2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ If it is true for any group, it must also be true for the cyclic group generated by x, which implies that powers of x are the only good candidates for the x_i. Which means you are certainly on the right track and moreover if you have insight into the subgroup structure of cyclic groups or use the fundamental theorem of f.g. abelian group, the result rolls out easily. (Although I suspect the purpose of this question is not to use that.) $\endgroup$ – Myself Aug 31 '14 at 22:27
  • $\begingroup$ @Myself Exactly, I'm just starting the course. In the next lecture we'll only start talking about cyclic groups and Lagrange's theorem. $\endgroup$ – Miguelgondu Aug 31 '14 at 22:28
  • 1
    $\begingroup$ Oh, btw, chinese remainder theorem might help here. To verify that $x = y \pmod n$, it it sufficient that $x = y \pmod{n_i}$ where $ n = \prod_i n_i$ with the n_i coprime. $\endgroup$ – Myself Aug 31 '14 at 22:29
  • $\begingroup$ @Myself That sounds useful indeed. Thanks. $\endgroup$ – Miguelgondu Aug 31 '14 at 22:29
1
$\begingroup$

(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 $x_1*x_2=x$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks, but $\gcd{(p_i, n)}$ would equal $n_i$, and the order would be $\frac{n}{n_i}$, not $n_i$. Or am I just confused? $\endgroup$ – Miguelgondu Aug 31 '14 at 22:24
  • $\begingroup$ @Miguelgondu oh duh, sorry. I did it backwards, and after thinking you were right from the start on that bit. I'll fix that. $\endgroup$ – Adam Hughes Aug 31 '14 at 22:26
1
$\begingroup$

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 $3$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.