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If I know that $$g^a \neq 1 \mod b$$ is that always true that if I will take a positive integer $c$ and count $(g^a)^c$, then $$(g^a)^c \neq 1 \mod b$$?

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No. It's never true when $g$ is a unit mod $b$, e.g. take $c=\varphi(b)$. –  blue May 20 at 18:27
    
So if $c = \phi(b)$ it is always true? –  Rop May 20 at 18:31
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The relation $x^{\varphi(b)}\equiv1$ mod $b$ is true for any unit $x$. In particular if $g$ is a unit then $g^a$ is a unit for any $a\in\Bbb Z$. –  blue May 20 at 18:32
    
@seaturtles I understand, thank you –  Rop May 20 at 18:34
    
@seaturtles If I can ask you one more question: if I know that $g^{\phi(b)/x} \neq 1 \mod b$, can I also say that always $(g^{\phi(b)/x})^c \neq 1 \mod b$? –  Rop May 20 at 18:50

2 Answers 2

Iff $\gcd(g^a,b) > 1$. Otherwise, take $c = \phi(b)$.

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Hint $\ $ In a finite commutative ring every element is a unit or a zero-divisor. By Lagrange's theorem, a unit has finite order (and the converse is clear), $ $ so the units are precisely the elements of finite order, so the elements you seek, those not of finite order, are precisely the zero-divisors.

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