---- The work in [[internal semi-direct product]] motivates the following definition. > [!definition] Definition. ([[external semi-direct product]]) Let $H$ and $K$ be abstract [[group]]s, and let $\phi:K \to \text{Aut}(H), \ \ k \mapsto \phi_{k}$ be a [[group homomorphism|homomorphism]]. The **(external) semi-direct product** $H \rtimes_{\phi} K$ is defined as the [[cartesian product]] $H \times K$, endowed with the [[binary operation|group law]] given by $(h_{1}, k_{1}) \cdot (h_{2},k_{2})=\big(h_{1}\phi_{k_{1}}(h_{2}), k_{1}k_{2}\big).$ > [!justification] > We need to show this is a [[group]]. The identity element is $(e_{H},e_{G})$. [[associative|Associativity]] holds because $\begin{align} (h_{1},k_{1}) \cdot \big( (h_{2},k_{2}) \cdot (h_{3},k_{3}) \big) = & (h_{1},k_{1}) \cdot (h_{2}\phi_{k_{2}}(h_{3}), k_{2}k_{3}) \\ = & (h_{1}\phi_{k_{1}}(h_{2}\phi_{k_{2}}(h_{3})), k_{1}(k_{2}k_{3}) )\\ = & (h_{1}\phi_{k_{1}}(h_{2})\phi_{k_{1}}(\phi_{k_{2}}(h_{3})), k_{1}k_{2}k_{3}) \\ = & \big(h_{1}\phi_{k_{1}}(h_{2})\phi_{k_{1}k_{2}}(h_{3}), (k_{1}k_{2})k_{3}\big) \\= & (h_{1}\phi_{k_{1}}(h_{2}),k_{1}k_{2})\cdot (h_{3},k_{3})\\ = & \big( (h_{1},k_{1}) \cdot (h_{2},k_{2}) \big)\cdot(h_{3},k_{3}). \end{align}$ The inverse of the element $(h,k)$ is $(\phi_{k}^{-1}(h^{-1}), k^{-1})$ , since \ $(h,k) \cdot (h,k)^{-1}:=(h,k) \cdot (h',k')=(h_{1}\phi_{k}(h'), kk')=(e_{H},e_{G})$ \ $\iff$ $h_{}\phi_{k}h'=e_{H}$ and $kk'=e_{G}$ \ $\iff$ $\phi_{k}(h')=h_{}^{-1}$ $k'=k^{-1}$ \ $\iff$ $h'=\phi_{k}^{-1}(h^{-1})$ and $k'=k^{-1}$ > [!note] Remark. > ![[external semidirect product induces internal and vice versa#^ec9bab]] > ![[external semidirect product induces internal and vice versa#^2b7719]] > [!justification] Motivation. How would we go about defining a product of two [[group]]s $H$ and $K$ with completely different group laws, and in a manner that respects the notion of an [[internal direct product of subgroups|internal product]]? Well, we would want the product to equal $G$, where >1. $G$ is a [[group]]; >2. $H_{G} \triangleleft G$ ($H_{G}$ denotes a [[subgroup]] of $G$ [[group isomorphism|isomorphic]] to $H$); >3. $H_{G} \cap K_{G}=\{ e \}$ ($K_{G}$ denotes a [[subgroup]] of $G$ [[group isomorphism|isomorphic]] to $K$); >4. $H_{G}K_{G}=G$. \ Proceed as follows. As a set, define $G:=H \times K$. The [[subgroup]] $H_{G}:={H \times \{e_{K}\}}$ is [[group isomorphism|isomorphic]] to $H_{G}$; likewise the [[subgroup]] $K_{G}:=\{ e_H \} \times K$ is [[group isomorphism|isomorphic]] to $H_{K}$. Under these definitions, $H_{G} \cap K_{G} =\{ e_{H} , e_{K}\}$, so $(3)$ is satisfied. \ Next, consider$H_{G}K_{G}=\{ (h,e_{K})\cdot (e_{H}, k): h \in H, k \in K \}.$ To satisfy $(4)$, we need $H_{G}K_{G}=G$. This can be accomplished by defining a [[binary operation|group law]] of $G$ having the property $(h,e_{K})\cdot (e_{H}, k)=(h,k)$. Then, $H_{G}K_{G}=\{ (h,k): h \in H, k \in K\}=G.$ If such a law exists, $(4)$ holds. \ So, how *should* we generally define $(h_{1},k_{1})\cdot (h_{2},k_{2})$ to include that special case? Begin by taking $(h_{1},k_{1})\cdot(h_{2},k_{2}):=\big((h_{1},e)\cdot(e,k_{1})\big) \cdot \big((h_{2},e), (e,k_{2})\big),$ but then since we want to make sure $H_{G}$ is a [[normal subgroup]] of $G$, we mimic the [[internal semi-direct product]] multiplication definition $(h_{1},e)\textcolor{LimeGreen}{(e,k_{1})(h_{2},e)(e,k_{1}^{-1})}(e,k_{1})(e,k_{2})$ and then enforce that $(e,k_{1})(h_{2},e)(e,k_{1}^{-1})$ has the form $\textcolor{LimeGreen}{(h,e)}$ (since we want $H_{G} \triangleleft G$). So, what is this [[conjugate|conjugation]] going to look like? Consider the [[inner automorphism]] [[conjugate|conjugation]] map; we showed in [[internal semi-direct product]] that that map $\phi:K \to \text{Aut}(K)$ given by $k \mapsto \phi_{k}$ is a [[group homomorphism|homomorphism]]. Thus, $\textcolor{LimeGreen}{(h,e)}=(\phi_{k_{1}}(h_{2}), e)$. So we obtain $(h_{1},e)\textcolor{LimeGreen}{(\phi_{k_{1}}(h_{2}),e)}(e,k_{1})(e,k_{2})$ which equals $(h_{1}\phi_{k_{1}}(h_{2}), k_{1}k_{2}).$ ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```