Left braces of size $p^2q^2$

TL;DR

The paper analyzes left braces of size $mn$ with $\gcd(m,n)=1$ under the condition that every solvable group of order $mn$ has a normal subgroup of order $m$, proving that each such brace is a semidirect product $B_1 \rtimes_{\tau} B_2$ of a brace of size $m$ by a brace of size $n$. It then provides a systematic method to classify braces of size $mn$ from the known classifications of braces of sizes $m$ and $n$, and applies this framework to describe all braces of size $p^2q^2$ for odd primes $p<q$ satisfying specific arithmetic constraints (notably including the case where $p$ is an odd Germain prime and $q=2p+1$). The approach hinges on analyzing the $ au$-actions of $(B_2,\cdot)$ via automorphisms of $(B_1,+,\cdot)$ and on the induced isomorphism criteria for semidirect products, yielding explicit lists of brace structures across all combinations of additive types for $B_1$ and $B_2$. Collectively, the results extend previous classifications and provide concrete, computable families of braces of size $p^2q^2$ under the stated hypotheses, with detailed multiplicative laws and symmetry considerations.

Abstract

We consider relatively prime integer numbers $m$ and $n$ such that each solvable group of order $mn$ has a normal subgroup of order $m$. We prove that each brace of size $mn$ is a semidirect product of a brace of size $m$ and a brace of size $n$. We further give a method to classify braces of size $mn$ from the classification of braces of sizes $m$ and $n$. We apply this result to determine all braces of size $p^2q^2$, for $p$ and $q$ odd primes satisfying some conditions which hold in particular for $p$ a Germain prime and $q=2p+1$.

Theorems & Definitions (10)

• Theorem 1
• proof
• Proposition 2
• proof
• Lemma 3
• Lemma 4
• Theorem 5
• Theorem 6
• Theorem 7
• Theorem 8