If $G = \left$ is a cyclic group of order $n= p^ka$ with $(p,a)=1$, can we write $x=cb$ with $c$ of order $p^k$ and $b$ of order $a$? If $G = \left<x\right>$ is a cyclic group of order $n= p^ka$ with $(p,a)=1$, can we write $x=cb$ with $c$ of order $p^k$ and $b$ of order $a$?
I have found that $c=x^{p^k}$ has order $a$ and $b=x^a$ have order $p^k$, but it doesnt seem like that gives $x=cb$.
Any help?
 A: If $x$ an element of order $ab$ with $au+bv=1,$ prove that the order of $x^{au}$ is $b$ (hence similarly, the order of $x^{bv}$ is $a$).
A: The reference to a a cyclic group and a prime power $p^k$ in the problem is unnecessary. What you are asking about is a property of every element of finite order in a group: all decompositions of the order of an element into two relatively prime parts lead to a decomposition of the element into a product of commuting elements with those two orders.
Theorem. Let $G$ be a group and $g \in G$ have order $N$.  Suppose $N = nn'$ where $\gcd(n,n') = 1$. Then there is a unique way to write $g = hh'$ where (i) $h$ has order $n$, (ii) $h'$ has order $n'$, and (iii) $h$ and $h'$ commute.
Proof.
Existence of $h$ and $h'$: we'll get such $h$ and $h'$ as powers of $g$. Since $\gcd(n,n') = 1$, we can write $1 = nm + n'm'$ for some integers $m$ and $m'$. Then
$$
g = g^1 = g^{nm + n'm'} = g^{nm}g^{n'm'} = g^{n'm'}g^{nm},  
$$
and we set $h = g^{n'm'}$ and $h' = g^{nm}$. Then $h$ and $h'$ commute since they are powers of $g$. Show the order of $g^{n'}$ is $n$, and then the order of $h = g^{n'm'}= (g^{n'})^{m'}$ is also $n$ (hint: $\gcd(m',n) = 1$). That the order of $h'$ is $n'$ is a similar argument.
Uniqueness of $h$ and $h'$: Suppose $g = hh'$ where properties (i), (ii), and (iii) hold.  Then $g^{n} = h^nh'^{n}$ by (iii), so $g^n = h'^{n}$ by (i). The exponent on powers of $h'$ only matter modulo $n'$, by (ii). Since $nm + n'm' = 1$ we have $nm \equiv 1 \bmod n'$, so $g^{nm} = h'^{nm} = h'^1 = h'$. That shows $h'$ must be given by the formula we used to define $h'$ in the existence part above. That $h = g^{n'm'}$ is a similar argument. $ \Box$
Remark. Without the commutativity condition on $h$ and $h'$, the uniqueness aspect can fail. For example, in $S_9$ let
$g = (123)(45)(6789)$. This has order $12 = 3 \cdot 4$.
Here are two different ways to write $g$ as a product $hh'$  where $h$ has order $3$ and $h'$ has order $4$: $h = (123)$, $h' = (45)(6789)$, and $h = (124)$, $h' = (2345)(6789)$. The first choice of $h$ and $h'$ commutes, while the second choice does not.
