3
$\begingroup$

This is exercise $10.E.4$ from Pinter:

Let $a$ and $b$ be elements of a group $G$.

Let ord($a$) = $m$ and ord($b$) = $n$.

Prove:

Let $a$ and $b$ commute.

If $m$ and $n$ are relatively prime, then ord($ab$) = $mn$.

(HINT: Use $10.E.2$.)

Here is $10.E.2$, which Pinter suggests that we use:

If $m$ and $n$ are relatively prime, then no power of $a$ can be equal to any power of $b$ (except for $e$).

I'm also going to use $10.E.1$:

If $a$ and $b$ commute, then ord($ab$) is a divisor of lcm($m$,$n$).

As well as $B.T6.i$ (Theorem $6.i$ from Appendix $B$. A fact from basic number theory.):

If gcd($m$,$n$) = 1 then lcm($m$,$n$) = $mn$

Let's begin.

We are given that $m$ and $n$ are relatively prime which means that:

$$\text{gcd}(m,n) = 1$$

By $B.T6.i$:

$$ \text{lcm}(m,n) = mn \tag{1} $$

By $10.E.1$:

$$ \text{ord}(ab)\ |\ \text{lcm}(m,n) $$

Substituting $(1)$:

$$ \text{ord}(ab)\ |\ mn $$

Which means that there is an integer $x$ such that:

$$ \text{ord}(ab) x = mn $$

This is so close! For the theorem to be true, we'd have to show that $x = 1$.

However, the original exercise statement says to use $10.E.2$.

  • Is that helpful in showing that $x = 1$?
  • Or is there some other completely different approach whereby $10.E.2$ is used?

UPDATE: Some comments regarding the answer below.

The proof uses the following fact:

If $a|c$ and $b|c$ then $lcm(a,b)|c$.

For this exercise, I wanted to only use theorems that had been presented in the book up to that point. And, I didn't seem to recall seeing this theorem. (If anyone spots this in Pinter, please comment below with the location.)

However, I did notice that the following similar fact is in Appendix B (REVIEW OF THE INTEGERS) as exercise B.9:

If $a|c$ and $b|c$ and $gcd(a,b) = 1$ then $ab|c$.

So yeah, it looks like the approach shown below is definitely a valid way to go if you want to stick to what's presented in the book.

$\endgroup$
1
$\begingroup$

$$\boxed{\textit{You can show $mn\ \big|\ \text{ord}(ab)$ using $10.E.2$:}}$$

Since $a$ and $b$ commute we can distribute common powers of $a$ and $b$: $$ e = (ab)^{\text{ord}(ab)} = a^{\text{ord}(ab)}b^{\text{ord}(ab)} $$ This implies $a^{\text{ord}(ab)} = b^{-\text{ord}(ab)}$ or, in other words, some power of $a$ equals some power of $b$. But $10.E.2$ says that since $m = \text{ord}(a)$ and $n = \text{ord}(b)$ are relatively prime, no power of $a$ can equal any power of $b$ unless those powers of $a$ and $b$ both evaluate to $e$.

So we must have $a^{\text{ord}(ab)} = e = a^m$ and $b^{-\text{ord}(ab)} = e = b^n$.

But then, $m\ \big|\ \text{ord}(ab)$. This is because $m$ is the order of $a$ i.e. the least positive integer power of $a$ that evaluates to $e$; so if any other positive integer power of $a$ evaluates to $e$ (like $\text{ord}(ab)$ here), then $m$ must divide that integer. For a similar reason, $n\ \big|\ \text{ord}(ab)$.

Therefore since both $m$ and $n$ divide $\text{ord}(ab)$, their least common multiple $\text{lcm}(m,n)$ must divide $\text{ord}(ab)$. But in our case $\text{lcm}(m,n) = mn$ so we finally have $mn\ \big|\ \text{ord}(ab)$.

Along with the result you already established $\text{ord}(ab)\ \big|\ mn$, this suggests $\text{ord}(ab) = mn$.

P.S. If you don't believe the bolded statements above, use Euclidean Division. E.g. to see $m\ \big|\ \text{ord}(ab)$, use Euclidean Division on the integer pair $\big(m, \text{ord}(ab)\big)$ to get $\text{ord}(ab) = km + r$ for some $k, r \in \Bbb Z$ with $0 \leq r < m$. Do you see why $r$ must be $0$?

$\endgroup$
  • $\begingroup$ Excellent and very clear presentation! I learned a few things from your proof. Thanks very much for explaining each part. I've updated my answer with some comments regarding your approach. $\endgroup$ – dharmatech Jul 26 '19 at 20:05

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