TITLE: Prove that a subgroup which contains half of all elements is a normal subgroup.
QUESTION [7 upvotes]: I have been going through some material on elementary group theory (Group Theory for Physicists, Christoph Ludeling) and in one place it states:
A subgroup which contains half of all elements, i.e. for which $|G|=2|H|$, is a normal subgroup.
Where a subgroup $H\subset G$ is called a normal subgroup if $gHg^{-1}=H~\forall g\in G$.
How would I go about proving this? Sorry if the answer exists somewhere, but I tried searching and could not find it.

REPLY [7 votes]: Not as slick but perhaps more direct (as given in Rotman):

It suffices to prove that if $h \in H$, then the conjugate
   $ghg^{-1}\in H$ for every $g \in G$.  If $g \in H$, then $ghg^{-1}\in
 H$, because $H$ is a subgroup.  If $g \notin H$, then $g = ah_0$,
   where $h_0\in H$ (for $G = H \cup aH$).  If $ghg^{-1} \in H$, we are
   done.  Otherwise, $ghg^{-1} = ah_1$ for some $h_1\in H$.  But $ah_1 =
 ghg^{-1} = ah_0hh_0^{-1}a^{-1}$.  Cancel $a$ to obtain $h_1 =
 h_0hh_0^{-1}a^{-1}$, contradicting $a \notin H$.

This was the proof I came up with before reading the slick proof in the text I was reading, seeing it appear in Rotman later made me feel less silly.