Bicrossed descent theory of exact factorizations and the number of types of groups of finite order

Details Bicrossed descent theory of exact factorizations and the number of types of groups of finite order Bicrossed descent theory of exact factorizations and the number of types of groups of finite order

2012-04-09

arxiv.org/abs/1204.1805v1

Mathematics

Group Theory

16 pages, submitted

Scientific paper

Let $A \leq G$ be a subgroup of a group $G$. A factorization $A$-form of $G$ is a subgroup $H$ of $G$ such that $G = A H$ and $A \cap H = {1}$. Let ${\mathcal F} (A, G)$ be the category of all factorization $A$-forms of $G$ and ${\mathcal F}^{sk} (A, G)$ its skeleton. The \emph{bicrossed descent} problem asks for the description and classification of all factorization $A$-forms of $G$. We shall give the full answer to this problem in three steps. Let $H$ be a given factorization $A$-form of $G$ and $(\triangleright, \triangleleft)$ the canonical left/right actions associated to the factorization $G = A H$. In the first step $H$ is deformed to a new $A$-form of $G$, denoted by $H_r$, using a certain map $r: H \to A$ called a descent map of the matched pair $(A, H, \triangleright, \triangleleft)$. Then the description of all forms is given: ${\mathbb H}$ is an $A$-form of $G$ if and only if ${\mathbb H}$ is isomorphic to $H_{r}$, for some descent map $r: H \to A$. Finally, the classification of forms proves that there exists a bijection between ${\mathcal F}^{sk} (A, G)$ and a combinatorial object ${\mathcal D} (H, A | (\triangleright, \triangleleft))$. Let $S_{n}$ be the symmetric group and $C_n$ the cyclic group of order $n$. By applying the bicrossed descent theory for the factorization $S_n = S_{n-1} C_n$ we obtain the following: $(1)$ any group $H$ of order $n$ is isomorphic to $(C_n)_r$, the $r$-deformation of the cyclic group $C_n$ for some descent map $r: C_n \to S_{n-1}$ of the canonical matched pair $(S_{n-1}, C_n, \triangleright, \triangleleft)$ and $(2)$ the number of types of isomorphisms of all groups of order $n$ is equal to $| {\mathcal D} (C_n, S_{n-1} | (\triangleright, \triangleleft)) |$.

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