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.

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 $xH = gH \iff x \in gH$.
