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 $$(g^a)^c \neq 1 \mod b$$?
Asked
Active
Viewed 56 times
0
-
2No. It's never true when $g$ is a unit mod $b$, e.g. take $c=\varphi(b)$. – anon May 20 '14 at 18:27
-
So if $c = \phi(b)$ it is always true? – Rop May 20 '14 at 18:31
-
1The 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 $a\in\Bbb Z$. – anon May 20 '14 at 18:32
-
@seaturtles I understand, thank you – Rop May 20 '14 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 '14 at 18:50
-
No. It's possible for $\phi(b)/x$ to not be divisible by the order of $g$ while $(\phi(b)/x)\cdot c$ is. I recommend learning modular arithmetic and elementary number theory through a textbook or lecture notes or some other source: you will be able to answer these on your own. – anon May 20 '14 at 18:52
-
@seaturtles Ok, I will. Thank you – Rop May 20 '14 at 18:59
2 Answers
1
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.
Bill Dubuque
- 272,048
