= Solution
Identify $S^{2b-1}$ with the unit sphere in $\mathbb C^b$, and put $\zeta=e^{2\pi i/a}$. The generator of the <cyclic group> $\mathbb Z/a$ acts by
$$
(z_1,\ldots,z_b)\longmapsto(\zeta z_1,\ldots,\zeta z_b).
$$
It preserves the sphere because $|\zeta|=1$. If its $k$th power fixes a point, then $(\zeta^k-1)z_j=0$ for every $j$. Some coordinate is nonzero, so $\zeta^k=1$, meaning $k\equiv0\pmod a$. Thus \b[the scalar root-of-unity action is free]. The case $a=1$ is the trivial <group> and also satisfies freeness. The quotient is a standard <lens space>.
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