Could someone please explain something about the following proof? It is based on the proof (https://math.stackexchange.com/q/84663).
If $H\leq G$ and $[G:H]=2$, prove that $H$ is normal in $G$.
Proof:
If $[G:H]=2$, then there are two distinct left cosets $H$ and $G\setminus H$. This is because left (right) cosets partition $G$. Let $x\in G$.
Case 1: $x\in H$. Then $xH=H=Hx$.
I don't understand where $H=Hx$ comes from. Is it similar to the proof for "if $x\in H$, then $xH=H$"?
For Case 2: $x\notin H$, the proof for $xH=G\setminus H$ is similar to that of "if $x\in H$, then $xH=H$" as well?
The proof I am basing this on is using "if right and left cosets are equal, then $H$ is normal", but can the same proof be applied for when index is only known and equal to $2$?
Best Answer
Actually, $xH=H$ when $x\in H$ has nothing to do with the index of $H$ in $G$. It is true for any subgroup, because the definition of a subgroup implies $xH\subset H$. The reverse inclusion is true, because any $h\in H$ can be written as $x(x^{-1}h)$, and $x^{-1}\in H $ if $x$ does, so $x^{-1}h\in H$.
Similar proof for $Hx=H$ if $x\in H$.
As to the case $x\notin H$, just observe that $xH\varsubsetneq H$ since $x\notin H$, so it is the other coset $G\setminus H$. For the same reason, $Hx= G\setminus H$, so $xH=Hx$.