Some congruences and conclusion.

Question

Let consider this situation:

$$ \gcd (b, m) = 1 \tag{$*$} \\ b^a \equiv 1 \\ b^c \equiv 1 $$ Assume that $ a \le c $ Whether it is reasonable to draw a conclusion that is $c = ak$ for some $k$? Why? Is the $(*)$ condition is necessary? Why?

Answer 1

1. First of all the (*) condition is necessary, because if $gcd(b,m)>1$ it means $b$ cannot have modulo inverse with respect to $m$. Thus $b^a \equiv 1$ would not be true for any $a$ as the inverse of $b$ would be $b^{(a-1)}$ which is not possible if $gcd(g,m) \neq 1$.
2. Secondly let $a$ be the smallest integer such that $b^a \equiv 1$, only then it is always true that, if $b^c \equiv 1$ then $c=ak$.

EDIT

The proof for second point is simple assume $a$ is the smallest integer such that $b^a \equiv 1$ , also we have another integer $c$ such that $b^c \equiv 1$. Let us assume $c=ak+r$ where $r$ is remainder of division of $c$ by $a$. Thus $r<a$. Now we have $$b^c \equiv 1$$ $$b^{(ak+r)} \equiv 1$$ as $b^a \equiv 1$ for some integer $k$, $b^{ak} \equiv 1 $ thus $$b^r \equiv 1$$ This is a contradiction as $r<a$ and we assumed $a$ was the smallest integer such that $b^a \equiv 1$. Thus by contradiction $r=0$ and $c=ak$.