b}^{{\mu }^{\prime }}. By Theorem 2.1, we get a brace H determined by h and \sigma.

Since some of these braces might be isomorphic, so we have to determine the repeated cases in the following text. We apply Theorem 2.1 and Proposition 2.1 to remove the repeated cases. We discuss it in following cases.

Case 1: s=r=0.

In this case, 1^{\sigma }=\left(\begin{array}{cc}1& 0\\ 0& 1+p^{n-2}t\end{array}\right). Since o(1^{\sigma })=o(1)=p, t\ne 0.

Subcase 1.1 \alpha =0

In this case, h:(x_1,x_2)\mapsto (x_1,x_2)\left(\begin{array}{c}0\\ \beta \end{array}\right), 1^{\sigma }=\left(\begin{array}{cc}1& 0\\ 0& 1+p^{n-2}t\end{array}\right), where t\beta \ne 0. Suppose

h_1:(x_1,x_2)\mapsto (x_1,x_2)\left(\begin{array}{r}0\\ 1\end{array}\right),1^{{\sigma }_1}=\left(\begin{array}{cc}1& 0\\ 0& 1+p^{n-2}\end{array}\right).

We take F=\left(\begin{array}{cc}1& 0\\ 0& t\beta \end{array}\right)\in \mathrm{A}\mathrm{u}\mathrm{t}(H,+). Then (x_1,x_2{)}^{h\sigma }F=F((x_1,x_2)F{)}^{h_1{\sigma }_1} for all (x_1,x_2)\in H. By Lemma 2.1, the brace determined by h and \sigma is isomorphic to the brace determined by h_1 and {\sigma }_1. Thus we only get a brace determined by h_1 and {\sigma }_1 in this case.

Now the \circ is defined as follows. \mathrm{\forall }(x_1,x_2),(y_1,y_2)\in H,

\begin{array}{rl}(x_1,x_2)\circ (y_1,y_2)& =(x_1,x_2{)}^{(y_1,y_2{)}^{h_1{\sigma }_1}}+(y_1,y_2)\\ & =(x_1,x_2){\left(\begin{array}{cc}1& 0\\ 0& 1+p^{n-2}\end{array}\right)}^{(y_1,y_2)\left(\begin{array}{r}0\\ 1\end{array}\right)}+(y_1,y_2)\\ & =(x_1,x_2)\left(\begin{array}{cc}1& 0\\ 0& 1+p^{n-2}{y}_2\end{array}\right)+(y_1,y_2)\\ & =(x_1,x_2+p^{n-2}{x}_2{y}_2)+(y_1,y_2)\\ & =(x_1+y_1,x_2+y_2+p^{n-2}{x}_2{y}_2).\end{array}

We get brace (1) in Theorem 3.2.

It can be calculated (1,0{)}^m=(m,0) and (0,1{)}^m=(0,m+p^{n-2}(\genfrac{}{}{0ex}{}{m}{2})) for m\in \mathbb{N}. Then o((1,0))=p, o((0,1))=p^{n-1} and ⟨(1,0)⟩\bigcap ⟨(0,1)⟩=⟨(0,0)⟩ in group (H,\circ ). Thus (H,\circ )=⟨(1,0),(0,1)⟩. It is easy to get (1,0)\circ (0,1)=(1,1)=(0,1)\circ (1,0). Then (H,\circ )\cong {\mathbb{Z}}_p\times {\mathbb{Z}}_{p^{n-1}}.

Subcase 1.2 \alpha \ne 0

In this case, h:(x_1,x_2)\mapsto (x_1,x_2)\left(\begin{array}{c}\alpha \\ \beta \end{array}\right), 1^{\sigma }=\left(\begin{array}{cc}1& 0\\ 0& 1+p^{n-2}t\end{array}\right), where t\alpha \ne 0. Suppose h_1:(x_1,x_2)\mapsto (x_1,x_2)\left(\begin{array}{c}1\\ 0\end{array}\right),1^{{\sigma }_1}=\left(\begin{array}{cc}1& 0\\ 0& 1+p^{n-2}\end{array}\right). We take F=\left(\begin{array}{cc}t\alpha & 0\\ t\beta & 1\end{array}\right)\in \mathrm{A}\mathrm{u}\mathrm{t}(H,+). Then (x_1,x_2{)}^{h\sigma }F=F((x_1,x_2)F{)}^{h_1{\sigma }_1} for (x_1,x_2)\in H. By Lemma 2.1, the brace determined by h and \sigma is isomorphic to the brace determined by h_1 and {\sigma }_1. Thus we only get a brace determined by h_1 and {\sigma }_1 in this case.

Now the \circ is defined as follows. \mathrm{\forall }(x_1,x_2),(y_1,y_2)\in H,

\begin{array}{rl}(x_1,x_2)\circ (y_1,y_2)& =(x_1,x_2{)}^{(y_1,y_2{)}^{h_1{\sigma }_1}}+(y_1,y_2)\\ & =(x_1+y_1,x_2+y_2+p^{n-2}{x}_2{y}_1).\end{array}

We get brace (2) in Theorem 3.2.

It can be calculated (1,0{)}^m=(m,0) and (0,1{)}^m=(0,m) for m\in \mathbb{N}. Then o((1,0))=p, o((0,1))=p^{n-1} and ⟨(1,0)⟩\bigcap ⟨(0,1)⟩=⟨(0,0)⟩ in group (H,\circ ). Thus (H,\circ )=⟨(1,0),(0,1)⟩. It is easy to get (1,0)\circ (0,1)=(1,1) and ((0,1)\circ (1,0)=(1,1+p^{n-2}). Thus (H,\circ ) is not abelian. It is easy to get (H,\circ )\cong {\mathrm{M}}_p(n-1,1).

Case 2: s=0 and r\ne 0

In this case, 1^{\sigma }=\left(\begin{array}{cc}1& p^{n-2}r\\ 0& 1+p^{n-2}t\end{array}\right), h:(x_1,x_2)\mapsto (x_1,x_2)\left(\begin{array}{c}\alpha \\ \beta \end{array}\right), where \alpha and \beta are not both 0. Suppose 1^{{\sigma }_1}=\left(\begin{array}{cc}1& p^{n-2}\\ 0& 1\end{array}\right). We take F=\left(\begin{array}{cc}r& 0\\ t& 1\end{array}\right)\in \mathrm{A}\mathrm{u}\mathrm{t}(H,+). Then F^{-1}{b}^{\sigma }F=b^{{\sigma }_1} for b\in H/\mathrm{S}\mathrm{o}\mathrm{c}(H). By Proposition 2.1 (2) , there is h_1 such that the brace determined by h and \sigma is isomorphic to the brace determined by h_1 and {\sigma }_1.

Now we can also assume h_1:(x_1,x_2)\mapsto (x_1,x_2)\left(\begin{array}{c}\alpha \\ \beta \end{array}\right), where \alpha and \beta are not both 0. Thus we only get braces determined by h_1 and {\sigma }_1 in this case.

Subcase 2.1: \alpha =0

In this case, h_1:(x_1,x_2)\mapsto (x_1,x_2)\left(\begin{array}{c}0\\ \beta \end{array}\right), 1^{{\sigma }_1}=\left(\begin{array}{cc}1& p^{n-2}\\ 0& 1\end{array}\right), where \beta \ne 0. Suppose h_2:(x_1,x_2)\mapsto (x_1,x_2)\left(\begin{array}{c}0\\ 1\end{array}\right). We take F=\left(\begin{array}{cc}{\beta }^{-1}& 0\\ 0& {\beta }^{-1}\end{array}\right)\in \mathrm{A}\mathrm{u}\mathrm{t}(H,+). Then h_2=F{h}_1 and F^{-1}{b}^{\sigma }F=b^{{\sigma }_1} for b\in H/\mathrm{S}\mathrm{o}\mathrm{c}(H). By Proposition 2.1 (1), the brace determined by h_1 and {\sigma }_1 is isomorphic to the brace determined by h_2 and {\sigma }_1. Thus we only get a brace determined by h_2 and {\sigma }_1 in this case.

Now the \circ is defined as follows. \mathrm{\forall }(x_1,x_2),(y_1,y_2)\in H,

\begin{array}{rl}(x_1,x_2)\circ (y_1,y_2)& =(x_1,x_2{)}^{(y_1,y_2{)}^{h_2{\sigma }_1}}+(y_1,y_2)\\ & =(x_1+y_1,x_2+y_2+p^{n-2}{x}_1{y}_2).\end{array}

Then we get brace (3) in Theorem 3.2. We also obtain (H,\circ )\cong {\mathrm{M}}_p(n-1,1) by a similar proof of Subcase 1.2.

Subcase 2.2: \alpha \ne 0 and \beta =0

In this case, h_1:(x_1,x_2)\mapsto (x_1,x_2)\left(\begin{array}{c}\alpha \\ 0\end{array}\right), 1^{{\sigma }_1}=\left(\begin{array}{cc}1& p^{n-2}\\ 0& 1\end{array}\right). Suppose h_2:(x_1,x_2)\mapsto (x_1,x_2)\left(\begin{array}{c}1\\ 0\end{array}\right). We take F=\left(\begin{array}{cc}{\alpha }^{-1}& 0\\ 0& {\alpha }^{-1}\end{array}\right)\in \mathrm{A}\mathrm{u}\mathrm{t}(H,+). Then h_2=F{h}_1 and b^{{\sigma }_1}F=F{b}^{{\sigma }_1} for b\in H/\mathrm{S}\mathrm{o}\mathrm{c}(H). By Proposition 2.1 (1), the brace determined by h_1 and {\sigma }_1 is isomorphic to the brace determined by h_2 and {\sigma }_1. Thus we only get a brace determined by h_2 and {\sigma }_1 in this case.

Now the \circ is defined as follows. \mathrm{\forall }(x_1,x_2),(y_1,y_2)\in H,

\begin{array}{rl}(x_1,x_2)\circ (y_1,y_2)& =(x_1,x_2{)}^{(y_1,y_2{)}^{h_2{\sigma }_1}}+(y_1,y_2)\\ & =(x_1+y_1,x_2+y_2+p^{n-2}{x}_1{y}_1).\end{array}

Then we get brace (4) in Theorem 3.2. We also obtain (H,\circ )\cong {\mathbb{Z}}_p\times {\mathbb{Z}}_{p^{n-1}} by a similar proof of Subcase 1.1.

Subcase 2.3: \alpha \ne 0 and \beta \ne 0

In this case, h_1:(x_1,x_2)\mapsto (x_1,x_2)\left(\begin{array}{c}\alpha \\ \beta \end{array}\right), 1^{{\sigma }_1}=\left(\begin{array}{cc}1& p^{n-2}\\ 0& 1\end{array}\right). Suppose h_2:(x_1,x_2)\mapsto (x_1,x_2)\left(\begin{array}{c}1\\ 1\end{array}\right). We take F=\left(\begin{array}{cc}\beta & 0\\ 0& {\alpha }^{-1}{\beta }^2\end{array}\right)\in \mathrm{A}\mathrm{u}\mathrm{t}(H,+). Then (x_1,x_2{)}^{h_1{\sigma }_1}F=F((x_1,x_2)F{)}^{h_2{\sigma }_1} for (x_1,x_2)\in H. By Theorem 2.1, the brace determined by h_1 and {\sigma }_1 is isomorphic to the brace determined by h_2 and {\sigma }_1. Thus we only get a brace determined by h_2 and {\sigma }_1 in this case.

Now the \circ is defined as follows. \mathrm{\forall }(x_1,x_2),(y_1,y_2)\in H,

\begin{array}{rl}(x_1,x_2)\circ (y_1,y_2)& =(x_1,x_2{)}^{(y_1,y_2{)}^{h_2{\sigma }_1}}+(y_1,y_2)\\ & =(x_1+y_1,x_2+y_2+p^{n-2}{x}_1(y_1+y_2)).\end{array}

Then we get brace (5) in Theorem 3.2. We also obtain (H,\circ )\cong {\mathrm{M}}_p(n-1,1) by a similar proof of Subcase 1.2.

Case 3: s\ne 0

Since \alpha s=0, \alpha =0. Since \alpha and \beta are not both 0, \beta \ne 0. Therefore,

1^{\sigma }=\left(\begin{array}{cc}1& p^{n-2}r\\ s& 1+p^{n-2}t\end{array}\right),h:(x_1,x_2)\mapsto (x_1,x_2)\left(\begin{array}{r}0\\ \beta \end{array}\right),

where s\beta \ne 0. Suppose 1^{{\sigma }_1}=\left(\begin{array}{cc}1& p^{n-2}{r}^{\prime }\\ 1& 1\end{array}\right), where r^{\prime }=rs. We take F=\left(\begin{array}{cc}{s}^{-1}& -p^{n-2}{s}^{-1}t\\ 0& 1\end{array}\right)\in \mathrm{A}\mathrm{u}\mathrm{t}(H,+). Then F^{-1}{b}^{\sigma }F=b^{{\sigma }_1} for b\in H/\mathrm{S}\mathrm{o}\mathrm{c}(H). By Proposition 2.1 (2), there is h_1 such that the brace determined by h and \sigma is isomorphic to the brace determined by h_1 and {\sigma }_1.

Now we can also assume h_1:(x_1,x_2)\mapsto (x_1,x_2)\left(\begin{array}{c}0\\ \beta \end{array}\right), where \beta \ne 0. Thus we only get braces determined by h_1 and {\sigma }_1 in this case.

Subcase 3.1: r^{\prime }=0

In this case, 1^{{\sigma }_1}=\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right), h_1:(x_1,x_2)\mapsto (x_1,x_2)\left(\begin{array}{c}0\\ \beta \end{array}\right), where \beta \ne 0. Suppose h_2:(x_1,x_2)\mapsto (x_1,x_2)\left(\begin{array}{c}0\\ 1\end{array}\right). We take F=\left(\begin{array}{cc}{\beta }^{-1}& 0\\ 0& {\beta }^{-1}\end{array}\right)\in \mathrm{A}\mathrm{u}\mathrm{t}(H,+). Then h_2=F{h}_1 and b^{{\sigma }_1}F=F{b}^{{\sigma }_1} for b\in H/\mathrm{S}\mathrm{o}\mathrm{c}(H). By Proposition 2.1(1), the brace determined by h_1 and {\sigma }_1 is isomorphic to the brace determined by h_2 and {\sigma }_1. Thus we only get a brace determined by h_2 and {\sigma }_1 in this case.

Now the \circ is defined as follows. \mathrm{\forall }(x_1,x_2),(y_1,y_2)\in H,

\begin{array}{rl}(x_1,x_2)\circ (y_1,y_2)& =(x_1,x_2{)}^{(y_1,y_2{)}^{h_2{\sigma }_1}}+(y_1,y_2)\\ & =(x_1+y_1+x_2{y}_2,x_2+y_2).\end{array}

Then we get brace (6) in Theorem 3.2. We also obtain (H,\circ )\cong {\mathbb{Z}}_p\times {\mathbb{Z}}_{p^{n-1}} by a similar proof of Subcase 1.1.

Subcase 3.2: r^{\prime } is a quadratic residue modulo p

Let r^{\prime }=a^2 and a\in {\mathbb{Z}}_p. In this case, 1^{{\sigma }_1}=\left(\begin{array}{cc}1& p^{n-2}{a}^2\\ 1& 1\end{array}\right) and h_1:(x_1,x_2)\mapsto (x_1,x_2)\left(\begin{array}{c}0\\ \beta \end{array}\right), where \beta \ne 0. Suppose 1^{{\sigma }_2}=\left(\begin{array}{cc}1& p^{n-2}\\ 1& 1\end{array}\right) and h_2:(x_1,x_2)\mapsto (x_1,x_2)\left(\begin{array}{c}0\\ 1\end{array}\right). We take F=\left(\begin{array}{cc}{a}^2\beta & p^{n-2}a(\genfrac{}{}{0ex}{}{a}{2})\beta \\ 0& a\beta \end{array}\right)\in \mathrm{A}\mathrm{u}\mathrm{t}(H,+). Then (x_1,x_2{)}^{h_1{\sigma }_1}F=F((x_1,x_2)F{)}^{h_2{\sigma }_1} for (x_1,x_2)\in H. By Theorem 2.1, the brace determined by h_1 and {\sigma }_1 is isomorphic to the brace determined by h_2 and {\sigma }_2. Thus we only get a brace determined by h_2 and {\sigma }_2 in this case.

Now the \circ is defined as follows. \mathrm{\forall }(x_1,x_2),(y_1,y_2)\in H,

\begin{array}{rl}(x_1,x_2)\circ (y_1,y_2)& =(x_1,x_2{)}^{(y_1,y_2{)}^{h_2{\sigma }_2}}+(y_1,y_2)\\ & =(x_1+y_1+x_2{y}_2,x_2+y_2+p^{n-2}{x}_2(\genfrac{}{}{0ex}{}{y_2}{2})+p^{n-2}{x}_1{y}_2).\end{array}

Then we get brace (7) in Theorem 3.2. We also obtain (H,\circ )\cong {\mathrm{M}}_p(n-1,1) by a similar proof of Subcase 1.2.

Subcase 3.3: r^{\prime } is a quadratic non-residue modulo p

Let \epsilon be a fixed quadratic non-residue modulo p. Then r^{\prime }{\epsilon }^{-1} is a quadratic residue modulo p. Assume r^{\prime }{\epsilon }^{-1}=a^2 and a\in {\mathbb{Z}}_p. In this case, 1^{{\sigma }_1}=\left(\begin{array}{cc}1& p^{n-2}\epsilon a^2\\ 1& 1\end{array}\right) and h_1:(x_1,x_2)\mapsto (x_1,x_2)\left(\begin{array}{c}0\\ \beta \end{array}\right), where \beta \ne 0. Suppose 1^{{\sigma }_2}=\left(\begin{array}{cc}1& p^{n-2}\epsilon \\ 1& 1\end{array}\right) and h_2:(x_1,x_2)\mapsto (x_1,x_2)\left(\begin{array}{c}0\\ 1\end{array}\right). We take F= \left(\begin{array}{cc}{a}^2\beta & p^{n-2}\epsilon a(\genfrac{}{}{0ex}{}{a}{2})\beta \\ 0& a\beta \end{array}\right)\in \mathrm{A}\mathrm{u}\mathrm{t}(H,+). Then (x_1,x_2{)}^{h_1{\sigma }_1}F=F((x_1,x_2)F{)}^{h_2{\sigma }_1} for (x_1,x_2)\in H. By Theorem 2.1, the brace determined by h_1 and {\sigma }_1 is isomorphic to the brace determined by h_2 and {\sigma }_2. Thus we only get a brace determined by h_2 and {\sigma }_2 in this case.

Now the \circ is defined as follows. \mathrm{\forall }(x_1,x_2),(y_1,y_2)\in H,

\begin{array}{rl}(x_1,x_2)\circ (y_1,y_2)& =(x_1,x_2{)}^{(y_1,y_2{)}^{h_2{\sigma }_2}}+(y_1,y_2)\\ & =(x_1+y_1+x_2{y}_2,x_2+y_2+p^{n-2}\epsilon x_2(\genfrac{}{}{0ex}{}{y_2}{2})+p^{n-2}\epsilon x_1{y}_2).\end{array}

Then we get brace (8) in Theorem 3.2. We also obtain (H,\circ )\cong {\mathrm{M}}_p(n-1,1) by a similar proof of Subcase 1.2.

In the end, we prove the eight braces in Theorem 3.2 are pairwise non-isomorphic braces. Since (H,\circ ) is abelian, one of braces (1), (4) and (6) is non-isomorphic to one of braces (2),(3),(5),(7),(8). Thus we only need to prove (1), (4), (6) are pairwise non-isomorphic braces and (2), (3), (5), (7), (8) are pairwise non-isomorphic braces.

We only prove (7) is not isomorphic to (8) as follows. The rest proof is similar to the following proof, so we omit it. Assume that (7) determined by h_7 and {\sigma }_7 is isomorphic to (8) determined by h_8 and {\sigma }_8. By Theorem 2.1, there is F\in \mathrm{A}\mathrm{u}\mathrm{t}(H,+) such that

(0,1{)}^{h_7{\sigma }_7}F=F((0,1)F{)}^{h_8{\sigma }_8}.

By Lemma 3.1 and Remark 3.1, we may let F=\left(\begin{array}{cc}{f}_{11}& p^{n-2}{f}_{12}\\ f_{21}& f_{22}\end{array}\right), where f_{11},f_{12},f_{21}\in {\mathbb{Z}}_p,f_{22}\in {\mathbb{Z}}_{p^{n-1}} and f_{11}{f}_{22}\not\equiv 0(\mathrm{mod}p).

By calculation we get that

\begin{array}{rl}(0,1{)}^{h_7{\sigma }_7}F& =\left(\begin{array}{cc}{f}_{11}& p^{n-2}{f}_{12}+p^{n-2}{f}_{22}\\ f_{11}+f_{21}& f_{22}+p^{n-2}{f}_{12}\end{array}\right),\\ F((0,1)F{)}^{h_8{\sigma }_8}& =\left(\begin{array}{cc}{f}_{11}& p^{n-2}\epsilon f_{11}{f}_{22}+p^{n-2}{f}_{12}\\ f_{21}+f_{22}^2& f_{22}+p^{n-2}\epsilon f_{21}{f}_{22}+p^{n-2}\epsilon f_{22}\left(\begin{array}{r}{f}_{22}\\ 2\end{array}\right)\end{array}\right).\end{array}

Since (0,1{)}^{h_7{\sigma }_7}F=F((0,1)F{)}^{h_8{\sigma }_8}, f_{11}\equiv f_{22}^2(\mathrm{m}\mathrm{o}\mathrm{d}p) and \epsilon f_{11}\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}p). Then \epsilon f_{22}^2\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}p). This is in contradiction with that \epsilon is a quadratic non-residue modulo p.□

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