Exercise 6.4.10

Proof.

(a)

As $p\ge 3$, there exist in ${𝔽}_p$ an element $2$ with $2\ne 0,2\ne 1$, so $(0,2)\in {𝔽}_p⋊{𝔽}_p^{\text{∗}}$, and also $(1,1)\in {𝔽}_p⋊{𝔽}_p^{\text{∗}}$. $$\begin{array}{lll}\hfill (0,2).(1,1)=(0+2\times 1,2\times 1)=(2,2)& & \hfill \\ \hfill (1,1).(0,2)=(1+1\times 0,1\times 2)=(1,2)& & \hfill \\ \hfill & & \hfill \end{array}$$

Since $2\ne 1$, $(0,2).(1,1)\ne (1,1).(0,2)$. So if $p\ge 3$, then ${𝔽}_p⋊F_p^{\text{∗}}$ is not Abelian.

(b)

By definition of the product in ${𝔽}_p\times {𝔽}_p^{\text{∗}}$, $(a,b)(c,d)=(\mathit{ac},\mathit{bd})=(\mathit{ca},\mathit{db})=(c,d)(a,b)$: ${𝔽}_p\times {𝔽}_p^{\text{∗}}$ is Abelian.

(c)

The sequence $$\{0\}\to {𝔽}_p\to {𝔽}_p\times {𝔽}_p^{\text{∗}}\to {𝔽}_p^{\text{∗}}\to \{1\}$$

is a short exact sequence (the first arrow is the injective map $x\mapsto (x,1)$, and the second one is the surjective map $(x,y)\mapsto y$). Actually, a direct product is a special case of semidirect product, where $\phi :G\to \mathrm{Aut}(h)$ is the trivial action defined by $\varphi (g)=1_H$ for all $g\in G$, so $\phi (g)(h)=g\cdot h=h$ for all $h\in H$. By part (a) and (b), these two extensions are not isomorphic.