If the correspodence $aHbH=abH$ defines a group operation on the set of left cosets of $H$ in $G$, then show that $H$ is normal in $G$.

Question

If the correspodence $aHbH=abH$ defines a group operation on the set of left cosets of $H$ in $G$, then show that $H$ is normal in $G$.

My attempt:

Let $x\in G$. Then we know that $xHx^{-1}H=H$.
This implies that for every $x\in G$, $\exists \,h\in H$ such that $xhx^{-1}H=H$.
This then implies that $\exists \,h\in H$ such that $xhx^{-1}\in H,\, \forall x\in G.$
Thus $xHx^{-1}\subseteq H,\, \forall x\in G.$
Thus $H$ is normal in $G$.

Is this correct? If you have a different approach, please post.

Answer 1

The first implication is right but you lost a sense of proriété The second implication is false, which allows you to recover the lost logical property the result is true.

So write better:

$xHx^{- 1} H = H \forall x \in G$

then $xhx^{- 1} H = H \forall x \in G$ and $\forall h \in H$

then $xhx^{- 1} \in H \forall x \in G$ and $\forall h \in H$

therefore H normal in G