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I appreciate helps for figuring out the problem of the following argument. I know it's certainly not the case that every subgroup of a group is normal. But I cannot find out where is this wrong.

Let $G$ be a group and let $H$ be a subgroup of it. Then $G$ acts on left cosets of $H$ by left multiplication. Let $ \Pi : G \to \mathrm{Sym}(G/H)$ be the homomorphism that denotes the action. I think $ H = \ker(\Pi) $, because for any left coset $aH$ and any $ h \in H$, we have $ h \cdot aH = aH $ since $ h a a^{-1} \in H$, and therefore $ H \subset \ker(\Pi) $. Also if $ g \in G \setminus H$, then $g \cdot 1H = gH \neq H$, so $ H = \ker(\Pi)$. And so $H$ is the kernel of a homomorphism so it's normal.

Thanks in advance.

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    $\begingroup$ The flaw is in the part "$h.aH = aH$ since $haa^{-1} \in H$". $\endgroup$
    – TheSilverDoe
    Dec 25, 2021 at 19:37
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    $\begingroup$ The point is $haH=aH$ if and only if $a^{-1}ha\in H$, which is basically the normality condition. $\endgroup$
    – user1729
    Dec 25, 2021 at 19:53
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    $\begingroup$ Uh yes, when deducing $ ha a^{-1} \in H $, I didn't taken caution in multiplying the inverse of $a$ from them correct side. $\endgroup$
    – Kooranifar
    Dec 25, 2021 at 20:03

2 Answers 2

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Why would $h.(aH)$ be equal to $aH$? If, say, $G=S_3$, $H=\{e,(1\ \ 2)\}$, and $a=(1\ \ 3)$, then$$aH=\{(1\ \ 3),(1\ \ 2\ \ 3)\}$$and, if $h=(1\ \ 2)(\in H)$,\begin{align}(1\ \ 2)(aH)&=(1\ \ 2)\{(1\ \ 3),(1\ \ 2\ \ 3)\}\\&=\{(1\ \ 3\ \ 2),(2\ \ 3)\}\\&\ne aH.\end{align}

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You've proved that every subgroup in an abelian group is normal. Not too bad.

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