A question on the order of an element involving relatively primes This question is based on an exercise that comes from the second chapter of Malik's Fundamentals of abstract algebra which states as follows (I paraphrase):

Let $(G, *)$ be a group and $x\in G$. Suppose $\circ(x) = n = n_1n_2\cdots n_k$, where for all $i\neq j$, $\gcd{(n_i, n_j)} = 1$ (that is, $n_i$ and $n_j$ are relatively primes). Show that there exists $x_i$ such that $\circ(x_i) = n_i$ for all $i = 1, 2, \dots, k$, $x=x_1*x_2*\cdots*x_k$ and $x_i*x_j = x_j*x_i$ for all $i$ and $j$.

In my attempt, I manage to get only so far: 
Proof: Let's try and define $x_i$ as a power of $x$ (that is, let $x_i = x^{p_i}$). If we want $\circ(x_i) = \circ(x^{p_i}) = n_i$, we need $$\circ(x^{p_i})=\frac{n}{\gcd{(n,p_i)}}=n_i$$ Which can be achieved by defining $$p_i := \frac{n}{n_i}, p_i \in \mathbb{Z}$$
Because $*$ is closed in $G$, we know $x_i \in G$. And, also, $$x_i*x_j = x^{p_i}*x^{p_j} = x^{p_i + p_j} = x^{p_j + p_i} = x^{p_j}*x^{p_i} = x_j*x_i$$
But I fail to demonstrate that $x=x_1*x_2*\cdots*x_k=x^{\frac{n}{n_1}}*x^{\frac{n}{n_2}}*\cdots*x^{\frac{n}{n_k}}$. I notice that it is as easy as demonstrating that $$\frac{n}{n_1} + \frac{n}{n_2} + \cdots + \frac{n}{n_k} = \alpha n + 1, \alpha \in \mathbb{Z}$$ Because then $x^{\alpha n + 1} = x^{\alpha n}*x = e*x = x$.
So my questions are:


*

*Am I on the right road?

*Any tips on proving that $x= x_1*x_2*\cdots*x_k$?
EDIT: Let $n=6$, for example. We can write $6$ as $2\cdot 3$, and $\frac{6}{2}+\frac{6}{3} = 3 + 2 = 6 \neq \alpha6+1$. So it looks like I'm not on the right track.
 A: (This got too long for the comment I was writing)
This is quite good. You just need to be a bit more careful with how you use notation, you don't even define these $p_i$ until later.
By the definition of order, each $x_i$ has order $n_i$, and since the $x_i$ commute, the order of their product is divisible by $n$, and is therefore equal to $n$. If you cannot see why $x_i*x_j=x_j*x_i$ note that both products are equal to $x^{p_i+p_j}=x^{p_j+p_i}$.
Now this doesn't quite give you $x=x_1*x_2*\ldots *x_k$, in fact, all you can really conclude with that is that if $\pi=x_1*x_2*\ldots *x_k$, then since $\langle\pi\rangle\subseteq\langle x\rangle$ and they have the same order, $\pi$ generates $\langle x\rangle$, not necessarily that $x=\pi$, but for that you can use the extended Euclidean algorithm to get tweaks of the $x_i$ with the same order properties such that the product is $x$.

In a specific example, let
$$G=\Bbb Z/3\Bbb Z\times\Bbb Z/2\Bbb Z\;(\cong \Bbb Z/6\Bbb Z)$$
then use the generator $x=(1,1)$ and $n_1=2, n_2=3$ so that $x_1=(0,1), x_2=(-1,0)$ then
$$x_1*x_2=(-1,0)+(0,1)=(-1,1)\ne x$$
since the group operation is component-wise addition.
On the other hand $x_1^1*x_2^2=x$, so that you can modify $x_1, x_2$ to produce $x_1', x_2'$ so that $x_1*x_2=x$.
A: The answer to your first question is : Yes, your reasonment is perfectly correct. 
The answer to your second question is: the statement $x=x_1*x_2*\ldots *x_k$ is erronous if we assume that the number of $x_i$ equals the number of $n_i$.
Consider the case where $\circ(x)=6$ then $n=n_1n_2$ where $n_1=2$ and $n_2=3$. As you stated correctly $x_1=x^3$ has order $2$ and $x_2=x^2$ has order 3. But $x \neq x_1x_2$ since that would mean that $x=x^3x^2=x^5$ which would imply that $x$ has order $2$ and not order $6$. But what we do have, resulting from Bézouts lemma is that $-1.3+2.2=1$ which implies $x=x_1^{-1}x_2x_2$, where each factor has order $2$ or $3$.
