- #1

fishturtle1

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## Homework Statement

1.31 Let ##p## be a prime and let ##q## be a prime that divides ##p - 1##.

a) Let ##a \epsilon F^*_p## and let ##b = a^{(p - 1)/q}##. Prove that either ##b = 1## or else ##b## has order ##q##. (Recall that order of ##b## is the smallest ##k \ge 1## such that ##b^k = 1## in ##F^*_p##. Hint. Use Proposition 1.30).

## Homework Equations

Proposition 1.30: Let ##p## be a prime and let ##a## be an integer not divisible by ##p##. Suppose that ##a^n \equiv 1 (\mod p)##. Then the order of ##a## modulo ##p## divides ##n##. In particular, the order of ##a## divides ##p - 1##.

## The Attempt at a Solution

Proof: Let ##a \epsilon F^*_p## and let ##b = a^{(p - 1)/q}##. Raising both sides to the qth power, we have ##b^q = a^{p - 1}##. Therefore, ##b^q \equiv a^{p - 1} (\mod p)##. Since ##a < p##, we can say ##p## does not divide ##a##. By Fermat's Little Theorem, we have ##a^{p - 1} \equiv 1 (\mod p)##. Since ##\equiv## is transitive, we have ##b^q \equiv 1 (\mod p)##. Thus, ##b = pk + 1##, where k is some integer. So ##b = 1## or ##b = pk + 1##, where ##p \neq 0##.

But I'm not sure how or if ##p \neq 0## implies ##q## is the order of ##b##.