Past exam of the short options course of the University of Oxford / 2017 / 2 / b / iii

Put $N=HK=⟨r⟩$, the normal subgroup of order $35$. By Cauchy's theorem for finite groups, $G$ contains an element $t$ of order two. Such an element cannot belong to the odd-order subgroup $N$. Thus $N\cap ⟨t⟩=\{1\}$ and $G=N⟨t⟩$, giving the semidirect productEvery automorphism of a finite cyclic group sends $r$ to $r^a$ for a unit $a$ modulo $35$. The relation $t^2=1$ forces the conjugation action of $t$ to square to the identity:The Chinese remainder theorem identifies a congruence class modulo $35$ with its pair of classes modulo $5$ and $7$. In each of these prime fields, the factorization $(a-1)(a+1)=0$ forces $a=1$ or $a=-1$. The choices are independent, giving exactly four homomorphisms from a finite cyclic group $C_2\to \mathrm{Aut}(C_{35})$. Explicitly,The factors $C_5$ and $C_7$ of $N$ are preserved by every automorphism, since each is its unique subgroup of that order. The four conjugation actions therefore give:

Here $D_n$ has group order $n$, as in the paper. The center of a group distinguishes these four groups of order seventy: its orders are $70,1,5,7$, respectively, so the four groups are pairwise nonisomorphic. Each construction is a semidirect product with $70$ elements, so every listed possibility occurs.