# Let $a$ and $b$ commute. If $m$ and $n$ are relatively prime, then ord($ab$) = $mn$. [duplicate]

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?

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.

• Jul 26, 2019 at 4:35
• Jul 26, 2019 at 11:36
• The 2nd $\Rightarrow$ in my answer in the dupe is essentially a proof of 10.E.2, so replace that by an invocation of 10.E.2 to obtain the hinted proof. This is a well-known proof and likely occurs here many times. Jul 26, 2019 at 14:47

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

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 $$0$$?

• 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. Jul 26, 2019 at 20:05
• can we do this question this way : is to consider various powers of $a^ib^j$. where $1 < i \leq m..1<j \leq n$ all of these are distinct and $a^mb^n=e$. so order of $ab$ is $\geq mn$... Apr 17, 2023 at 16:45
• is their any proof of this : m=ord(a) and n=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 Apr 17, 2023 at 17:05
• @SophieClad You can post the questions on the site; comment sections are too small for answers Apr 18, 2023 at 0:37