# $\zeta$ is a p-th primitive untary root iff $-\zeta$ if a 2p-th primitive unitary root, with $p$ an odd prime

$$\Rightarrow$$ $$(-\zeta)^{2p}=(-\zeta^p)^{2}=1$$ and if $$i \in[{1,2p-1}]$$ there is $$k \in[0,p-1]$$ such that $$i=2k+1$$ if is odd and $$2k$$ if is even so $$(-\zeta)^{i}=(-\zeta)^{2k}=(\zeta^k)^{2}\neq 1$$ since $$\zeta$$ is p root and $$p> 2$$ ($$p$$ is prime and odd).

but i dont know if this is right, if it is then the odd case will follow to.

My question is indeed in how to prove $$\Leftarrow$$, if $$(-\zeta)^{i}\neq 1$$ $$\forall i \in[{1,2p-1}]$$, how can i use this to prove $$(\zeta)^{i}\neq 1$$ $$\forall i \in[{1,p-1}]$$? the negative sign is what confusing me

Answering your question on the reverse implicatuon: If z is a 2p primitive root then $$z^p=-1$$ since -1 is the only primitive square root. Then $$(-z)^p=1$$ since $$p$$ is odd. And $$(-z)^j\ne 1$$ for $$0 lest $$z^{2j}=1$$, contradicting that $$2p$$ is the least positive power of z that equals 1.
On the forward implication, if some power k of (-z) less than 2p were unity, then (a) if that power is p then since p is odd $$z^p=-1$$, contradiction; (b) otherwise k and therefore 2k is prime to p and $$z^{2k}=1$$, contradiction.
I used z for $$\zeta$$ throughout to save keystrokes.
• I can't see why the contradiction on the reverse implication tell me that $z$ is a $p$ root, and on the item b), i guess that some information about $k$ is missing. – Eduardo Silva Oct 31 '18 at 23:25