= Solution
Put $N=HK=\langle r\rangle$, 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 $N\cap\langle t\rangle=\{1\}$ and $G=N\langle t\rangle$, giving the <semidirect product>
$$
\boxed{G\cong C_{35}\rtimes C_2.}
$$
Every <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:
$$
trt^{-1}=r^a,\qquad a^2\equiv1\pmod{35}.
$$
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> $C_2\to\operatorname{Aut}(C_{35})$. Explicitly,
$$
\boxed{a\equiv1,6,29,34\pmod{35}.}
$$
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:
* $a=1$: the action is trivial on both factors, so $G\cong C_{35}\times C_2\cong\boxed{C_{70}}$, by <cyclicity of a product of two finite cyclic groups>.
* $a=34$: the action inverts both factors, so $\langle r,t\mid r^{35}=t^2=1,\ trt^{-1}=r^{-1}\rangle$ is the <dihedral group> $\boxed{D_{70}}$.
* $a=6$: it is the identity modulo $5$ and inversion modulo $7$, so the $C_5$ factor commutes with the other factors and $G\cong\boxed{D_{14}\times C_5}$.
* $a=29$: it is inversion modulo $5$ and the identity modulo $7$, giving $G\cong\boxed{D_{10}\times C_7}$.
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.
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