模群代数的单幂自同构与Lazard李代数的同调
Unipotent automorphisms of modular group algebras and the homology of the Lazard Lie algebra
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中文总结 AI 辅助
本文证明指数为p且类至多p-2的有限p-群的模群代数单幂自同构与群自同构在关系模上作用一致，并给出U(H)的Lazard李代数刻画，计算了类2群的U(H)大小。
中文摘要 AI 辅助
设$p$为奇素数，$H$为有限$p$-群。Hertweck和Soriano通过关系模$V$的子空间参数化$H$的中心Frattini扩张，将扩张的同构类与$\mathrm{Out}(H)$的轨道相匹配，并将关联的小群环的同构类与$\mathrm{Out}(\mathbb{F}_p H)$的轨道相匹配。两者之间的差异由$\mathbb{F}_p H$的规范化单幂自同构诱导的位移群$U(H)\leq GL(V)$度量；这是特征2中模同构问题反例背后的机制。我们证明，对于指数为$p$且类至多为$p-2$的$H$，每个这样的自同构在$V$上的作用恰好与$H$的自同构相同，且$U(H)=\exp\Omega\bigl(\mathrm{Der}_{>0}(\mathfrak{l}(H))\bigr)$，其中$\mathfrak{l}(H)$是$H$的Lazard李代数，$\mathrm{Der}_{>0}$是其提升下中心滤过的导子，$\Omega$是导子在$H_2(\mathfrak{l}(H))$上的自然作用。因此，$U(H)=1$当且仅当每个正次数的导子在$H_2(\mathfrak{l}(H))$上平凡作用。我们计算了所有类为2且指数为$p$的群的$U(H)$，对于Heisenberg群$\mathrm{Heis}(p^n)$得到$p^{2n(n-1)}$。
英文摘要
Let $p$ be an odd prime and $H$ a finite $p$-group. Hertweck and Soriano parametrize the central Frattini extensions of $H$ by subspaces of a relation module $V$, matching isomorphism classes of extensions with orbits of $\mathrm{Out}(H)$ and isomorphism classes of the associated small group rings with orbits of $\mathrm{Out}(\mathbb{F}_p H)$. The discrepancy between the two is measured by the group $U(H)\leq GL(V)$ of displacements induced by the normalized unipotent automorphisms of $\mathbb{F}_p H$; it is the mechanism behind the counterexamples to the modular isomorphism problem in characteristic $2$. We prove that for $H$ of exponent $p$ and class at most $p-2$ every such automorphism acts on $V$ exactly as an automorphism of $H$ does, and that $U(H)=\expΩ\bigl(\mathrm{Der}_{>0}(\mathfrak{l}(H))\bigr)$, where $\mathfrak{l}(H)$ is the Lazard Lie algebra of $H$, $\mathrm{Der}_{>0}$ its derivations raising the lower central filtration, and $Ω$ the natural action of derivations on $H_2(\mathfrak{l}(H))$. Hence $U(H)=1$ precisely when every derivation of positive degree acts trivially on $H_2(\mathfrak{l}(H))$. We compute $U(H)$ for all groups of class $2$ and exponent $p$, obtaining $p^{2n(n-1)}$ for the Heisenberg groups $\mathrm{Heis}(p^n)$.
