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Let $G$ be a non-abelian finite group. If $a\in G$ has order $2$ and $b\in G$ has order $3$, what can we say about the order of $ab$?

I know that, in the abelian case, $o(ab)=6$. In the non-abelian case, can we say $o(ab)\geq k$ for some $k>1$? My take would be that $k\geq 2$ since, if we take the relation $ba=ab^{-1}$ then $baba=e$, where $e$ is the identity.

What do you think?

Adrian Leverkuhn
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  • Well, the only way $k≥2$ could be false is if $k=1$ which would mean that $ab=e$, which is not possible. – lulu Aug 03 '21 at 12:45
  • I think the order of $ab$ could be anything $\ge2$ – a1402 Aug 03 '21 at 12:45
  • thanks @DietrichBurde, what about a lower bound on the order? Do you know if 2 is correct? – Adrian Leverkuhn Aug 03 '21 at 12:47
  • Note: the problem specifies that $G$ is finite, so the order could not be infinite. – lulu Aug 03 '21 at 12:47
  • @a1402 do you think it is correct that the lower bound on the order is 2? – Adrian Leverkuhn Aug 03 '21 at 12:51
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    Yes - see Dietrich Burde's answer here – a1402 Aug 03 '21 at 12:52
  • @a1402 Thank you, I was just searching my answer here. In the above question, we can just take $G=S_3$ and $a$ a transposition and $b$ a $3$-cycle. Then $ab$ cannot have order $6$, for example, because $S_3$ is not cyclic. – Dietrich Burde Aug 03 '21 at 12:54
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    I am not convinced that the duplicate really answers the question. The answer (as given by @a1402) is that $ab$ can have order $k$ for any $k \ge 2$. In fact, given any integers $m,n,k \ge 2$, there is a finite nonabelian groups with elements $a,b$ such that $a$, $b$ and $ab$ have orders $m,n$ and $k$, respectively. – Derek Holt Aug 03 '21 at 13:15
  • @DerekHolt You are right, the link of a1402 is the correct duplicate and exactly says what you have commented above. I cannot change the duplicates. Could you do this? – Dietrich Burde Aug 03 '21 at 13:22

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