For a proof of the following theorem, I am wondering if it can be modified in the beginning steps so that it is the same for the proof of Lagrange's theorem.
Theorem: If $N$ is a subgroup of a group $G$ of index $2$, then $N$ is a normal subgroup of $G$.
It is in the beginning portion of the proof that I have some questions. The beginning of the proof I have in mind goes as follows:
"Let $a\in G$, since $[G:N]=2$, then there are two cosets of $G$. So we can further assume that $a\notin N$, hence $aN\neq N$ and so $N\cap aN=\emptyset$..."
Instead of having $N\cap aN=\emptyset$, can't we have $aN\cap bN=\emptyset$ where $a, b\in G$ and $a\neq b$, and would the proof still work. The reason I ask is that in the proof of Lagrange theorem, if given a finite group of order $n$ and a subgroup $H$ of $G$ of order $k$ then $|H|\mid |G|$. In the proof of Lagrange theorem, we can decompose $G$ into the union of its disjoint $k$ cosets: $G=a_1H \cup a_2H \cup \dots \cup a_kH$ and $H$ mpas bijectively to each $a_iH$ for $i=1,2,\dots k$, we also have $|G|=|a_1H|+|a_2H|+\dots+|a_kH|$.
Applying the same type of reasoning to the theorem, if $a\notin N$, then since $[G:N]=2$, then don't we have $G=aN \cup bN$ with $a\neq b$, $aN\cap bN=\emptyset$, and $|G|=|aN|+|bN|$.
Thank you in advance.
Since $[G:N]=2$, there is exactly two cosets of $N$ in $G$. One of them must be $eN=N.$ We have
$$n\in N\iff nN=N.$$
(Can you prove this?)
This is equivalent to
$$a\notin N\iff aN\neq N.$$
If $b\neq a$, $a\notin N$, and $aN\cup bN=G$, then, since cosets are disjoint, we must have $b\in N$, so that $bN$ is simply $N$. This holds for any such $b$; therefore, there is no loss of generality in assuming $b=e$, in which case we are back to your first proof of the theorem in question.