semidirect product

《近世代数》 · 第 59 页

设 $G$ 是一个群，$N\trianglelefteq G$，$H<G$，如果 $G=NH$，且 $N\cap H=\{e\}$，那么称 $G$ 可分解成它的正规子群 $N$ 与子群 $H$ 的半直积，记做 $G=N\rtimes H$。

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《Abstract Algebra》 · 第 176–177 页 · Section 5.5「Semidirect Products」，Theorem 10 及其后 Definition

Theorem 10. Let $H$ and $K$ be groups and let $\varphi$ be a homomorphism from $K$ into $\operatorname{Aut}(H)$. Let $\cdot$ denote the (left) action of $K$ on $H$ determined by $\varphi$. Let $G$ be the set of ordered pairs $(h,k)$ with $h\in H$ and $k\in K$ and define the following multiplication on $G$: $(h_1,k_1)(h_2,k_2)=(h_1k_1\cdot h_2,k_1k_2)$. (1) This multiplication makes $G$ into a group of order $|G|=|H||K|$. (2) The sets $\{(h,1)\mid h\in H\}$ and $\{(1,k)\mid k\in K\}$ are subgroups of $G$ and the maps $h\mapsto(h,1)$ for $h\in H$ and $k\mapsto(1,k)$ for $k\in K$ are isomorphisms of these subgroups with the groups $H$ and $K$ respectively: $H\cong\{(h,1)\mid h\in H\}$ and $K\cong\{(1,k)\mid k\in K\}$. Identifying $H$ and $K$ with their isomorphic copies in $G$ described in (2) we have (3) $H\trianglelefteq G$ (4) $H\cap K=1$ (5) for all $h\in H$ and $k\in K$, $khk^{-1}=k\cdot h=\varphi(k)(h)$. […] Definition. Let $H$ and $K$ be groups and let $\varphi$ be a homomorphism from $K$ into $\operatorname{Aut}(H)$. The group described in Theorem 10 is called the semidirect product of $H$ and $K$ with respect to $\varphi$ and will be denoted by $H\rtimes_\varphi K$ (when there is no danger of confusion we shall simply write $H\rtimes K$).