Let $r$, $s$, and $t$ be positive integers greater than $1$. How to prove that there exists a finite group $G$ having elements $x$ and $y$ such that $x$ has order $r$, $y$ has order $s$, and $xy$ has order $t$?
Thanks in advance.
For any $r$, $s$, $t>1$, then there's a group such that $x$ has order $r$, $y$ has order $s$ while $xy$ has order $t$?
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$\begingroup$
abstract-algebra
group-theory
finite-groups
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0Oh thanks, that's very helpful. – 2017-02-04
1 Answers
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The following theorem, proved in Milne's lecture notes on group theory, shows this result:
THEOREM 1.64 For any integers $m,n, r > 1$, there exists a finite group $G$ with elements $a$ and $b$ such that $a$ has order $m$, $b$ has order $n$, and $ab$ has order $r$.
The proof uses the groups $\mathbb{F}_q^{\times}$ and $SL_2(\mathbb{F}_q)$. Other proofs have been given implicitly here on MSE, e.g., here, or here.
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0Thank you very much. That proof seems quite simple, shame that I haven't come up with that. – 2017-02-04
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0You are welcome. I don't think it is a shame. It needs a good idea. – 2017-02-04
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0I review that proof and have problems with the existence of $q$. How exactly can we guarantee that there is a prime $q$ such that $2mnr\ |\ q-1$? – 2017-02-05