Past exam of the short options course of the University of Oxford / 2015 / 2 / b

For the given group homomorphism $\phi :H\to \mathrm{Aut}(N)$, the external semidirect product has underlying Cartesian product $N\times H$ and operationThe notation $\mathrm{Aut}(N)$ denotes the automorphism group of $N$. Closure follows because $\phi (h)$ is an automorphism of $N$. To check associativity, calculate both bracketings:They agree because $\phi (hk)=\phi (h)\phi (k)$. The identity is $(e_N,e_H)$, using $\phi (e_H)={\mathrm{id}}_N$, and the two-sided inverse isIndeed multiplication in either order gives the identity, using $\phi (h)\phi (h^{-1})={\mathrm{id}}_N$. This verifies the group axioms.

For the specified group presentation, take $N=C_9=⟨X⟩$ and $H=C_3=⟨Y⟩$. The rule $\alpha (X)=X^4$ defines an automorphism because $4$ is coprime to $9$. Furthermore $4^3\equiv 1(\mathrm{mod}9)$, so $Y\mapsto \alpha$ defines the required group homomorphism $C_3\to \mathrm{Aut}(C_9)$. In the resulting semidirect product $S$, the canonical generators satisfyConsequently the group presentation defines a surjective group homomorphism $G\to S$, sending $x$ to $X$ and $y$ to $Y$. The underlying Cartesian product gives $|S|=27$, whereas part (a) gives $|G|\le 27$. Surjectivity forces equality, so the map is an isomorphism:This also proves that the $27$ normal forms in part (a) are distinct and that $x$ has element order exactly $9$.