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These are all the groups of group order $70$ up to group isomorphism, with $D_n$ denoting a dihedral group of order $n$. The Sylow theorems make the Sylow subgroups of orders five and seven unique: the possible counts divide fourteen and ten, respectively, and only one has the required residue. These normal subgroups of coprime order commute, giving a normal subgroup $N\cong C_{35}$. Cauchy's theorem for finite groups provides an element of order two outside $N$, so $G=N⋊C_2$.

The automorphism of a finite cyclic group defining this semidirect product is a unit $a$ modulo $35$ with $a^2\equiv 1$. The Chinese remainder theorem makes its restrictions modulo five and seven independently $+1$ or $-1$. The resulting exponents are $1,34,6,29$, giving the four displayed groups in order. Their center of a group orders are $70,1,5,7$, so no two are isomorphic. Every case exists as the specified semidirect product.

Let $H,K$ be the normal subgroups of orders $3,5$. Their intersection is a subgroup of both, so Lagrange's theorem makes its order divide both $3$ and $5$; hence $H\cap K=\{1\}$. For $h\in H$ and $k\in K$, normality shows that $hk{h}^{-1}{k}^{-1}$ belongs to both $H$ and $K$. It must be the identity, so $hk=kh$. This proves the needed instance of normal subgroups of coprime order commute.

Choose nonidentity $h\in H$ and $k\in K$. Their orders are $3$ and $5$, since these orders divide the respective prime subgroup orders. If $(hk{)}^m=1$, commutativity gives $h^m=k^{-m}\in H\cap K$, so $3\mid m$ and $5\mid m$. Conversely $(hk{)}^{15}=1$. Thus the product of commuting elements of coprime order gives an element $hk$ of order $15$.

For the counterexample, take the dihedral group of symmetries of a regular pentagon:Its ten elements are $r^j$ and $s{r}^j$, $0\le j<5$. Nonidentity rotations have order $5$, and $(s{r}^j{)}^2=r^{-j}{r}^j=1$, so all reflections have order $2$. There is no element of order $10$.

The Sylow theorems also give $n_5\mid 14$ and $n_5\equiv 1(\mathrm{mod}5)$. Of $1,2,7,14$, only $1$ is congruent to one modulo five. Thus $H$ is a normal subgroup of $G$ as well.

By Lagrange's theorem, $H\cap K=\{1\}$, since its group order divides both $5$ and $7$. For $h\in H$ and $k\in K$, the group commutator $hk{h}^{-1}{k}^{-1}$ lies in $H$ by the normal subgroup property of $H$, and in $K$ by the normal subgroup property of $K$. It is therefore the identity element, so all elements of $H$ commute with all elements of $K$. This is the fact that normal subgroups of coprime order commute.

The subgroup $HK$ consequently has $|H||K|=35$ elements: the expression $hk$ is unique because $H\cap K=\{1\}$. Since $H$ and $K$ have prime orders, they are cyclic groups. Write $H=⟨h⟩$ and $K=⟨k⟩$. The product of commuting elements of coprime order has order $35$, so $hk$ generates $HK$. Finally, $HK$ is a normal subgroup because both its factors are normal subgroups; equivalently, it has subgroup index two. Therefore