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Why must a subgroup $H$ be a normal subgroup of $G$ so that the group operation $Hg Hf = H(gf)$ is well-defined?

In my notes I have written:

the operation must be well-defined so if $Hg_1 = Hg_2$ and $Hf_1 = Hf_2$ we must have $Hf_1g_1 = Hf_2g_2$.

But then I have written (which I do not understand):

So we have $g_2g_1^{-1} \in H$ and $f_2f_1^{-1} \in H$ (why is this) and we require $g_2f_2f_1^{-1}g_1^{-1}$ (why is this?).

If someone could explain this last line I'd be very grateful

Partey5
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  • See https://math.stackexchange.com/questions/14282/why-do-we-define-quotient-groups-for-normal-subgroups-only/14315#14315 – Arturo Magidin May 07 '19 at 17:04
  • If $Hg_1=Hg_2$ then $Hg_1g_1^{-1}=H=Hg_2g_1^{-1} $ so $g_2g_1^{-1}\in H$ – J. W. Tanner May 07 '19 at 17:06
  • Your note is about $Hg_2f_2 = Hg_1f_1$ instead of $Hf_1g_1 = Hf_2g_2$. Your first comment looks at $(Hf_1)(Hg_1)$ compared to $(Hf_2)(Hg_2)$, but your comment looks at $(Hg_2)(Hf_2) = (Hg_1)(Hf_1)$. – Arturo Magidin May 07 '19 at 17:08
  • Note that $Hx = Hy$ if and only if $xy^{-1}\in H$, if and only if $yx^{-1}\in H$. So $Hg_1=Hg_2$ if and only if $g_2g_1^{-1}\in H$. And $Hf_1=Hf_2$ if and only if $f_2f_1^{-1}\in H$. For $H(g_2f_2)$ to equal $H(g_1f_1)$, you need $(g_2f_2)(g_1f_1)^{-1}\in H$, which is the same as $g_2f_2f_1^{-1}g_1^{-1}\in H$. – Arturo Magidin May 07 '19 at 17:09

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Let's say the operation is well defined and let $g\in G,x\in H$. In this case we have:

$Hgxg^{-1}=HgHxHg^{-1}=HgHeHg^{-1}=Hgeg^{-1}=He=H$

Hence $gxg^{-1}\in H$ for all $g\in G,x\in H$. And this implies $H\trianglelefteq G$. So if $H$ is not normal in $G$ then there is no chance the operation will be well defined.

Mark
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