if $b^k$ is a primitive root, then $b$ is a primitive root

Question

Any hints or strategies would be greatly appreciated:

If $m$ is an integer and $b^k$ is a primitive root mod $m$, then $b$ is a primitive root mod $m$.

I am reviewing material from my elementary number theory course. I had a true/false question that I think is true (by process of elimination of how many "trues" the problem set was supposed to have). But I and am absolutely stuck on trying to prove it.

Welcome to the site, and please use Latex to format your equations. (Put them between a pair of $'s). — user111187, Jun 14 '14 at 21:42
Have you tried contraposition? Instead of "$b^k$ primitive $\implies$ $b $ primitive", try proving "$b $ not primitve $ \implies $ $b^k$ not primitive". — zo0x, Jun 14 '14 at 21:48
Text is Rosen. Definition is: If $r$ and $n$ are relatively prime integers with $n > 0$ and if $ord_n r = phi(n)$, then $r$ is called a primitive root mod $n$. — Meghan K., Jun 14 '14 at 22:07
http://math.stackexchange.com/questions/134109/gk-is-a-primitive-element-modulo-m-iff-gcd-k-varphim-1 — lab bhattacharjee, Jun 15 '14 at 04:37
Many thanks to all responders for ideas and help. I think I almost have a proof of the contrapositive. — Meghan K., Jun 16 '14 at 15:17

score 2 · Answer 1 · answered Jun 14 '14 at 21:47

If powers of $b^k$ generate every coprime residue class mod m, then powers of $b$, which powers are a superset of the first set of powers, also generate every coprime residue class mod m. So $b$ is also a primitive root. (A coprime residue class mod m is an invertible one, i.e., $\{a|(a,m)=1\}$.)

if $b^k$ is a primitive root, then $b$ is a primitive root

1 Answers1