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$\begingroup$

$H$ is not normal group, take $x \in G$ such that $Hx \ /xH$. That is, there is $h ∈ H$ such that $hx \not \in xH$. Then $hxH \not = H$, and, if the same definition were to work, supposedly $hH ∗ xH = (hx)H \not = xH$ But, on the other hand, since $hH = eH$, $hH ∗ xH = eH ∗ xH = (ex)H = xH$ That is, if H is not normal, this apparent definition is in fact not well-defined. ($H$ is a subgroup of $G$)

What is $Hx \ /xH$?

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    Are you sure it's not a misprint of $Hx \not\color{red}{=} xH$?2012-08-19
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    Another rationale: it wouldn't make grammatical sense to say "[noun] is such that [noun]."2012-08-19
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    That's what I thought. Maybe misprint in the book....2012-08-19
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    @Lucy: your accounts have been merged. Please register to avoid problems with creating duplicate accounts in the future.2012-08-19

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I think it's just a misprint of $Hx \not\color{red}{=} xH.$

Rationale: I assume this is a proof by contradiction that $H$ is a normal subgroup. But the definition of a normal subgroup requires that $Hx = xH.$ So it makes sense to start the proof by contradiction by assuming that $Hx \neq xH$, and then showing a contradiction from there.

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    If you're wondering; it's `\not\color{red}{=}` :)2012-08-19
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    Prove that it's _not_ a normal subgroup? But if we assume $Hx \ne xH$ and we find contradiction, that implies that $Hx = xH$ and so it _is_ a normal subgroup!2012-08-19
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    @mike4ty4 it's a typo "prove it's a normal subgroup". Thanks.2012-08-19