What does order of $a$ in multiplicative group modulo $p^k$ imply for its order in $p^{k+1}$? More specifically, I would like to prove, assuming  $p$ is a prime bigger than 2 and both $a$ and $k$ are positive integers:
If $a$ mod $p^k$ has an order $r$ in the multiplicative group mod $p^k$, and $r$ is not divisible by $p$, then the order of $a$ mod $p^{k+1}$ in multiplicative group mod $p^{k+1}$ is either $r$ or $rp$.
When $r$ is divisible by $p$, then the new order is $rp$.
 A: No takers, so let me try myself (with help from the two comments).
Assume that $r$ is the smallest integer such that $p \not\mid r$ and $a^r=1\bmod p^k$, implying that
$$(a\bmod p^k)^r=1\bmod p^k$$
and
$$(a\bmod p^k)^r=1+q \cdot p^k \bmod p^{k+1},$$
where $q$ is an integer between $0$ and $p-1$.
Similarly,
$$a\bmod p^{k+1} = (a \bmod p^k+j \cdot p^k)\bmod p^{k+1}$$
where $j$ is an integer between $0$ and $p-1$.
When $j$ meets $q+j \cdot r \cdot (a\bmod p^k)^{r-1}=0 \bmod p$ (note that exactly one such $j$ does) then
\begin{align*}
a^r 
&=(a\bmod p^k+j \cdot p^k)^r \\
&=(a\bmod p^k)^r+r \cdot j \cdot p^k \cdot (a\bmod p^k)^{r-1}\\
&=1+q \cdot p^k+r \cdot j \cdot p^k \cdot (a\bmod p^k)^{r-1} \\
&=1 \bmod p^{k+1}
\end{align*}
implying that $r$ is also the order of $a$ in the multiplicative group $\bmod p^{k+1}$.
When $j$ does not meet the same equation, then $r$ is not the order of $a$ in the new group, but $r \cdot p$ is, since now (replacing $r$ by $r \cdot p$ in the previous argument)
\begin{align*}
(1+q \cdot p^r)^p &+ r \cdot p \cdot j \cdot p^k \cdot (1\bmod p^k)^{r-1} \\
&= 1+r \cdot q \cdot p^k+r \cdot p \cdot j \cdot p^k \cdot (1\bmod p^k)^{r-1} \\
&=1 \bmod p^{k+1}
\end{align*}
When $p \mid r$ then $a^r=1+q \cdot p^k\bmod p^{k+1}$ which implies that $r$ must be further multiplied by $p$ to get $a^{r\cdot p}=1 \bmod p^{k+1}$. Note that this time $q=0$ would imply that our $r$ can be divided by $p$ and still achieve $a^{r/p}=1\bmod p^k$, contradicting the smallest $r$ assumption.
