# A primitive root modulo $p^k$ is primitive modulo $p^{k+1}$,for $k\geq 2$.

I am a graduate student of Mathematics.I am stuck with the following number theory problem:

Let $$p$$ be an odd prime.Prove that any primitive root modulo $$p^k$$ is a primitive root modulo $$p^{k+1}$$, for any $$k\geq 2$$.

I am a newcomer in number theory and I do not know how to proceed.Can please someone provide a solution to this problem?I have tried to proceed as follows:

Let $$\alpha$$ be a primitive root modulo $$p^k$$.

So,$$(\alpha)=U(p^k)$$

Which implies $$\alpha^{\phi(p^k)}\equiv 1 (\mod p^k)\implies \alpha^{p^{k-1}(p-1)}\equiv 1(\mod p^k)$$,thus $$\alpha^{p^{k-1}(p-1)}=1+mp^k,$$for some $$m\in \mathbb Z$$,so $$\alpha^{{p^k}(p-1)}\equiv 1 (\mod p^{k+1})$$.Then I got stuck.Can someone help me how to proceed?

• The proof should be in any number theory textbook that discusses primitive roots. See Theorem 4 of this post, for example. Oct 19, 2021 at 4:47
• @GregMartin I did not find it in any book.Please help me to show that the order of $\alpha$ is $p^k(p-1)$ modulo $p^{k+1}$. Oct 19, 2021 at 5:10
• $\alpha^{p^{k-1}(p-1)}\equiv 1 (\mod p^k)$ and $\alpha^t\equiv 1 (\mod p^{k+1})$ where $t$ is the order of $\alpha$ modulo $p^{k+1}$,so $\alpha^t\equiv 1 (\mod p^k)$,ehnce $p^{k-1}(p-1)|t$. Oct 19, 2021 at 7:09
• Also $\alpha^{p^k(p-1)}\equiv 1 (\mod p^{k+1})\implies t|p^k(p-1)$,So,possible values of $t$ are $p^{k-1}(p-1)$ and $p^k(p-1)$. Oct 19, 2021 at 7:11
• Now how do I eliminate the first case and establish that the second one is actually the order. Oct 19, 2021 at 7:12