# For any integers $m,n>1$ , 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 ?

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 ?

• @yes , but I said any integer $m,n>1$ ... Commented Dec 28, 2014 at 10:08
• @SouvikDey: in that case $m=2$, what is th problem ? Commented Dec 28, 2014 at 10:21
• @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 ... Commented Dec 28, 2014 at 10:28
• @SouvikDey: I got your point. Interesting question. Commented Dec 28, 2014 at 10:32
• @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). Commented Dec 28, 2014 at 11:35

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 $ab \in A * B$.