Exercise 6.4.8

Proof. By definition of an action by group homomorphisms, there exists a group homomorphism $\phi :G\to \mathrm{Aut}(H)$ such that for all $(g,h)\in G\times H$,

so $h\mapsto g\cdot h$ is a group automorphism of $H$ for all $g\in G$.

(a)

$I$ If $h,h^{\prime }\in H,g,g^{\prime }\in G$, then $g\cdot h^{\prime }\in H$, thus $(h(g\cdot h^{\prime }),g{g}^{\prime })\in H\times G$ , so this law defines a binary operation on $H\times G$.

$A$ Let $(h,g),(h^{\prime },g^{\prime }),(h^{″},g^{″})\in H\times G$. Then

$$\begin{array}{llll}\hfill ((h,g).(h^{\prime },g^{\prime })).(h^{″},g^{″})& =(h(g\cdot h^{\prime }),g{g}^{\prime }).(h^{″},g^{″})& \hfill & \\ \hfill & =(h(g\cdot h^{\prime })((g{g}^{\prime })\cdot h^{″}),g{g}^{\prime }{g}^{″})& \hfill & \\ \hfill & =(h(g\cdot h^{\prime })(g\cdot (g^{\prime }\cdot h^{″})),g{g}^{\prime }{g}^{″}).& \hfill & \end{array}$$

The last equality is true because $G$ acts on $H$.

$$\begin{array}{llll}\hfill (h,g).((h^{\prime },g^{\prime }).(h^{″},g^{″}))& =(h,g)((h^{\prime }(g^{\prime }\cdot h^{″}),g^{\prime }{g}^{″})& \hfill & \\ \hfill & =(h(g\cdot (h^{\prime }(g^{\prime }\cdot h^{″}))),g{g}^{\prime }{g}^{″})& \hfill & \\ \hfill & =(h(g\cdot h^{\prime })(g\cdot (g^{\prime }\cdot h^{″})),g{g}^{\prime }{g}^{″}).& \hfill & \end{array}$$

The last equality is true because $G$ acts on $H$ by group homomorphism.

Therefore $((h,g).(h^{\prime },g^{\prime })).(h^{″},g^{″})=(h,g).((h^{\prime },g^{\prime }).(h^{″},g^{″}))$: the law is associative.

$N$ Write $e_H,e_G$ the identity of $H$ and the identity of $G$.

$$\begin{array}{lll}\hfill (f,g).(e_H,e_G)=(f(g\cdot e_H),g{e}_G)=(f{e}_H,g{e}_G)=(f,g),& & \hfill \\ \hfill (e_H,e_G).(f,g)=(e_H(e_G\cdot f),e_{G}g)=(e_{H}f,e_{G}g)=(f,g).& & \hfill \end{array}$$

So $(e_H,e_G)$ is the identity of $H⋊G$, which we will write now $(1,1)$.

$S$

Analysis: if $(h^{\prime },g^{\prime })$ is the inverse of $(h,g)$, then $(h(g\cdot h^{\prime }),g{g}^{\prime })=(1,1)$. Thus $g^{\prime }=g^{-1}$, and $g\cdot h^{\prime }=h^{-1}$, therefore $h^{\prime }=g^{-1}\cdot (h^{-1})$.

Synthesis: we show that $(g^{-1}\cdot (h^{-1}),g^{-1})$ is the inverse of $(h,g)$ :

$$\begin{array}{llll}\hfill (h,g).(g^{-1}\cdot (h^{-1}),g^{-1})& =(h(g\cdot (g^{-1}\cdot (h^{-1}))),g{g}^{-1})& \hfill & \\ \hfill & =(h(g{g}^{-1}\cdot (h^{-1})),1)& \hfill & \\ \hfill & =(h{h}^{-1},1)=(1,1)& \hfill & \\ \hfill (g^{-1}\cdot (h^{-1}),g^{-1})(h,g)& =((g^{-1}\cdot (h^{-1}))(g^{-1}\cdot h),g^{-1}g)& \hfill & \\ \hfill & =(g^{-1}\cdot (h^{-1}h),1)& \hfill & \\ \hfill & =(g^{-1}\cdot 1,1)=(1,1)& \hfill & \end{array}$$

Every element of $H⋊G$ has an inverse.

$H⋊G$ is a group.

(b)

Let $\psi :\{\begin{array}{ccc}\hfill H⋊G\hfill & \hfill \to \hfill & \hfill G\hfill \\ \hfill (h,g)\hfill & \hfill \mapsto \hfill & \hfill g\hfill \end{array}$.

$\psi ((h,g).(h^{\prime },g^{\prime }))=\psi (h(g\cdot h^{\prime }),g{g}^{\prime })=g{g}^{\prime }=\psi (h,g)\psi (h^{\prime },g^{\prime })$. $\psi$ is so a group homomorphism.

As every $g$ in $G$ is the image of $(1,g)\in H⋊G$ by $\psi$, $\psi$ is surjective.

$\ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )=\{(h,g)\in H⋊G|g=e\}=H\times \{e\}$.

(c)

Let $\chi :\{\begin{array}{ccc}\hfill H\hfill & \hfill \to \hfill & \hfill H\times \{e\}\hfill \\ \hfill h\hfill & \hfill \mapsto \hfill & \hfill (h,e)\hfill \end{array}$.

$\chi (h)\chi (h^{\prime })=(h,e)(h^{\prime },e)=(h(e\cdot h^{\prime }),e)=(h{h}^{\prime },e)=\chi (h{h}^{\prime })$ : $\chi$ is a group homorphisme from $H$ on the subgroup $H\times \{e\}$ of $H⋊G$.

$\chi (h)=(e,e)⟺h=e$, thus $\chi$ is injective, and surjective since every element of $H\times \{e\}$ is of the form $(h,e)=\chi (h)$ : $H\simeq H\times \{e\}$.

Therefore the sequence $\{e\}\to H\to H⋊G\to G\to \{e\}$ is a short exact sequence, so $H⋊G$ is an extension of $H$ by $G$.