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Let $G$ be a group and let $a,b \in G$ s.t $O(a)=m$ and $O(b)=n$ and $ab=ba$. Then $O(ab)=lcm( m,n)$.

My attempt: since $ab=ba$ then $HK=KH$

$ |HK|=O(H)O(K)/O(H \cup K)$

$l=(mn)/O(H \cap K)$ $\Longrightarrow O(H \cap K). l=(mn)$ $ O(H \cap K) | gcd(m,n)$ $ \therefore mn=O(H \cap K).l \leq gcd(m,n).lcm (m,n)=mn$ $ \Longrightarrow O(H \cap K)=gcd(m,n)$ and $l=lcm(m,n)$ as this is the only possibility

What is wrong with this proof as if $b=a^{-1}$ then it doesnt work.

Abhinav Singh
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    You are correct, the result itself doesn't work as $o(ab)=1\neq lcm(m,n)$. – cr001 Nov 24 '15 at 07:48
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    There are other cases where the result isn't correct: let $a$ be an element of order $2m$ in $C_{2m}$ and let $b=a$. Then $ab=ba=a^2$ and $O(ab)=O(a^2)=m\neq\mathrm{lcm}(2m, 2m)$. – Steven Stadnicki Nov 24 '15 at 07:55
  • This might help you. – Nizar Nov 24 '15 at 07:58
  • $ |HK|=O(H)O(K)/O(H \cup K)$

    $l=(mn)/O(H \cap K)$ $\Longrightarrow O(H \cap K). l=(mn)$ $ O(H \cap K) | gcd(m,n)$ $ \therefore mn=O(H \cap K).l \leq gcd(m,n).lcm (m,n)=mn$ $ \Longrightarrow O(H \cap K)=gcd(m,n)$ and $l=lcm(m,n)$ as this is the only possibility

    What is wrong with this proof as if $b=a^{-1}$ then it doesnt work.

     – Abhinav Singh Nov 24 '15 at 08:00
  • @StevenStadnicki Why does this happen when o(a)=o(b). – Abhinav Singh Nov 24 '15 at 08:09
  • To sum up, "let $a,b \in G$ s.t $O(a)=m$ and $O(b)=n$ and $ab=ba$. Then $O(ab)=lcm( m,n)$" is wrong but "let $a,b \in G$ s.t $O(a)=m$ and $O(b)=n$ and $ab=ba$. Then $O(ab)$ divides $lcm( m,n)$" is true. – Did Nov 24 '15 at 08:15
  • Psiible dublicate of http://math.stackexchange.com/questions/67180/order-of-product-of-two-elements-in-a-group – Nizar Nov 24 '15 at 09:09

2 Answers2

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$o(a)$ is equal to $o(\langle a\rangle)=o(H).$

But, $o(ab)$ is not necessarily the order of $\langle a\rangle.\langle b\rangle=HK$.

You may track which argument is not valid in your proof.

The statement in the title is wrong. It should be

If $o(a)=m$, $o(b)=n$ and if $ab=ba$ then $o(ab)$ divides $lcm(o(a),o(b))$.

Groups
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If o(a) = m, o(b) = n and

i) ab = ba

ii) < a > I < b > = {e} 

   Then      O(ab) = l.c.m (m, n)

P. S. : I = intersection

Drish
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