The product of all elements in $G$ cannot belong to $H$ 
Let $G$ be a finite group and $H\leq G$ be a subgroup of order odd such that $[G:H]=2$. Therefore the product of all elements in $G$ cannot belong to $H$.

I assume $|H|=m$ so $|G|=2m$. Since $[G:H]=2$ so $H\trianglelefteq G$ and that; half of the elements of the group are in $H$. Any Hints? Thanks.
 A: HINTS: 


*

*The product of elements of $H$ is in $H$.

*If $a,b\in G\setminus H$, $ab\in H$.

*$|G\setminus H|$ is odd.
A: For some fixed $g\in G-H$, we have $G = H \cup g H$ (disjoint).
Then $(\prod_{a \in G} a)H = \prod_{a \in G} aH = \prod_{a \in G-H} aH=(gH)^m=gH$.

By the way, how do you define $\prod_{a \in G} a$ unambiguusly if $G$ is not necessarily abelian?
A: Take the two different cosets of H in G as {H, gH}, g is not H.
Order of g is 2 in G/H. If gh1, gh2 are in gH; then their product is in H, since there is no element in common in H and gH; take h1 = 1 and; gh1*gh2 = h2 which is in H.
So the product of all elements of G is  ghi*ghj..*hi*hj..  = k * ghi {for some k in H}
As H if odd order, we can rename product as gh0*(gh1*gh2*...gh2n) * k
As product of 2 elements of gH are in H, so the product (gh1*gh2*...gh2n) is in H.
So, the product is gh0*(gh1*gh2*...gh2n*k) is not in H.
A: Consider the image of the product under the quotient map $G\to G/H\cong C_2$.
A: Since the index is $2$ and $|H|$ is odd. That means $G$ will have an even number of elements and exactly half of those elements are in $H$ and half are in $gH$ (both odd number of elements).
Consider $$G = \{ g_1, g_2, ... , g_{k}, h_1, h_2, ..., h_{k}\}$$ and let $$x = g_1 \times g_2 \times ... \times g_k \times h_1 \times h_2 \times ... \times h_k.$$
The order that we multiply the $g_i$s and $h_i$s don't matter. And since each time we have $h_i$ we stay in $H$ and we switch from $gH$ to $H$ or $H$ to $gH$ when we have $g_i$, and we switch an odd number of times for the $g_i$, hence we always end up in $gH.$
Lastly, we know that $xH = gH \iff x \in gH$.
