If $a$ and $b$ commute and $\text{gcd}\left(\text{ord}(a),\text{ord}(b)\right)=1$, then $\text{ord}(ab)=\text{ord}(a)\text{ord}(b)$.

Question

Prove if $\operatorname{ord}(a)=m$, $\operatorname{ord}(b)=n$, and $\operatorname{gcd}(m,n)=1$, then $\operatorname{ord}(ab)=mn$.

I was reading this and was thinking how this proof would look like. I tried to do it and am not sure if this is correct. Here is what I did:

If $a$ and $b$ commute then $(ab)^{mn} = a^{mn} * b^{mn} = (a^m)^n * (b^n)^m = 1 * 1 = 1.$ So $ord(ab) | mn$.

Now, take $k = \operatorname{ord}(ab) = m^\prime * n^\prime$ where m' is relatively prime with $n$ and $n'$ is relatively prime with $m$. By the result above $m' | m$ and $n' | n$. Now we have $((ab)^k)^{m/m'} = 1$ since $(ab)^k=1$. But on the other hand, by the commutativity:

$$((ab)^k)^(m/m') =$$

$$((ab)^(m' * n'))^(m/m') =$$

$$ a^{n'm} * b^{n'm} = $$

$$(a^m)^{n'} * b^{n'm'}=$$

$$b^{n'm'} = 1$$

This implies that $n' * m'$ is divisible by $n$. But $m'$ is relatively prime with $n$, so we must have $n' = n.$ By symmetry, $m' = m$. So $ord(ab) = mn$.

Just to say this again, I want to prove if $a$ and $b$ commute and $m$ and $n$ are relatively prime then

$$ord(ab)=mn.$$

The comments are vague. I guess this must be the answer then and there probably is not another way to do this.

Answer 1

Proof $\rm\ (ab)^k\! = 1\Rightarrow a^k\! = b^{-k}\! =\color{#c00}c \in \langle a\rangle\!\cap\!\langle b\rangle\Rightarrow ord\,c\mid m,n\,$ $\rm \Rightarrow\, ord\,c\mid(m,n)\!=\!1\,\Rightarrow\, \color{#c00}{c\! =\! 1},\,$ thus $\rm\ a^k\! = 1 = b^k\,$ thus $\rm\, m,n\mid k\:\Rightarrow\:\ell \!=\! lcm(m,n)\mid k.\ $ Conversely $\rm\:m,n\mid \ell \:\Rightarrow\:(ab)^\ell\! = 1.$

Answer 2

Hint/roadmap:

If $|a|$ and $|b|$ are coprime and $a$ and $b$ commute, then $|ab|=|a||b|$. In particular, this holds for all pairs of elements in abelian groups.

This follows from these facts:

1. if $ab=ba$, then $(ab)^x=a^xb^x$, $\hspace{17pt}$(note: this is the step which fails without commutativity)

2. if $g$ is an element in any group $G$ and $g^x=\operatorname{id}_G$, then $|g|$ divides $x$,

3. if $|a|$ and $|b|$ are coprime, the smallest number divisible by both $|a|$ and $|b|$ (the least common multiple, one might say) is $|a||b|$.

From these facts, one may show that the smallest number $x$ for which $(ab)^x$ is the identity is $|a||b|$, hence $|ab|=|a||b|$.

Answer 3

$ord_pa=m\iff a^m\equiv1\pmod p$ and $ord_pb=n\iff b^n\equiv1\pmod p$ where $n$ is any integer

$\implies a^{lcm(m,n)}\equiv1\pmod p,b^{lcm(m,n)}\equiv1\pmod p$

$\implies (ab)^{lcm(m,n)}\equiv1\pmod p\implies ord_p(ab)$ divides lcm$(m,n)$

Conversely, let $ord_p(ab)=h$ and $(m,n)=d$ and $\frac mM=\frac nN=d$

As $(ab)^h\equiv1\pmod p\implies (ab)^{mh}\equiv1,$ $\implies (a^m)^h\cdot b^{mh}\equiv1$

$\implies b^{mh}\equiv1\implies n$ divides $mh$ as $ord_pb=n$

$\implies Nd$ divides $Mdh \implies N$ divides $Mh \implies N$ divides $h$ as $(M,N)=1$

Similarly, $M$ divides $h\implies $lcm$(M,N)$ divides $h=ord_p(ab)$

But $ord_p(ab)$ divides lcm$(m,n)\implies ord_p(ab)=$ lcm$(m,n)$

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

Answer 4

Say $a_1$, $\ldots$, $a_n$ commuting elements of a group $G$, $\operatorname{ord}(a_i) = m_i$, and $(m_i, m_j) = 1$ for $i\ne j$. Then $$\operatorname{ord}(a_1 \cdots a_n) = m_1 \cdots m_n$$

Indeed we have $$(a_1 \cdots a_n) ^{m_1 \cdots m_n} = 1$$

Then, take any $m$ such that $(a_1 \cdots a_n)^{m} = 1$. Raising to power $m_2 \cdots m_n$ we get $a_1^{m m_2 \cdots m_n} = 1$, so $m m_2 \cdots m_n \vdots m_1$. Now use $(m_1, m_2 \cdots m_n) = 1$. We conclude $m \vdots m_1$. Similarly $m \vdots m_i$ for all $i$. Again use $m_i$ relatively prime and get $m \vdots m_1 \cdots m_n$

$\bf{Added:}$ There is still something to be said about the order of the product of commuting elements $a_1$, $\ldots$, $a_n$. Assume that $p$ is a prime number such that the $p$-adic valuation $v_p(\cdot)$ achieves its maximum $e$ at a unique $a_i$. Then the order of $a_1 \cdots a_n$ is divisible by $p^{e}$. Thus, if for each $p$ dividing some $a_i$, the maximum of $v_p$ is achieved exactly once, then the order of the product of $a_i$'s is the $\operatorname{lcm}$ of the orders of $a_i$'s.

For a proof: we know that the order of the product divides the $m\colon = \operatorname{lcm}(m_i)$. So consider

$$(a_1 \cdots a_n) ^{\frac{m}{p}}$$

After opening the brackets we will have exactly one factor that is not $1$ above. Hence the order is divisible by the largest power of $p$.

Note: if $v_p(\cdot)$ achieves its maximum at least twice then the only conclusion is $v_p(m)\le \max_{1\le i \le n} (v_p(m_i))$

Answer 5

Suppose $(ab)^t=a^t b^t=1$. Then $t/mn$. Note that $t$ cannot be a multiple of $m$ and not of $n$, since then $a^tb^t=b^t \neq 1$. Now suppose $t$ is neither a multiple of $n$ nor a multiple of $m$. Then let $n=p_{1}^{\alpha_1}...p_k^{\alpha_k}$ and $m=q_1^{\beta_1}...q_n^{\alpha_n}$. Then $t$ takes a proper piece $r$ of $n$ and a proper piece $s$ of $m$.Since $x^t=y^{-t}$, $ord (x^t)=ord (y^{-t})=ord (y^t)$, but $ord (x^t)$ is what $r$ lacks from $n$(which are a product of $p_{k}$'s), while $ord (y^t)$ is a product of $q_i$'s, which is a contradiction because $(n,m)=1$.

Answer 6

Recall the following elementary group theory,

$\quad$ (For all $m \gt 0$) $ $ [$\alpha^m =e \text{ iff } \text{ord}(\alpha)$ divides $m$]

$\quad$ (For all integer $s,t \gt 1$) $ $ [$\text{ord}(a) = st \text{ implies } \text{ord}(a^s) = t$]

which we use without mention in the following proof.

Proposition: If $ab = ba$ and $\operatorname{gcd}(\operatorname{ord}(a),\operatorname{ord}(b))=1$ then
$\operatorname{ord}(ab)=\operatorname{ord}(a)\operatorname{ord}(b)$.
Proof(sketch):
We can prove this using Fermat's method of descent.

To begin, observe that

$$ (a\cdot b)^{\text{ord}(a) \cdot \text{ord}(b)} = a^{\text{ord}(a) \cdot \text{ord}(b)} \cdot b^{\text{ord}(a) \cdot \text{ord}(b)} = e$$

Logic quickly hones in on the case we must descend from:

$\tag 1 1 \lt \text{ord}(ab) \lt \text{ord}(a) \cdot \text{ord}(b) \; \land \; \text{ord}(ab) \mid \,\bigr{(}\text{ord}(a) \cdot \text{ord}(b)\bigr{)}$

It is a simple argument to show that $\text{ord}(ab)$ can't be a prime number, and that is key since we'll need at least two factors in $\text{ord}(ab)$ so we can keep descending in $\text{(1)}$. The argument analyzes the expressions

$$ (a^p \cdot b^p)^{\text{ord}(b)} = a^{p\cdot\text{ord}(b)} \cdot b^{p\cdot\text{ord}(b)} = a^{p\cdot\text{ord}(b)} =e$$

$$ (a^p \cdot b^p)^{\text{ord}(a)} = a^{p\cdot\text{ord}(a)} \cdot b^{p\cdot\text{ord}(a)} = b^{p\cdot\text{ord}(a)} =e$$

We'll leave the details to the reader.

Let $p$ be a prime factor of $\text{ord}(ab)$ (there has to be at least two now) and then descend by setting $a' = a^p$ and $b' = b^p$, returning to

$\quad 1 \lt \text{ord}(a'b') \lt \text{ord}(a') \cdot \text{ord}(b') \; \land \; \text{ord}(a'b') \mid \,\bigr{(}\text{ord}(a') \cdot \text{ord}(b')\bigr{)}$

with $\text{ord}(a') \cdot \text{ord}(b') \lt \text{ord}(a) \cdot \text{ord}(b)$.

In the descent $\text{ord}(a'b') = \frac{\text{ord}(ab)}{p}$ and $\text{ord}(a') \cdot \text{ord}(b') = \frac{\text{ord}(a) \cdot \text{ord}(b)}{p}$.

This concludes our proof by contradiction sketch.