$(\mathbb{Z}/77 \mathbb{Z})^{\times} \cong \mathbb{Z}/10 \mathbb{Z} \times \mathbb{Z}/6 \mathbb{Z}$ - Group Let $A$ a ring with unit element $1 \ne 0$ let $A^{\times}=\{a \in A: a$ invertible$\}$. Show that $(\mathbb{Z}/77 \mathbb{Z})^{\times} \cong  \mathbb{Z}/10 \mathbb{Z} \times \mathbb{Z}/6 \mathbb{Z}$ as a group.
I am stucked on this problem for a while. Is anyone could help me at this point? or just a theorem allow me to conclude?

Hint: When $p$ is a prime, we have that $(\mathbb{Z}/p\mathbb{Z})^\times \cong (\mathbb{Z}/(p-1)\mathbb{Z})$. Notice that $(\mathbb{Z}/77\mathbb{Z}) \cong (\mathbb{Z}/7\mathbb{Z})\times (\mathbb{Z}/11\mathbb{Z})$. 
Can you show that  $$(\mathbb{Z}/7\mathbb{Z}\times \mathbb{Z}/11\mathbb{Z})^\times \cong (\mathbb{Z}/7\mathbb{Z})^\times\times (\mathbb{Z}/11\mathbb{Z})^\times \cong (\mathbb{Z}/6\mathbb{Z}) \times (\mathbb{Z}/10 \mathbb{Z})$$
