Every periodic orbit on a star-shaped level is a finite cover of a prime periodic orbit with the same trajectory

Let $S=H^{-1}(1)\subset\mathbb{R}^4$ be a strictly star-shaped level and $Q=(x,T)$ a periodic orbit of $X_H$ on $S$. Then $x$ has a least positive period $T_0$, and $T=kT_0$ for some integer $k\ge1$. In other words, $Q_0=(x,T_0)$ is a prime periodic orbit and $Q$ is its $k$-fold cover.

Proof idea. The periods of $x$ form an additive subgroup $G\subset\mathbb{R}$ containing $T>0$. If $G$ were dense, continuity would make $x$ constant, so $X_H(x)=x'=0$. But $X_H\neq0$ on $S$, because $dH(x)\,x>0$ there. So $G=\mathbb{Z}T_0$ is cyclic with $T_0>0$. A return $x(t)=x(0)$ with $0<t<T_0$ would make $t$ a period, by uniqueness of solutions of $\dot y=X_H(y)$, which contradicts minimality.