 Solution

Assume the $g$ cycle has least period $n$ and nonzero periodic-orbit multiplier $P=(g^n{)}^{\prime }(\hat{x})$. It then avoids zero. Combining the iterate identity with the periodic-orbit multiplier-sign formula givesThus $P>0$ makes $\hat{x}$ return after $n$ iterates, whereas $P<0$ gives $f_L^n(\hat{x})=-\hat{x}$. In the latter case oddness on the noncritical orbit impliesTo prove the least periods, note that a $g$ cycle cannot contain two distinct points $x$ and $-x$: evenness would give them the same image, but $g$ permutes its distinct cycle points bijectively. Therefore the $n$ absolute values in the cycle are distinct. Since $|f_L^k(\hat{x})|=|g^k(\hat{x})|$, any return under $f_L$ must occur at a multiple of $n$. For $P<0$ the first such return fails because $\hat{x}\ne 0$, so $2n$ is minimal. Hence the period transfer from a quadratic map to its Lorenz sign lift isThere are two symmetry-related $f_L$ cycles in the positive case and one cycle containing both signs in the negative case. Away from zero $f_L^{\prime }(x)=2|x|$, so their periodic-orbit multipliers are $|P|$ and $|P{|}^2$, respectively. A stable $g$ cycle therefore gives stable lifted cycles in both cases.