Basic group theory 01
- Grigorios Kostakos
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Basic group theory 01
Let a group $G$ and $x,a,b\in{G}$, such that: \[x=a\,b=b\,a \quad {\text{and}} \quad a^{p}=1=b^{q}\,,\] where $p$ and $q$ are relative prime numbers. Prove that exists relative prime numbers $n$ and $m$, such that \[a=x^{n}\quad {\text{and}} \quad b=x^{m}\,.\]
Grigorios Kostakos
Re: Basic group theory 01
So we got $ \exists k , l \in \mathbb{Z} $ such that $ kp+lq=1 $. So $ kp$ and $lq $ are relatively prime (Bezout ) .
$$ (ab)^{lq}=a^{1-kp}b^{lq}=a a^{-kp}=a=x^{lq}\\
(ab)^{kp}=a^{kp}b^{1-ql}=b=x^{pk} $$
$$ (ab)^{lq}=a^{1-kp}b^{lq}=a a^{-kp}=a=x^{lq}\\
(ab)^{kp}=a^{kp}b^{1-ql}=b=x^{pk} $$
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