Number Theory | There is a primitive root modulo every power of an odd prime!!

We prove that for an odd prime p, there is a primitive root modulo p^n for all natural numbers n.

[georgesadler]

Professor Penn, thank for a fantastic lecture on primitive roots modulo every power of an odd prime.

[PunmasterSTP]

Every power of an odd prime? More like "Amazing lectures that are mind-blowing every time!" 🤯

[rikpaul]

Thanks for uploading such a good content..

[tonk]

Pliz Explain it more clearly ..just dont go without explaining the steps....it will helps thr viwers a lot...

[prathikkannan]

Can someone pls explain the phi(p) | k part.
Here is what i brainstormed so far.

Lets first consider ord_p(a) | k = ord_p(a) | ord_p^n(a).
I am unfamiliar with this result so here is how it can be derived.

Let ord_p(a) = m. This means a^m = 1(mod p), or a^m = 1 + pz, z is an integer
Let ord_p^n(a) = q. This means a^q = 1(mod p^n) = 1(mod p), or a^q = 1+py, y is an integer

Subtracting the equations gives a^q - a^m = p(y -z) or a^q = a^m(mod p)
By a previous result, this implies q = m(mod ord_p(a)) --> q = m(mod m)
This means m|q.

So we have shown ord_p(a) | ord_p^n(a)
its also known ord_p(a) | phi(p)

so how do i show phi(p) | ord_p^n(a) ?