So I have to prove: Let $p\ $ be a odd number, $\alpha\ $ natural number and r primitive root of $p^{\alpha}$, so $\ o_{p^{\alpha}}r=\phi(p^{\alpha})$
$r+p^{\alpha}$ is a primitive root of $2p^{\alpha}$.
So i have to proof:
$$o_{2p^{\alpha}}(r+p^{\alpha})=\phi(2p^{\alpha})$$
But i don't know how to connect the given data, and that part i need to proof, any help would be apreciated. Thank you in advance.
First, note that $$\varphi(p^\alpha)= p^{\alpha} \left( 1 - \tfrac 1 p \right) = 2 \left(1 - \tfrac 1 2 \right)p^{\alpha} \left( 1 - \tfrac 1 p \right) = \varphi(2 p^\alpha) .$$ So we just need to prove that if $r$ has order $\varphi(p^\alpha)$ in the multiplicative group of units modulo $p^\alpha$, then $r + p^\alpha$ has order $\varphi(p^\alpha)$ in the multiplicative group of units modulo $2p^\alpha$.
Let's consider the powers of $r + p^\alpha$ modulo $2p^\alpha$. For any $n \in \{1, 2, \dots, \varphi(p^\alpha) - 1 \}$, it is impossible for $(r + p^\alpha)^n$ to be congruent to $1$ modulo $2p^\alpha$.
[For if $(r + p^\alpha)^n \equiv 1 {\rm \ mod \ } 2p^\alpha$, then by expanding $(r + p^\alpha)^n $ using the binomial theorem, we get that $r^n \equiv 1 {\rm \ mod \ } p^\alpha$, which we know to be false for $n \in \{1, 2, \dots, \varphi(p^\alpha) - 1 \}$.]
Hence the order of $r+p^\alpha$ modulo $2p^\alpha$ must be at least $\varphi(p^\alpha)$. But since the order of the group of units modulo $2p^\alpha$ is precisely $\varphi(2p^\alpha) = \varphi(p^\alpha)$, we see by Lagrange's theorem that the order of $r+p^\alpha$ modulo $2p^\alpha$ must in fact be equal to $\varphi(2p^\alpha) = \varphi(p^\alpha)$, i.e. $r + p^\alpha$ is a generator of the group of units modulo $2p^\alpha$.
