Vtgyjfy

I have some questions about moving from $mathbb Z_{c^k}$ into $mathbb Z_{c^{k+1}}$-specifically, with regard to the order of elements.

Suppose $a$ (which is coprime to $c$) has multiplicative order $b$ in $mathbb Z_{c^k}$.

Now if $a$ has multiplicative order $p$ in $mathbb Z_{c^{k+1}}$, then, $c^{k+1}mid (a^p-1)implies c^kmid(a^p-1)implies bmid p$

So $p=bn$ for some positive integer $n$.

$$require{cancel}a^bequiv 1mod c^kimplies a^{bn}equiv 1mod c^{k+1}$$

$$(a^b)^nequiv 1mod c^{k+1}implies ((a^b-1)+1)^nequiv 1 mod c^{k+1}$$

$$sum_{i=0}^nBiggl(binom{n}{i}(a^b-1)^i1^{n-i}Biggr)equiv 1 mod c^{k+1}$$

Since $c^k|(a^b-1):$
$$binom{n}{0}(a^b-1)^01^{n-0}+binom{n}{1}(a^b-1)^11^{n-1}+sum_{i=2}^ncancelto{0}{Biggl(binom{n}{i}(a^b-1)^i1^{n-i}Biggr)}equiv 1mod c^{k+1}$$

$$1+n(a^b-1)equiv 1mod c^{k+1}$$

So, finally:
$$n(a^b-1)equiv 0mod c^{k+1}$$

This gives us that $n|c$. More specifically, if $g.c.d.Bigl(frac{a^b-1}{c^k},cBigr)=d$, then $n=frac{c}{d}$.

Now, looking at something like the base-$a$ Wieferich primes, there are clearly examples when $nneq c$ (technically, Wieferich primes don't care specifically about the multiplicative order of the bases, but it happens frequently enough that $p-1$ is the base's order).

I have two main questions. For any non-perfect power, $c$, can we always find an $a$ and a $k$ such that, if $ operatorname{ord}(a)=b$ in $mathbb Z_c$ and $ operatorname{ord}(a)=n$ in $Z_{c^k}$, $nneq bc^{k-1}$? And, second, (with $c$ still being a non-perfect power$neq2$) is it possible that, if, for some given $a$, that $ operatorname{ord}(a)=b$ in $mathbb Z_c$ and $ operatorname{ord}(a)=bc$ in $mathbb Z_{c^2}$, but for some $k>2$, $ operatorname{ord}(a)neq bc^{k-1}$ in $mathbb Z_{c^k}$?

Edit: I think I have an answer to the first question-it looks to be a bit more general.

If we look at the set of elements in $mathbb Z_{c^{k+1}}$ that are invertible, we have a group under the operation of multiplication- denote it by $V_{c^{k+1}}$. Of course, these are exactly the elements that are coprime to $c^{k+1}$ so the order of this group is $$phi(c^{k+1})=c^{k+1}prod_{p:p=prime,pmid c^{k+1}}Bigl(1-frac{1}{p}Bigr)=c^{k+1}prod_{p:p=prime,pmid c}Bigl(1-frac{1}{p}Bigr)=\frac{c^{k+1}}{prod_{p:p=prime,pmid c}p}prod_{p:p=prime,pmid c}(p-1)=dprod_{p:p=prime,pmid c}(p-1)$$

where $d$ is clearly an integer. Now, consider any prime factor, $q$, of $p_{min}-1$ (where $p_{min}$ is the smallest prime that divides $c$).

$$qmid (p_{min}-1)midbiggl(prod_{p:p=prime,pmid c}(p-1)biggr)mid phi(c^{k+1})=Bigl|V_{c^{k+1}}Bigr|$$

So $qmid Bigl|V_{c^{k+1}}Bigr|$. But by Cauchy's theorem for groups, this means that $V_{c^{k+1}}$ has an element of order $q$- denote it by $x$. Clearly the multiplicative of order of $x$ in $mathbb Z_{c^{k+1}}$ is also $q$. Referring back to the way $q$ was defined, it's clear that $q$ and $c$ are coprime (since $q$ must be less than any prime factor of $c$). But, if $x$ has order $b$ in $mathbb Z_{c^k}$, then the order of $x$ in $mathbb Z_{c^{k+1}}$ is (as shown before) of the form $bn$ where $n$ is some divisor of $c$.

That means $bn=q$. Since $q$ and $c$ are coprime, $n=1$, and $b=q$, i.e., the order of $x$ is the same in $mathbb Z_{c^{k+1}}$ as it is in $mathbb Z_{c^k}$.

The reason this seems more general is that it works for any $k$. Of course, there's a small flaw when it comes to $c$ being even, i.e. that $p_{min}-1=2-1=1$ always. But, the above argument could still easily be extended to any even numbers where the $2^{nd}$ smallest prime divisor isn't a Fermat prime (just let $q$ be some prime other than $2$ that divides the second smallest prime-1). Even if the second smallest prime divisor is Fermat, we'll still have $operatorname{ord}(a)=2bneq bc^k$ (I just can't guarantee an element with the same order in $mathbb Z_{c^k}$ and $mathbb Z_{c^{k+1}}$ exists). That seems to just leave $c=2$. Still have no clue how to deal with the second question.

Edit 2: The above argument (in the Edit) can be tweaked a bit- once we have an $x$ of order $q$ in $mathbb Z_{c^{k+1}}$, instead of returning to $mathbb Z_{c^k}$ and examining $x$'s order there, we can look directly at $mathbb Z_c$ instead. If the order of $x$ in $mathbb Z_c$ is $b$, then by extending the reasoning presented above, the order of $x$ in $mathbb Z_{c^{k+1}}$ is $bN$, where $N$ is a divisor of $c^k$. That means $bN=q$ and (since $q$ and $c$ are coprime) $b=q$ but now $b$ is the order of $x$ in $mathbb Z_c$. This reasoning can be extended so that for every $1leq hleq k+1$, the order of $x$ in $mathbb Z_{c^h}$ is the same. This seems to be suggesting that the answer to the second question is generally no- but that would be to assume that elements that do not have order $bc^k$ in $mathbb Z_{c^{k+1}}$ are only constructed by the method in the first Edit (it also isn't true in the case when $c=2$ just by the fact that all odd numbers have order 1 in $mathbb Z_2$).