$(\mathbb{Z}/77 \mathbb{Z})^{\times} \cong \mathbb{Z}/10 \mathbb{Z} \times \mathbb{Z}/6 \mathbb{Z}$ - 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?

Since $\Bbb Z/77\Bbb Z \cong \Bbb Z/11\Bbb Z \times \Bbb Z/\Bbb 7Z$ then 
$$(\Bbb Z/77\Bbb Z)^\times \cong (\Bbb Z/11\Bbb Z)^\times \times (\Bbb Z/7\Bbb Z)^\times$$
Show that $(\Bbb Z/11\Bbb Z)^\times \cong \Bbb Z/10\Bbb Z$ amd $(\Bbb Z/7\Bbb Z)^\times \cong \Bbb Z/6\Bbb Z$. You can check that $[7]_{11}$ is a generator of $(\Bbb Z/11\Bbb Z)^\times$ and $[3]_7$ is a generator of $(\Bbb Z/7\Bbb Z)^\times$.
