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Let $G$ be a group, and $a,b\in G$. Suppose $\operatorname{ord}(a)=m, \operatorname{ord}(b)=n$, and that $ab=ba$. Prove that there is an element $c\in G$ such that $\operatorname{ord}(c)=\operatorname{lcm}(m,n)$.

Let $A=\operatorname{lcm}(m,n)$. I consider $(ab)^A=a^Ab^A=1$, so the order of $ab$ divides $A$. What can we do to find an element with order exactly $A$?

user26857
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Mika H.
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  • It would be useful if you divide this into cases : when $m=n$ $m,n$ are co prime $m$ and $n$ have same primes in its factorization.. I am not able to write down all possibilities but that would be one way of looking at your question... Suppose $m$ and $n$ are coprime i am sure it would be not difficult to find element of order $mn$ –  Jan 12 '14 at 16:30
  • I think you need to use the least part of the least common multiple. – Viktor Vaughn Jan 12 '14 at 16:30
  • @PrahladVaidyanathan I think it's not duplicate. In your link you have an abelian group; here we only have $ab=ba$. – Mika H. Jan 12 '14 at 17:03

3 Answers3

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Theorem $\ $ A finite abelian group $\rm\:G\:$ has an lcm-closed order set, i.e. with $\rm\: o(X) = $ order of $\rm\: X$

$\rm\quad\quad\quad\quad\quad\quad\ X,Y \in G\ \ \Rightarrow\ \ \exists\ Z \in G\!:\,\ o(Z) = lcm(o(X),o(Y))$

Proof $\ \ $ By induction on $\rm\ o(X)\ o(Y)\:.\ $ If it is $\,1\,$ then trivially $\rm\,Z = 1.\,$ $\ $ Otherwise

write $\rm\ o(X)\ =\ AP,\: \ o(Y) = BP',\ \ \ P'\!\mid P = p^m > 1,\ \ $ prime $\rm\: p\:$ coprime to $\rm\: A,B$

Then $\rm\: o(X^P) = A,\ \ o(Y^{P\,'}) = B\:.\ $ By induction there's a $\rm\: Z\:$ with $\rm \: o(Z) = lcm(A,B)$

so $\rm\ o(X^A\: Z)\: =\: P\ lcm(A,B)\: =\: lcm(AP,BP')\: =\: lcm(o(X),o(Y)).\ \ $ QED

Bill Dubuque
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Let $m=\prod_{i=1}^t p_i^{m_i}$ and $n=\prod_{i=1}^t p_i^{n_i}$ where $p_1,p_2,\cdots,p_t$ are distinct primes and $m_i,n_i \geq 0$.

Furthermore, may assume that $m_i < n_i$ if $1 \leq i \leq l$ and $m_i \geq n_i$ if $l+1 \leq i \leq t$.

Let $c=\prod_{i=1}^l p_i^{m_i},d=\prod_{i=l+1}^t p_i^{n_i}$.

We can conclude that $o(a^c)=\prod_{i=l+1}^t p_i^{m_i}$ and $o(b^d)=\prod_{i=1}^l p_i^{n_i}$, clearly that $(o(a^c),o(b^d))=1$ and $a^cb^d=b^da^c$. Hence the order of $a^cb^d$ is $\prod_{i=1}^l p_i^{n_i} \cdot \prod_{i=l+1}^t p_i^{m_i}=[m,n]$, which completes the proof.

Remark: The method above can also be used to prove the existence of an element in $G$ with order $(m,n)$.

Xucheng Zhang
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Since $a$ and $b$ commute, we know that $(ab)^k = a^kb^k$ for any integer $k$.

As you pointed out, letting $A = \text{lcm}(m,n)$, we know that the order of $ab$ is at most $A$ (and divides $A$).

So, since we don't have any group elements running around besides products of $a$'s and $b$'s, let's try showing that we must have ord($ab$)=$A$. That is, no smaller divisor of $A$ will do the trick. Raising $ab$ to a power that's a multiple of $m$, we see

$(ab)^{mk} = ((ab)^m)^k = b^k$ = 1 if and only if $n$ divides $k$.

A similar argument should tell you about a relationship between $m$ and $k$, and this should give you want you want. Think about it, and I can provide more details if necessary.

pjs36
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    This seems to imply that $ab$ works, but it doesn't. Just take $a = b = (12) \in S_3$ – Prahlad Vaidyanathan Jan 12 '14 at 16:41
  • This is fair. Can't we assume that $a \neq b$ since otherwise, we can simply take $a$ to be the element of order $A$? – pjs36 Jan 12 '14 at 17:58
  • Assuming lcm($n,m$) $\neq n,m$ is still not sufficient. Consider the case that $G$ is the cyclic group of order $N$ generated by $c$ and we take $a=c^{N/n}$, $b=c^{N/m}$: then ord($ab$) $=$ ord($c^{(\frac{N}{n}+\frac{N}{m})}$) $= \frac{N}{g}$, where $g=$ gcd($N,\frac{N}{n}+\frac{N}{m}$), and this may not be lcm($n,m$). In particular, taking $N=120$, $n=10$, and $m=6$, we find lcm($n,m$)=30, while ord($ab$)=ord($c^{32}$)$=15$. – jawheele Mar 10 '19 at 20:56