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Classification of right nilpotent $\mathbb{F}_p$-braces of cardinality $p^5$

Snehashis Mukherjee

Abstract

In this article the right nilpotent $\mathbb{F}_p$-braces of cardinality $p^5$ has been classified. We use the connection between nilpotent $\mathbb{F}_p$-braces of cardinality $p^5$ and nilpotent pre-Lie algebras of the same order, building on the known relationship between pre-Lie algebras and braces. Leveraging insights from the classification of nilpotent pre-Lie algebras over $\mathbb{F}_p$, we aim to provide a comprehensive classification of right nilpotent $\mathbb{F}_p$-braces of cardinality $p^5$.

Classification of right nilpotent $\mathbb{F}_p$-braces of cardinality $p^5$

Abstract

In this article the right nilpotent -braces of cardinality has been classified. We use the connection between nilpotent -braces of cardinality and nilpotent pre-Lie algebras of the same order, building on the known relationship between pre-Lie algebras and braces. Leveraging insights from the classification of nilpotent pre-Lie algebras over , we aim to provide a comprehensive classification of right nilpotent -braces of cardinality .
Paper Structure (19 sections, 27 theorems, 96 equations)

This paper contains 19 sections, 27 theorems, 96 equations.

Table of Contents

  1. Introduction
  2. Preliminaries
  3. Nilpotency Index
  4. Nilpotency Index
  5. pre-Lie algebra $A$ with $(A,+)\cong C_p^5$ and generated by one element as a pre-Lie algebra
  6. $A^{[7]} \neq 0$ and $A^{[8]}=0$
  7. $A^{[7]}=0$ and $A^{[6]}\neq 0$.
  8. $A^{[6]}=0$ and $A^{[5]}\neq 0$.
  9. $A^{[5]}=0$ and $A^{[4]}\neq 0$.
  10. pre-Lie algebra $A$ with $(A,+)\cong C_p^5$ and generated by two elements as a pre-Lie algebra
  11. $A^{[2]} \neq 0$ and $A^{[3]}=0$.
  12. $A^{[3]} \neq 0$ and $A^{[4]}=0$.
  13. $A^{[4]} \neq 0$ and $A^{[5]}=0$.
  14. $A^{[4]}=A^{[5]}\neq 0$ and $A^{[6]}=0$.
  15. pre-Lie algebra $A$ with $(A,+)\cong C_p^5$ and generated by three elements as a pre-Lie ring
  16. ...and 4 more sections

Key Result

Theorem 2.0.1

so5 Let $(A,+,\circ)$ be a left brace. If $m, n$ are natural numbers and $A^n = A^{(m)} = 0$, then $A^{[s]} = 0$ for some natural number $s$.

Theorems & Definitions (45)

  • Theorem 2.0.1
  • Theorem 2.0.2
  • Example 3.1
  • Theorem 3.1.1
  • proof
  • Theorem 4.0.1
  • proof
  • Lemma 4.1.1
  • proof
  • Lemma 4.1.2
  • ...and 35 more