# If $ab=ba$ what are the possible orders of $ab$?

The question is:
Let $$a$$ and $$b$$ be two elements of in a finite group $$G$$, say $$o(a) = m$$ and $$o(b) = n$$. If $$ab = ba$$, determine all possible values of $$o(ab)$$.(Assume $$o(x)=$$the order of x)

Proof. Assume $$(m,n)=d$$, $$m=dm'$$ and $$n=dn'$$, in the way that $$(m',n')=1$$. \begin{align} &d = p_1^{\alpha_1}\dots p_a^{\alpha_a}q_1^{\beta_1}\dots q_b^{\beta_b}t_1^{\gamma_1} \dots t_c^{\gamma_c}\\ &n' = p_1^{\alpha_1'}\dots p_a^{\alpha_a'}.n'', (n'',d)=1, \alpha_i' \ge 1\\ &m' = q_1^{\beta_1'}\dots q_b^{\beta_b'}.m'', (m'',d)=1, \beta_i' \ge 1 \end{align} In fact, we break $$d$$ to three parts. First, primes that $$n'$$ posses. Second, primes that $$m'$$ posses. We use the fact that these two numbers do not have any primes in common too.($$(n',m')=1$$).
We want to find for what $$r$$ we have: $$(ab)^r=1$$ $$$$ab=ba \rightarrow (ab)^r=a^rb^r=1\implies a^r=(b^{-1})^r$$\tag{1}\label{eq1}$$

We know $$b$$ has order n then $$b^{-1}$$ has the same order too. If we have $$a^r=(b^{-1})^r$$, so the we should have $$o(a^r)=o((b^{-1})^r)$$. Moreover, We know order of elements in the cyclic group $$\langle a \rangle$$ and $$\langle b^{-1} \rangle$$. $$$$\begin{cases} o(a^r)=\frac{m}{(m,r)}\\[2ex] o((b^{-1})^{r})=\frac{n}{(n,r)} \end{cases} \xrightarrow{\eqref{eq1}} \frac{m}{(m,r)}=\frac{n}{(n,r)}\implies m.(n,r)=n.(m,r)$$\tag{2}\label{eq2}$$ $$$$m.(n,r)=n.(m,r) \rightarrow dm'.(dn',r)=dn'(dm',r)\implies m'.(dn',r)=n'(dm',r)$$\tag{3}\label{eq3}$$

As a result, we should have $$m' |(dm',r)$$, and this means that $$m' |r$$. With the symmetry:

$$$$\begin{cases} m' |r\\[2ex] n' |r \end{cases} \xrightarrow{(m',n')=1} n'm'|r \implies r=n'm'r'$$\tag{4}\label{eq4}$$ $$$$m'.(dn',r)=n'(dm',r) \xrightarrow{ r=n'm'r'} (d,r'm')=(d,r'n')$$\tag{5}\label{eq5}$$

Lemma 1:$$r|dn'm'$$.
Proof. As we have $$(ab)^{dn'm'}=a^{nm'}b^{mn'}=1_G$$. Thus $$o(ab)|dn'm'$$.$$\blacksquare$$

As we proved that $$r=n'm'r'$$, and by the use of Lemma 1: $$$$r=n'm'r'|dn'm' \rightarrow r'|d$$\tag{6}\label{eq6}$$ Consequently, we can write $$r'$$ in the following format: $$$$r'=p_1^{\zeta_1}\dots p_a^{\zeta_a}q_1^{\eta_1}\dots q_b^{\eta_b}t_1^{\mu_1} \dots t_c^{\mu_c}\\ \zeta_i \le \alpha_i , \eta_i \le \beta_i,\mu_i \le \gamma_i \\$$\tag{7}\label{eq7}$$ Now we want to use the $$\eqref{eq5}$$ to prove that $$\zeta_i=\alpha_i$$ and $$\eta_i=\beta_i$$. $$$$\begin{cases} (d,r'm')= p_1^{\min(\alpha_1,\zeta_1)} \dots p_1^{\min(\alpha_a,\zeta_a)} q_1^{\min(\beta_1,\eta_1+\beta_1^{'})} \dots q_b^{\min(\beta_b,\eta_b+\beta_b^{'})} t_1^{\mu_1} \dots t_c^{\mu_c}\\[2ex] (d,r'n')= p_1^{\min(\alpha_1,\zeta_1+\alpha_1^{'})} \dots p_1^{\min(\alpha_a,\zeta_a+\alpha_a^{'})} q_1^{\min(\beta_1,\eta_1)} \dots q_b^{\min(\beta_b,\eta_b)} t_1^{\mu_1} \dots t_c^{\mu_c}\\ \end{cases} \xrightarrow{\eqref{eq5}}$$\tag{8}\label{eq8}$$

$$$$\begin{cases} \min(\alpha_i,\zeta_i+\alpha_i')=\min(\alpha_i,\zeta_i)= \zeta_i; \forall 1 \le i \le a\\[2ex] \min(\beta_j,\eta_j+\beta_j')=\min(\beta_j,\eta_j) = \eta_j\; \forall 1 \le j \le b \end{cases} \xrightarrow{\beta_j',\alpha_i' \ge 1} \begin{cases} \alpha_i=\zeta_i\\[2ex] \beta_j =\eta_j \end{cases}$$\tag{9}\label{eq9}$$ In the above equation we use the facts that $$r'|d$$ and $$(m'',d)=(n'',d)=1$$. At the end we can conclude the following statement: If $$ab=ba$$, $$o(a)=m$$ and $$o(b)=n$$. If $$o(ab)=r$$ then $$r$$ is the following format: \begin{align} &d = p_1^{\alpha_1}\dots p_a^{\alpha_a}q_1^{\beta_1}\dots q_b^{\beta_b}t_1^{\gamma_1} \dots t_c^{\gamma_c}\\ &n' = p_1^{\alpha_1'}\dots p_a^{\alpha_a'}.n'', (n'',d)=1, \alpha_i' \ge 1\\ &m' = q_1^{\beta_1'}\dots q_b^{\beta_b'}.m'', (m'',d)=1, \beta_i' \ge 1\\ &\implies r=n'm'p_1^{\alpha_1}\dots p_a^{\alpha_a}q_1^{\beta_1}\dots q_b^{\beta_b}t_1^{\mu_1} \dots t_c^{\mu_c}, \mu_i \le \gamma_i \end{align}

Questions:

1. Is this correct or am I missing something?
2. Are any other restrictions that can be added?
3. Is there any way to check this orders exist or not?
• The proof is much more easier. Just consider that $(ab)^l = a^lb^l$. If $l = lcm(m,n)$, then $(ab)^l=a^lb^l=$. Then $o(ab) | lcm(o(a), o(b))$ Commented Aug 20, 2021 at 15:11
• Don't use ^{'} in math mode: it makes it hard to read. The prime automatically becomes an "exponent", and normally there is no reason to further raise it and shrink it. Compare a', which renders as $a'$, with a^{'}, which yields $a^{'}$. Even multiple primes are not a problem: a'' yields $a''$. Commented Aug 20, 2021 at 15:16
• Also, \min will produce $\min$, in roman typeface, instead of math italic Commented Aug 20, 2021 at 15:28
• Not sure the duplicate target actually answers the full question: for one thing, it explicitly does not establish sufficiency. Commented Aug 20, 2021 at 16:03
• @ Shaun math.stackexchange.com/questions/1554431/… just reached the same answer. Commented Aug 20, 2021 at 16:26

I'm having a hard time following your argument/notation. It is simpler if we work one prime at a time.

Say $$m=p^{\rho}$$, $$n=p^{\sigma}$$, and without loss of generality assume that $$0\leq\rho\leq \sigma$$.

It is clear that the order of $$ab$$ divides $$p^{\sigma}$$, since $$a^{p^{\sigma}}=b^{p^{\sigma}}=1$$.

Claim 1. If $$\rho\lt\sigma$$, then the order of $$ab$$ is $$\mathrm{lcm}(m,n) = p^{\sigma}$$.

Proof. Suppose that $$(ab)^{p^t} = e$$, with $$t\leq \sigma$$. Then $$b^{p^t}=a^{-p^t}$$. Now, the order of $$a^{p^t}$$ is $$\max(p^{\rho-t},1)$$, while the order of $$b^{p^t}$$ is $$p^{\sigma-t}$$. Since $$\rho-t\lt\sigma-t$$ and $$1\leq p^{\sigma-t}$$, the only way we can have $$\max(p^{\rho-t},1)=p^{\sigma-t}$$ is if $$1=p^{\sigma-t}$$, which means $$t=\sigma$$, as required. $$\Box$$

You can even achieve this within the same cyclic group: take $$C_{p^{\sigma}}$$, cyclic group of order $$p^{\sigma}$$ generated by an element $$z$$; take $$b=z$$, and $$a=z^{p^{\sigma-\rho}}$$.

Claim 2. If $$\rho=\sigma$$, then the order of $$ab$$ may be any of $$p^t$$, $$0\leq t\leq \sigma$$.

Proof. Fix $$t$$ with $$0\leq t\leq \sigma$$. Let $$C_{p^t}$$ be the cyclic group of order $$p^t$$ with generator $$x$$, and let $$C_{p^{\sigma}}$$ be the cyclic group of order $$p^{\sigma}$$ with generator $$y$$. Let $$a=(x,y)$$ and $$b=(1,y^{-1})$$. Then both $$a$$ and $$b$$ have order $$p^{\sigma}$$, and $$ab=(x,1)$$ has order $$p^t$$, as required. $$\Box$$

Now, if I follow your notation correctly: the $$p_i$$ are the primes that occur in both $$n$$ and $$m$$, but occur in $$m$$ to a strictly smaller power than they do in $$n$$. The $$q_i$$ are the primes that occur in both $$n$$ and $$m$$, but occur in $$n$$ to a strictly smaller power than they do in $$m$$. And the $$t_i$$ are the primes that occur in both, to the exact same power. Meanwhile $$n''$$ consists of primes that occur in $$n$$ but not in $$m$$ (so, to a "different power"), and symmetrically with $$m''$$.

So, doing it one prime at a time as I do above, you get that your only leeway is in the power of the $$t_i$$, and those may be any quantity between $$0$$ and the largest power that divides $$\gcd(n,m)$$.

This is exactly what you have at the end. I can't quite follow all your calculations (too many indices, subindices, etc, too little time), but the answer to 2 is "No, that's the only restrictions you get", and the answer to 3 is "You construct explicit examples."

I've shown above how to construct specific examples for each prime, so you can then just take the direct product of each of the examples to get an example for orders $$n$$ and $$m$$.

• Thanks. Yes it has lots of indices. I could not find an easier way to write it. As I understand, we can make groups that the possible order works in it. Even though we cannot be sure for the specific group that we are working on has that order or not. Commented Aug 24, 2021 at 13:29
• @Janbazif From just the information that $a$ and $b$ commute, and their orders, this is the best you can say. Other information may help. For example, knowing about $\langle a\rangle\cap\langle b\rangle$ may help. Commented Aug 24, 2021 at 14:17