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We are given a free action of an abelian finite group on $S^3$. Let $L$ denote the quotient space and let an element $\alpha \in \pi_1 L =G$ be given. Does there exist a cyclic homotopy $h_t:L \to L$ such that $H_0=H_1={\rm id}_L$ and the loop $h_t(x_0)$ represents the class $\alpha$?

This is true for $L=L(p,q)$ lens space. In fact $L(p,q)$ is given by the action $k\circ (z_1,z_2)= (exp(2\pi k/p)z_1, exp(2\pi kq/p)z_2)$ hence the twist of $S^3$ given by $h_t(exp(2\pi t/p)z_1, exp(2\pi t q/p)z_2)$ induces the loop representing the generator of $\pi_1 (L(p,q))= \mathbb Z_p$.

On the other hand this is not true for nonabelian $G$, since in general any Jiang space has a commutative homotopy group.

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    $\begingroup$ It follows from the elliptization theorem that the only free actions of an abelian group on $S^3$ are cyclic groups, with action conjugate to one of the ones you describe. $\endgroup$
    – mme
    Aug 13, 2019 at 13:18

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