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The Left-Nilpotent Residual and Its Supplements in Finite Skew Braces

Abstract

Let X be a finite skew brace, and let L∞(X) be the final term of its left series. We introduce the left-nilpotent residual RL(X) and prove that $ \RL(X)=\Id_X\bigl(L_\infty(X)\bigr). $ Thus X/RL(X) is the largest left-nilpotent quotient of X. For skew braces of nilpotent type, we relate RL(X) to γ∞(X,·) and obtain conditions ensuring that L∞(X) is an ideal, including the case where |X| is cube-free. Our main results concern supplements to RL(X). We prove existence results using coprime action and Sylow and Hall theory, and give an example of order $18$ whose residual has no proper supplement. Finally, iterating RL yields the largest perfect subskew brace of X, and X is KSV-solvable if and only if this subskew brace is zero.

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BibTeX
@misc{ercan2026leftnilpotent,
  title = {{The Left-Nilpotent Residual and Its Supplements in Finite Skew Braces}},
  author = {Gülin Ercan and İsmail Ş. Güloğlu},
  howpublished = {arXiv},
  year = {2026},
  month = sep,
  doi = {10.48550/arxiv.2609.06994},
  eprint = {2609.06994},
  archivePrefix = {arXiv},
  url = {https://arxiv.org/abs/2609.06994},
}
RIS
TY  - UNPB
TI  - The Left-Nilpotent Residual and Its Supplements in Finite Skew Braces
AU  - Gülin Ercan
AU  - İsmail Ş. Güloğlu
PY  - 2026
DA  - 2026/09/07
PB  - arXiv
DO  - 10.48550/arxiv.2609.06994
UR  - https://arxiv.org/abs/2609.06994
AB  - Let X be a finite skew brace, and let L_∞(X) be the final term of its left series. We introduce the left-nilpotent residual RL(X) and prove that $ \RL(X)=\Id_X\bigl(L_\infty(X)\bigr). $ Thus X/RL(X) is the largest left-nilpotent quotient of X. For skew braces of nilpotent type, we relate RL(X) to γ_∞(X,·) and obtain conditions ensuring that L_∞(X) is an ideal, including the case where |X| is cube-free. Our main results concern supplements to RL(X). We prove existence results using coprime action and Sylow and Hall theory, and give an example of order $18$ whose residual has no proper supplement. Finally, iterating RL yields the largest perfect subskew brace of X, and X is KSV-solvable if and only if this subskew brace is zero.
ER  -