TITLE: $H$ is a subgroup of index $2$ in finite group $G$, then every left coset of $H$ is a right coset as well
QUESTION [7 upvotes]: I know this means that there are two cosets. I also know that one must be H itself.
This means that the remaining coset(s) must be equal for right and left. Also since there is only one possible coset, this means that for all elements, a,b in G, then $aH = bH$ and $Ha = Hb$, since each element in G acting on the subgroup must produce the same set. This means $aH = Ha$ I believe. From there $H = aHa^{-1}$, which given the information I am not sure if this good or how to use it. Any help would good.

REPLY [2 votes]: For a problem like this one, it helps to know the principle that
Any two left (right) cosets are either disjoint or equal.
This may be proved as follows:  suppose $g_1H$ and $g_2H$ are left cosets of $H$ in $G$, and $g_1H \cap g_2 H \ne \emptyset$; then there must be some $h_1, h_2 \in H$ with
$g_1 h_1 = g_2 h_2; \tag 1$
thus
$g_1 = g_2 h_2 h_1^{-1}; \tag 2$
i.e., we have
$h = h_2 h_1^{-1} \in H \tag 3$
with
$g_1 = g_2 h; \tag 4$
now if 
$g_1 h_3 \in g_1H, \tag 5$
then by (4),
$g_1 h_3 = g_2 h h_3 \in g_2 H, \tag 6$
which shows that
$g_1 H \subset g_2 H; \tag 7$
likewise, the roles of $g_1$ and $g_2$ may reversed in the above to show that
$g_2 H \subset g_1 H; \tag 8$
from (7) and (8) we conclude that
$g_1 H = g_2 H. \tag 9$
A similar demonstration works for right cosets.  We note that this Principle holds for any group $G$ and subgroup $H$, whether finite or not, and whether $[G:H]$ is finite or not.  
We exploit said Principle in the present question as follows:  if $e \in G$ is the identity element, then clearly
$eH = He = H, \tag{10}$
and if
$g \in G \setminus H, \tag{11}$
then
$gH \ne H, \tag{12}$
since
$g = ge \in gH, \tag{13}$
but by (11),
$g \notin H; \tag{14}$
therefore,
$gH \ne H, \tag{15}$
and it thus follows from our Principle that
$gH \cap H = \emptyset. \tag{16}$
Since $[G:H] = 2$, it follows that $H = eH$ and $gH$ are the only two left cosets of $H$ in $G$; thus 
$G = H \cup gH; \tag{17}$
in a similar manner we may also see that
$G = H \cup Hg, \; H \cap Hg = \emptyset, \tag{18}$
and we conclude from (16)-(18) that
$gH = Hg. \tag{19}$
We see that the above demonstration uses notions of basic set theory as well as the fact that there are only two distinct left or right cosets, which are disjoint, and one of which is always $eH = He = H$.  And obviously, a good amount of logic.