Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 / B / B2

Given $N⊴G$ and $H\le G$, the conditions are $\boxed{G=NH,N\cap H=\{1\}}$. Every element then has a unique expression $nh$. The extra datum is the action $\alpha :H\to \mathrm{Aut}(N)$ given by conjugation, ${\alpha }_h(n)=hn{h}^{-1}$. On pairs $(n,h)$ the reconstructed multiplication isFor example $S_3=⟨(123)⟩⋊⟨(12)⟩$. The order-three subgroup is normal, the intersection is trivial, and their sizes multiply to six. Conjugation by $(12)$ inverts $(123)$, so the action is nontrivial. Consequently this is not the direct product of these subgroups: $C_3\times C_2$ is Abelian, whereas $S_3$ is not.

Compose the second map first. Direct substitution givesThe result is affine because the product matrix is invertible. The identity is $(I,0)$ and the inverse is $(A^{-1},-A^{-1}b)$. Associativity follows from composition of maps, or by expanding this product three times. Distinct pairs give distinct maps, since evaluating at zero gives $b$ and evaluating the difference on basis vectors gives all columns of $A$. Thus these pairs faithfully describe a group.

The map $A\mapsto (A,0)$ is an injective group homomorphism, so its image is the required copy of ${\mathrm{GL}}_n(F)$. Translations are the subgroup $N=\{(I,b):b\in F^n\}$, isomorphic to the additive group of $F^n$. For any affine map,Thus conjugation preserves translations, proving $N⊴G$.

Every pair factors as $(A,b)=(I,b)(A,0)$, and a linear map which is also a translation fixes zero and has linear part $I$, hence is the identity. Thus $G=NH$ and $N\cap H=\{1\}$. The conjugation formula in (c) says that the action needed for reconstruction is simply the usual matrix action $A:t\mapsto At$ on the additive group $F^n$. Therefore

There are $p^n$ choices for the translation. Choose the columns of an invertible matrix successively: the first is any nonzero column, the second avoids its span, and at stage $j+1$ one avoids a $j$-dimensional span of size $p^j$. ConsequentlyThe successive linear independence conditions are exactly the criterion that the columns form a basis.