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A non-abelian group of order $ 6 $ is isomorphic to $ S_3 $

i have taken from

A non-abelian group of order $ 6 $ is isomorphic to $ S_3 $

as i did n't undersatnds the enter image description here red line not getting in my head

pliz help me,,,,,,

jasmine
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1 Answers1

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You have $x^2=y^{-2}$, hence $x^2$ is generated by $y$ and hence $x^2 \in \langle y \rangle$. So if for example we assume $x^3=e$ then we get $e=x^3=xx^2 \in x\langle y \rangle$. But we have a problem here because the identity does not belong to any left coset of $\langle y \rangle$ which is not $\langle y \rangle$ itself. That is a contradiction. Same idea for $x^k$ with any odd $k$.

Mark
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  • @ Mark.....one more doubts why identity doesnot belong to any left coset of ??.. – jasmine Jul 28 '18 at 21:19
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    Let's assume it does belong to $x\langle y \rangle$. Then we have $e=xy^k$ for some k. Then we can multiply both sides by the inverse of $y^k$ and get $y^{-k}=x$ and hence $x \in \langle y \rangle$ which is a contradiction. – Mark Jul 28 '18 at 21:24
  • thanks u @Mark. – jasmine Jul 28 '18 at 21:28