# Where am I wrong that I can prove every subgroup is normal?

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.

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