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For any integers $m,n$ , both greater than $1$ , does there exist a group $G$ with elements $a,b \in G$ such that $o(a)=m , o(b)=n$ but $ab$ has infinite order ?

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    $\begingroup$ @yes , but I said any integer $m,n>1$ ... $\endgroup$
    – Souvik Dey
    Commented Dec 28, 2014 at 10:08
  • $\begingroup$ @SouvikDey: in that case $m=2$, what is th problem ? $\endgroup$
    – mesel
    Commented Dec 28, 2014 at 10:21
  • $\begingroup$ @mesel: but this doesn't assure we can always find a group with required elements for arbitrary integers $m,n >1$ ; there is a result stating "For any integers $m,n,r $, all greater than $1$ , there is a finite group $G$ with elements $a,b\in G$ such that $o(a)=m,o(b)=n,o(ab)=r$ perhaps you can see my motivation ... $\endgroup$
    – Souvik Dey
    Commented Dec 28, 2014 at 10:28
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    $\begingroup$ @SouvikDey: I got your point. Interesting question. $\endgroup$
    – mesel
    Commented Dec 28, 2014 at 10:32
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    $\begingroup$ @SouvikDey The word `any' is very ambiguous. Although we understand what you meant, mesel's interpretation of your question was valid. A better wording would "Let $m$ and $n$ be given integers greater than $1$. Does there exist ...". I often advise people to avoid the word "any" completely when writing mathematics (although I don't always follow my own advice). $\endgroup$
    – Derek Holt
    Commented Dec 28, 2014 at 11:35

1 Answer 1

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Take free product $A*B$ of the cyclic groups $A = \langle a\rangle, B = \langle b\rangle$ of order $m$ and $n$ respectively and thoughtfully look at the element $ab \in A * B$.

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