Commutator of a group is identity..? If $H$  is a normal subgroup of $G$, $[G:H]=2$. For $x, y \notin H$ then $xyx^{-1}y^{-1}$=id?
Is this true in general?
If so how do we prove this?
 A: This isn't true in general. Consider $G=S_5$, the symmetric group of $5$ elements and let $H=A_5$, then $[G:H]=2$, but consider $x=(1 2 3 4 5)$, $y=(123)$, then 
$$(1234)(13)(1432)(13)=(13)(24)\neq e.$$
A: No, this is not true in general.
Consider the symmetric group $S_3 = \lbrace (), (1,2), (2,3), (1,3), (1,2,3), (1,3,2)\rbrace$, of order 6.
The cyclic subgroup (say) $H = \lbrace (), (1,2,3), (1,3,2)\rbrace$ is normal in $S_3$ and $[S_3 : H] = 2$.
If we take $x = (1,2),y = (2,3) \notin H$, then $xy = (1,2)(2,3) = (1,2,3)$ but $yx = (2,3)(1,2) = (1,3,2) \ne (1,2,3)$.
Thus, $xy \ne yx \Rightarrow xyx^{-1}y^{-1} \ne id$.
Ref: http://groupprops.subwiki.org/wiki/Symmetric_group:S3
A: This question has already an answer, actually it can be showed that it is true for only abelain groups.
Assume $[G:H]=2$ and $xyx^{-1}y^{-1}=e$ for $x,y\notin H$. 
So, let $r\notin H$ then $C_G(r)$ must contains every element not in $H$, i.e $G-H\subset C_G(r)$ and it can not be equal as left hand side is not a group. 
So, $$\dfrac {|G|}{2}<C_G(r)\implies C_G(r)=G\implies r\in Z(G)$$
But it means that $G-H\subset Z(G)$ with same reasoning we must have $G=Z(G)$
So, $G$ must be abelian group.
