elementary number theory - Bezout's identity for cyclic groups

Let $\bar{a},\bar{b} \in \mathbb{Z}/k$, s.t. $gcd((a,k),(b,k))=1$. Does the Bezout lemma imply the existence of $x,y \in \mathbb{Z}$, s.t. $x\bar{a}+y\bar{b}$ is a generator of $\mathbb{Z}/k$?

Answer

Yes it does since it implies that there exists x and y such that $ax + by \equiv 1 [k]$ and 1 generates $\Bbb{Z}/k\Bbb{Z}$.