If $H$ is subgroup $G$ of index 2 then $H$ is normal subgroup of $G$ If $G$ is a group and $H$ is a subgroup of index $2$ in $G$, prove that $H$ is a normal subgroup of $G$.
Proof: We know that $i_G(H)=2$ $\Rightarrow$ subgroup $H$ has two cosets in $G$. One of the cosets is $He=H.$ Since the cosets decompose the group $G$ $\Rightarrow$ another coset is $H^c$ , complement of $H$.
It's easy to prove that if $g\in H$ $\Rightarrow$ $Hg=gH=H$. And also if $g\in H^c$ $\Rightarrow$ $gH=Hg=H^c$.
Let's consider two cases:
1) If $g\in H\subset G$ $\Rightarrow$ $gHg^{-1}=g(Hg^{-1})=gH=H$ $\Rightarrow$ $H$ is a normal subgroup of $G$.
2) If $g\in H^c\subset G$ $\Rightarrow$ $gHg^{-1}=g(Hg^{-1})=gH^c$ since $g^{-1}\in H^c$. Let's prove that $gH^c\subset H$. Taking any element $z\in gH^c$ $\Rightarrow$ $z=gh'$ for some $h'\in H^c$ and since $H^c=g^{-1}H$ $\Rightarrow$ $h'=g^{-1}h''$ for some $h''\in H$ $\Rightarrow$ $z=gh'=gg^{-1}h''=h''\in H$ $\Rightarrow$ $gHg^{-1}=gH^c\subset H$ so $H$ is a normal subgroup of $G$.
Remark: Please do not duplicate it since I would like to know the correctness of my solution. Is this proof correct? Please check it out and assess my solution. However, my proof is not short.
 A: The proof has the ingredients, but they're quite mixed up.
Especially in points 1 and 2, you should not end them with “$\Rightarrow$ $H$ is a normal subgroup of $G$”. It's the combination of the two parts that shows normality. A proof written like this would get no positive score, because it shows lack of understanding what you need to prove.
Anyway, besides this important point, the proof is unnecessarily long. After you prove that


*

*for $g\in H$, we have $gH=Hg$

*for $g\in H^c$, we have $gH=Hg$


the proof is complete, because you have $gH=Hg$ for all $g\in G$, so $gHg^{-1}=Hgg^{-1}=H$.
Why is $gH=Hg$ for $g\notin H$? There are just two right cosets and two left cosets. In both cases one coset is $H$, so the other one is $H^c$.
A: Yet another possible approach is the following.

Lemma. Let $f$ be the group operation, $H \lneq G$ and $H^c:=G\setminus H$. Then:
$$f(H^c \times H^c)=H\iff[G:H]=2 \tag 1$$
Proof.
Take $a \in H^c$ (such element exists, because $H^c\ne\emptyset$ by hypothesis); then $a^{-1}\in H^c$ and $h \in H \Rightarrow ah\in H^c$. Therefore:
$$h\in H \Rightarrow h= a^{-1}ah \in f(H^c\times H^c) \tag 2$$
whence:
$$H \subseteq f(H^c\times H^c) \tag 3$$
This was in general. Now: 
$\Leftarrow$) If $[G:H]=2$, then $a\in H^c\Rightarrow H^c=aH$; therefore, $\forall b \in f(H^c,H^c) \Rightarrow \exists h_1,h_2 \in H\mid b=ah_1ah_2$. By contrapositive, suppose $ah_1a\in H^c$; then $H^c\ni(ah_1a)(ah_1a)^{-1}=ah_1aa^{-1}h_1^{-1}a^{-1}=e$: contradiction. Then, $ah_1a\in H$ and thence $b=(ah_1a)h_2\in H$. Therefore, $f(H^c,H^c)\subseteq H$, and by $(3)$:
$$f(H^c,H^c)= H\tag 4$$
$\Rightarrow$) Coversely, if $f(H^c\times H^c)=H$, then $a,b\in H^c \Rightarrow a^2\in H \wedge ab\in H \Rightarrow a^2H=abH \Rightarrow aH=bH$, i.e. there's one coset only, further than $H$ itself: $[G:H]=2$.
$\Box$
Corollary. If $[G:H]=2$, then $H\lhd G$.
Proof. If $g\in H$, then trivially $ghg^{-1} \in H, \forall h \in H$. If $g\in H^c$, then $gh\in H^c, \forall h \in H$, and -by the Lemma- $ghg^{-1}\in H, \forall h \in G$. $\space\Box$
A: I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
This problem is the same problem as Problem 2 on p.53 in Herstein's book.
I solved this problem as follows:

Let $g\in G$.
Let $h\in H$.
If $g\in H$, then $ghg^{-1}\in H$.
If $g\notin H$, then $(ghg^{-1})(g^{-1})^{-1}=gh\in gH$.
Since $gH\cap H=\emptyset$, $(ghg^{-1})(g^{-1})^{-1}=gh\notin H$.
So, $ghg^{-1}\notin Hg^{-1}$.
Since $i_G(H)=2$, $G=H\cup Hg^{-1}$.
So, $ghg^{-1}\in H$.
Therefore $H$ is a normal subgroup of $G$.


By the way, the index of $A_n$ in $S_n$ is $2$.
So, $A_n$ is a normal subgroup of $S_n$.

