Show that this dephasing operation is Markovian

[Pete5876]

Show that this dephasing operation is Markovian
$$E_t(\rho)=\left(\frac{1+e^{-t}}{2}\right)\rho + \left(\frac{1-e^{-t}}{2}\right)Z\rho Z$$

The operation is supposed to be Markovian when
$$E_g E_h = E_{g+h}$$

So this is what I get when I apply this multiplication
$$
E_g E_h=\frac{1+e^{-g}}{2}\frac{1+e^{-h}}{2} \rho^2 + \frac{1-e^{-g}}{2} \frac{1-e^{h}}{2}Z\rho ZZ\rho Z + \frac{1+e^{-g}}{2} \frac{1-e^{-h}}{2} = \rho Z \rho Z + \frac{1+e^{-g}}{2} \frac{1-e^{-h}}{2} Z \rho Z \rho
$$
We can lose ZZ because
$$ZZ=I$$
but I can't see what else can be done. ρ after all only commutes with Z if it's diagonal but we can't assume that for ρ in general?

 

[haruspex]

You have misunderstood the question. $E_gE_h$ means the composition of those functions.
The algebra might be a bit simpler using hyperbolic trig - if you know the formulae for $\cosh(x+y)$ etc.