I am not entirely sure if this is true, but if it is, I would be done with a very important proof. Let $a$, $b$ and $d$ be pairwise coprime. Prove that: $$|(\mathbb{Z}/ab\mathbb{Z})^*/<d>_{ab}| = |(\mathbb{Z}/a\mathbb{Z})^*/<d>_a|\times|(\mathbb{Z}/b\mathbb{Z})^*/<d>_b|$$
Where $<d>_k$ is the cyclic group generated by $d$ in the corresponding multiplicative group $(\mathbb{Z}/k\mathbb{Z})$. Of course it suffices to show that: $$(\mathbb{Z}/ab\mathbb{Z})^*/<d>_{ab} \cong ((\mathbb{Z}/a\mathbb{Z})^*/<d>_a) \times ((\mathbb{Z}/b\mathbb{Z})^*/<d>_b)$$
Can someone indicate the method of proof or give a counter example.
Unfortunately it is not true. You can use the Chinese Remainder Theorem to show that $\left(\mathbb{Z}/ab\mathbb{Z}\right)^\times \cong \left(\mathbb{Z}/a\mathbb{Z}\right)^\times \times \left(\mathbb{Z}/b\mathbb{Z}\right)^\times$. But in general, the quotient groups you mention are not isomorphic, and the equality you would like to hold is not true. Here is a counter-example:
Let $a=7$, $b=11$, $d=692$. Then $|Z_7^*/\left<692\right>|=|Z_7^*/\left<6\right>|=6/2=3$. $|Z_{11}^*/\left<692\right>|=|Z_{11}^*/\left<10\right>|=10/2=5$. But $|Z_{77}^*/\left<692\right>|=|Z_{77}^*/\left<76\right>|=6\times 10/ 2 = 30$.