 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 byIt 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 $k\equiv 0(\mathrm{mod}a)$. Thus 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.