For an odd prime $p$, prove that any primitive root $r$ of $p^n$ is also a primitive root of $p$
So I have assumed $r$ have order $k$ modulo $p$ , So $k|p-1$.Then if I am able to show that $p-1|k$ then I am done .But I haven't been able to show that.Can anybody help me this method?Any other type of prove is also welcomed.
For any $a$ relatively prime to $p^n$, there is an integer $k$ such that $r^k\equiv a \pmod{p^n}$ and hence such that $r^k \equiv a\pmod{p}$. Thus $r$ is a primitive root of $p$.
Note that an integer $r$ with $\gcd(r,p)=1$ is a primitive root modulo $p^k$ when the smallest $b$ such that $r^b\equiv1\bmod p^k$ is $b=p^{k-1}(p-1)$.
Suppose that $r$ is not a primitive root modulo $p$, so there is some $b<p-1$ such that $r^b\equiv 1\bmod p$.
In other words, there is some integer $t$ such that $r^b=1+pt$.
Then of course we have that $p^{n-1}b<p^{n-1}(p-1)$, and $$r^{p^{n-1}b}\equiv 1\bmod p^n$$ because of
the binomial theorem.
(mouse over to reveal spoilers)
$\begin{eqnarray}{\bf Hint}\ \ \ \ \rm a\ \ is\, \ coprime\, \ to\ \ {\bf\color{#C00}p}\,\ &\Longrightarrow&\rm\ \ \ a\ \ is\ \ coprime\ \ to\ \ {\bf\color{#90f}{p^n}}\\ & & \qquad\qquad{ \Downarrow}\\ \smash{\rm r^k\equiv a\ \ (mod\,\ {\bf\color{#C00}p})}\, &\smash{\overset{\ \ \rm\color{#0a0}{CP}}\Longleftarrow}&\rm \smash{\exists\,k\!:\ r^k\equiv a\ \ (mod\,\ {\bf\color{#90f}{p^n}})} \end{eqnarray}$
where we employed the result $\,\rm \color{#0a0}{CP} = $ Congruences Persist mod factors of the modulus.
Exercise: If $X$ is a generating set for a group $G$ and $\phi:G\to H$ a homomorphism, then $\phi(X)$ is a generating set for the image of $G$ under $\phi$; in particular this means that if $\phi$ is surjective/onto then $\phi(X)$ is a generating set for $H$.
Now consider $G=(\Bbb Z/p^n\Bbb Z)^\times$ and $H=(\Bbb Z/p\Bbb Z)^\times$ with $\phi$ the mod $p$ map (see that it's well-defined and onto!) with a singleton generating set $X=\{a\}$ for $G$.
If r is a primitive root of prime p, we can show that either r is a primitive root of $p^2$ or (r+c.p) is a primitive root of $p^2$ where (c,p)=1.
We can also show that if g is a primitive root of both p and $p^2$, it is also a primitive root of $p^k$ where k is a natural number.
Clearly, there exists one r+c.p where p|c or (c,p)=1 which is a primitive root of both p and $p^2$ =>r+c.p is a primitive root of $p^k$ where k is a natural number.
Each of $\phi(p-1)$ primitive root of p, will have associated (p-1) primitive roots which are congruent(mod p), but in-congruent(mod $p^2$) of $p^2$, p(p-1) primitive roots of $p^3$ and $p^{k-2}(p-1)$ primitive roots of $p^k$ where k≥3.
Each primitive root of $p^k$ will be associated to p primitive roots of $p^{k+1}$ where k > 1.
Using this, if $ord_{(p^k)}a = d$ where k is a natural number, we can show that $ord_{(p^{k+1})}a = d$ or $pd$
Now, if r is a primitive root of $p^{k+1} => ord_{(p^{k+1})}a = \phi(p^{k+1})=p^k(p-1)$
=>$ ord_{(p^{k})}a = p^k(p-1)\ or\ p^{k-1}(p-1) $
=>But, $ ord_{(p^{k})}a ≤\ \phi(p^k) =p^{k-1}(p-1)$
=>$ ord_{(p^{k})}a = p^{k-1}(p-1)=\phi(p^k) $ => r is a primitive root of $p^k$ for all natural number k≥1
Let the order of $r$ modulo $p$ be $k$ . So, $$r^k\equiv_p 1$$ which means $r^k = 1 + ap\, \,$ where $\, \,p \nmid a$
Now, raise both sides to power $p$ and use binomial theorem to get $$ r^{pk} \equiv_{p^2} 1$$
Repeat this $n-2$ more times to get $$ r^{p^{n-1}k} \equiv_{p^n} 1$$ But $r$ is a primitive root of $p^n$, so, $$\phi(p^n) \mid p^{n-1}k$$ where $\phi$ is Euler's totient function. Thus, we have
$$p-1 \mid k \Rightarrow k = p-1$$
