$\left| G:H \right|=2 \implies H$ is normal in $G$ I have a question to the proof of this claim: "Let $G$ be a group. Then $\left| G:H \right|=2 \implies H$ is normal in $G$. "
Here is an excerpt of the proof:
Let $G= g_1H\biguplus g_2H.$ Since $e_GH=H ,$ where $e_G=$ identity of $G$, is one of the left cosets of $H$ in $G$, then we may assume $g_1H=H$ ....
Here is my question: Given any $G = \biguplus_{i\in I}g_iH$, why is it necessary that $e_G = g_i,$ for some $i\in I ?$
Thank you. 
 A: Because $|G:H|=2$, we know there are only 2 distinct cosets. As the proof does, call the cosets $g_1H$ and $g_2H$. Since cosets are disjoint or equal and $H$ has index $2$ in $G$, we know that $G=g_1H \cup g_2H$. Since $e_G \in G$, we know that $e_G\in g_1H \cup g_2H$. Then $g_1 \in g_1H$ or $g_2H$. Of course, it cannot be in both because distinct cosets (as these are) are disjoint. Since it doesn't matter which it is in, $g_1H$ or $g_2H$, we suppose it is in one or the other. Most often the book will say, "without loss of generality, assume $e_G \in g_1H$". This just a fancy way of saying you might as well call the coset with $e_G$ in it $g_1H$. But then $g_1H=e_GH=H$.
Your question phrases it in general, when you have a union of more than two cosets, why does $e_G$ have to be in one of them. But the reason is because $e_G \in G$. If the union of some number of distinct cosets is $G$, then $e_G$ must be in one of them (because their union is $G$, the reason it is in only one is because cosets are either disjoint or equal and our union is of all distinct cosets). The "for some $i$" part just means that it is in one of them--could be $g_1H$ or $g_{24}H$, but certainly $g_iH$ for some $i$.
A: $\forall x\in gH$, we have $xH=gH$(check it) so, if $e\in gH$ for some $g\in G$, then 
$$H=eH=gH$$
$\forall g\in G\setminus H$, then 
$$G=H\cup gH=H\cup Hg$$
so
$$gH=Hg$$
If $g\in H$, $gH=Hg$. It follows that $H$ is normal in $G$
