# Completing the Proof: If $a$ and $b$ commute, then ${\rm ord}(ab)$ is a divisor of ${\rm lcm}(m,n)$

This is a very simple question, but I am not too familiar with the deeper theory behind least common multiple (lcm) and greatest common divisor (gcd) so I'll fire away.

To provide context, in the Pinter's "A Book of Abstract Algebra", Chapter 10 Exercise E1 asks the reader to prove the following:

Given that $$a,b \in G$$ and $${\rm ord}(a)=m$$ and $${\rm ord}(b)=n$$, prove that if $$a$$ and $$b$$ commute, then $${\rm ord}(ab)$$ is a divisor of $${\rm lcm}(m,n)$$

I worked my way through the problem and demonstrated that if $${\rm ord}(ab)=q\neq 1$$, then $$q | m$$ and $$q|n$$.

To complete this, I need to show that $$q\ | \ {\rm lcm} (m,n)$$.

As I said to begin with, I am not very familiar with gcd's and lcm's...but I do know their relationship. Specifically, $${\rm lcm}(m,n)=\frac{mn}{{\rm gcd}(m,n)}$$

Using this, I will move forward as follows:

Rewrite $$\frac{mn}{{\rm gcd}(m,n)}$$ as either $$m*\frac{n}{{\rm gcd}(m,n)}$$ or $$n*\frac{m}{{\rm gcd}(m,n)}$$. I will only consider the first case (as the second case should work out similarly without loss of generality).

In $$m*\frac{n}{{\rm gcd}(m,n)}$$, the second factor $$\frac{n}{{\rm gcd}(m,n)}$$ will always be equal to an integer between $$1$$ and $$n$$ (by definition of gcd). Therefore, we have:

$${\rm lcm}(m,n) = m *a$$ where $$a \in \mathbb Z$$.

We know that if $$q | m$$, $$\exists x$$ s.t. $$q*x=m$$.

Multiplying both sides by $$a$$, we get $$q*x*a=m*a$$, which can be rewritten as $$q*z = m*a$$ where $$z=x*a$$.

Therefore, $$q |m*a$$ . Also, looking at the other case ($$n*\frac{m}{{\rm gcd}(m,n)}$$), we would get $$q | b*n$$.

Combining these two statement, I conclude that $$q \ | \ {\rm lcm}(n,m)$$, which means that $${\rm ord}(a,b) \ | \ {\rm lcm}(m,n)$$ .

Is this correct?

## 3 Answers

Below I show how this is essentially the same as the well-known proof that since $$\ell = {\rm lcm}(m,n)$$ is a common denominator of $$\,j/m,\,k/n\,$$ it is also a denominator for their sum. We write the proof in multiplicative $$(*)$$ and additive form $$(+)$$ where $$H$$ is a subgroup of $$G\ \$$ ($$H = \{1\}\,$$ in the OP).

Since we have $$\,m,n\mid \ell\,$$ we can write $$\,m\hat m = \ell = n\hat n\,$$ for some integers $$\,\hat m,\, \hat n,\$$ hence

$$(*)\ \ \ \ \ a^{\large m},\,\color{#c00}{b^{\large n} \in H}\,\Rightarrow\,(a *b)^{\large \ell}\! =\, (a^{\large m})^{\large \hat m} *\, (\color{#c00}{b^{\large n}})^{\large \hat n}\!\in\:\! H * \color{#c00}H \subseteq H$$

$$(+)\ \ \ \ ma,\color{#c00}{nb \in H}\,\Rightarrow\, \ell(a\!+\!b) = \hat m(ma)\! +\! \hat n(\color{#c00}{nb})\in H\!+\!\color{#c00}H\subseteq H,\$$ e.g

$$\,\ \ \ a = \dfrac{j}m,\ b = \dfrac{k}n\,\Rightarrow\, \ell(a\!+\!b)\in \Bbb Z + \Bbb Z\subseteq \Bbb Z\ \,$$ for $$\,\ H = \Bbb Z\subset \Bbb Q$$

i.e. $$\,\ell = {\rm lcm}(m,n)$$ is a denominator for $$\,a\!+\!b\,$$ since it is a common denominator of $$\,a\,$$ and $$\,b$$.

• Note: ee used commutativity to infer $\ (a * b)^{\ell} = a^{\ell} *b^{\ell}\$ and $\ \ell(a+b) = \ell a + \ell b.\ \ \$ Commented Nov 20, 2019 at 3:45

Much more complicated than it needs to be. For any group element $$g, g^n=1 \iff \operatorname{order}(g) \vert n$$. Note that $$k = \operatorname{lcm}(m, n)$$ is a multiple of both $$m$$ and $$n$$, so $$a^k=b^k=1$$. Thus, $$(ab)^k=a^kb^k=1 \Rightarrow \operatorname{order}(ab) \vert k = \operatorname{lcm} (m, n).$$

Let $$l={\rm{lcm}}(m,n)$$, We have $$l=rm=sn$$ for some integer $$r,s$$. we also have $$ab=ba$$, then $$(ab)^l=a^lb^l=(a^m)^r(b^n)^s=e$$. Thus $${\rm{ord}}(ab)\mid l$$.