Problem statement: Let $H$ be a subgroup of $G$ such that $[G:H] = 3$ (i.e. the index) and such that there is a $x \in G \setminus H$ for which $xH = Hx$. Prove that $H$ is a normal subgroup of $G$.
Attempt: I was just trying to use the definition. We need to prove that $\forall g \in G: gH = Hg$.
So let $g \in G$. (i) If $g \in H$, then $gH = H = Hg$ and $H$ is normal.
(ii) Suppose $g \notin H$. There are two more cosets to which $g$ can belong. We know there exists a $x \in G \setminus H$ such that $xH=Hx$. This means $xH \neq H$. If then $g \in xH$, then $gH = xH = Hx = Hg$ and so $H$ is normal.
(iii) Now suppose $g \notin H$ and $g \notin xH$. This is the part I don't know how to accomplish. What can I say about this case?
Maybe I shouldn't use the definition at all for this problem? Not sure. Help is appreciated.
Best Answer
Let $x\in G\setminus H$ such that $xH=Hx$. As $x\notin H$ we have $xH\neq H$. As $[G:H]=3$ there exists a $y\in G$ such that the left cosets are
$$ \{ H, xH, yH \}.$$
Note that $H, Hx=xH$ are right cosets. As $[G:H]=3$ there is only one coset missing. As both the left and the right cosets form a partition of $G$ we conclude
$$ Hy=yH.$$
Thus, the left and the right cosets coincide and hence $H$ is normal.
Added: Let me elaborate a bit on why $Hy=yH$. We know that $y\notin H$ and $y\notin xH=Hx$. Thus $Hy$ forms another right coset. As $[G:H]=3$, we get that the right cosets of $H$ are $\{ H, Hx, Hy \}$. As both the right and the left cosets form a partition of $G$ we get
$$ Hy = G \setminus (H \cup Hx) = G\setminus (H \cup xH) = yH.$$
(we used as well the fact $Hx=xH$).