Past exam of the options II course of the University of Oxford / 2010 / B / B2

With $N◃G$, the internal semidirect product conditions are $G=NH$ and $N\cap H=\{1\}$. Every element then has a unique expression $nh$. The extra data needed to reconstruct the group are the conjugation actionOn the set $N\times H$, the external semidirect product $N{⋊}_{\varphi }H$ has multiplicationThe map $(n,h)\mapsto nh$ identifies it with $G$. A direct product results when this action is trivial.

For example, $S_3=A_3⋊⟨(12)⟩\cong C_3⋊C_2$. The two subgroups intersect trivially and their orders multiply to six, so they generate $S_3$. Conjugation by the transposition inverts the three-cycle. This action is nontrivial, and $S_3$ is nonabelian whereas $C_3\times C_2$ is abelian, so this is not their direct product.

Use composition from right to left. Applying $(C,d)$ and then $(A,b)$ givessoThe product remains an affine transformation because $AC$ is invertible. Composition of functions is associative. The identity is $(I,0)$, and solving $w=Av+b$ gives the inverseThus these transformations form the affine group. The representation by $(A,b)$ is unique: evaluating at zero determines $b$, and then evaluating on the standard basis determines $A$.

Embed ${\mathrm{GL}}_n(F)$ by $A\mapsto (A,0)$. The multiplication and inverse formulas show that this is an injective group homomorphism, so its image $H$ is a subgroup. The translations form $N=\{(I,c):c\in F^n\}$, with $(I,c)(I,d)=(I,c+d)$ and inverse $(I,-c)$. Hence $N$ is isomorphic to the additive group of $F^n$.

For an arbitrary affine transformation,Conjugation therefore preserves $N$, proving that $N$ is normal in $G$.

Every affine transformation factors asso $G=NH$. A translation fixing zero has zero translation vector, so $H\cap N=\{(I,0)\}$. Together with normality from part (c), these are the internal semidirect product conditions. ThusThe reconstruction data are the additive group $F^n$, the linear group ${\mathrm{GL}}_n(F)$, and its standard action $A:c\mapsto Ac$ on translations. In pair notation $(b,A)$, the product is $(b,A)(d,C)=(b+Ad,AC)$, equivalent to the formula in part (b).

There are $p^n$ possible translation vectors. An invertible matrix is specified by an ordered basis of ${\mathbb{F}}_p^n$. Its first column can be any nonzero vector, giving $p^n-1$ choices. After choosing $i$ independent columns, their span has $p^i$ elements, so the next column has $p^n-p^i$ choices. Therefore