= Solution
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 $hkh^{-1}k^{-1}$ 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 \b[an element $hk$ of order $15$].
For the counterexample, take the <dihedral group> of symmetries of a regular pentagon:
$$
\boxed{D_5=\langle r,s\mid r^5=s^2=1,\ srs=r^{-1}\rangle}.
$$
Its ten elements are $r^j$ and $sr^j$, $0\le j<5$. Nonidentity rotations have order $5$, and $(sr^j)^2=r^{-j}r^j=1$, so all <reflections> have order $2$. \b[There is no element of order $10$.]
Back to article page
