If $|G|=2n$ where $n$ is odd. Then $\prod_{g\in G}g\notin H$ If $|G|=2n$ where $n$ is odd and $H$ is a subgroup of order $n$. Then $\underset{g\in G}{\prod}g\notin H$
pf: Since $G$ has even order, it must have a element of order $2$, then $\underset{g\in G}{\prod}g$ has even order. So $\underset{g\in G}{\prod}g\notin H$ as the order of everything in $H$ must divide $n$. Is this correct?
 A: Lemma Put $x=\prod_{g \in G} g$ and let $\sigma$ be a bijection (not necessarily an automorphism) of the set $G$. Write $x_{\sigma}=\prod_{g \in G}\sigma(g)$. Then $x \notin H$ if and only if $x_{\sigma} \notin H$.
Proof Since $|G:H|=2$, $H$ is normal and $G/H$ is abelian. Hence in $G/H$, where the order of multiplication does not matter, we have $\bar{x}=\prod_{g \in G} \bar{g}=\prod_{{g \in G,g^2=1}} \bar{g}=\prod_{g \in G} \overline{\sigma(g)}$, since the odd order elements of $G$ must lie in $H$ and do no longer appear in the product when modding out. Hence $\bar{x}=\overline{x_{\sigma}}$, that is $x=x_{\sigma}h$ for some $h \in H$ and the Lemma follows.
Now let us proceed with the proof of the question above. Let $t \in G$ be an element of order $2$ (Cauchy's Thereom assures its existence!), put $P=\{1,t\}$. Note that $t \notin H$, $G=HP$ and $H \cap P=1$. If $H=\{1, h_2, h_3, \dots, h_n\}$, we can write $G=\{1,h_2,h_3, \dots, h_{n}, t, th_2, \dots, th_n\}=\{1,h_2^{-1},h_3^{-1}, \dots, h_{n}^{-1}, t, th_2, \dots, th_n\}$. Now take the  product in the last set of all the elements of $G$ as follows: $(t \cdot 1) \cdot (th_2 \cdot h_2^{-1}) \cdots (th_n \cdot h_n^{-1})=t^n=t$, since $n$ is odd and $t$ has order $2$. Because $t \notin H$, we are done by the Lemma.
Note Surprisingly, in general one can show that if $G$ is a finite group, $P \in Syl_2(G)$ and $P$ is non-cyclic or trivial (that is, $|G|$ is odd), then the set of all possible products of elements of $G$ (each taken exactly once of course) equals $G'$, the commutator subgroup. If $P$ is cyclic, then this set equals the coset $tG'$ where $t$ is the unique element of order $2$ of $P$.
A: That answer is not a valid proof. The product of the elements does not need to have the order related to one element in the product. For example, you can multiply two elements with order 2 to get an element of order 3 or 5 or any number.
Hint: This fact I think must have come up before this exercise: if a subgroup H that has half as many elements as G, then H is a normal subgroup of G.
A: Given the writing of the product, one can argue that the group actually is commutative (otherwise it would not be well-defined: the product would depend on the order in which the elements are multiplied). Hence, I do not have to care about normality, as everything is automatic. 
Let $\pi$ denote the canonical surjective group homomorphism $G\rightarrow G/H$, that is the projection of $G$ onto the quotient $G/H$. This quotient is a group that contains two elements, so it is isomorphic to $\{-1, 1\}$. 
What we need to show is that the element $\underset{g\in G}{\prod}g$ is not $1$ in the quotient $G/H$. So we need to compute its image by $\pi$ and check that it not $1$. 
Given that $\pi$ is a group homomorphism, it is clear that 
$$\pi\left(\underset{g\in G}{\prod}g\right)=\underset{g\in G}{\prod}\pi(g)$$
To compute this product, we may separate the terms whether their value is $1$ or $-1$ under $\pi$ (this is because $\pi$ is surjective, so that $G=\pi^{-1}\{1\}\sqcup\pi^{-1}\{-1\}$ as sets). 
$$\underset{g\in G}{\prod}\pi(g)=\underset{g\in \pi^{-1}\{1\}}{\prod}\pi(g)\times\underset{g\in \pi^{-1}\{-1\}}{\prod}\pi(g)=\underset{g\in \pi^{-1}\{1\}}{\prod}1\times\underset{g\in \pi^{-1}\{-1\}}{\prod}-1=(-1)^{|\pi^{-1}\{-1\}|}$$
So now, all we need to do is to understand what $|\pi^{-1}\{-1\}|$ is (the cardinality of $\pi^{-1}\{-1\}$). Actually, this is $n$. Indeed, it is a known result that every cosets of a subgroup have the same cardinality, that is precisely the cardinality of the said subgroup. 
So we may now conclude 
$$\pi\left(\underset{g\in G}{\prod}g\right)=(-1)^{|\pi^{-1}\{-1\}|}=(-1)^n=-1$$
because $n$ is odd.
NB: Note that if $G$ is not abelian (and we assume that the product is defined after an implicit choice of the order), the above proof works just as well because the quotient, $G/H$, which is isomorphic to $\{1,-1\}$, is abelian. Hence, the manipulations I did using $\pi$, in particular the separation of the product into two parts, still holds.  
One needs to justify however that $H$ is a normal subgroup of $G$. This is a result that is true for any subgroup of index $2$. A proof may be found here.
NB2: to justify in general that every cosets of a subgroup have the same cardinality, one can use the pretty straightforward argument. Say, $G$ is a finite group with subgroup $H$, and consider $gH$ a left coset of $H$. Then we have a map $H\rightarrow gH$ that sends an element $h\in H$ to $gh$. This map is bijective because it has an inverse, that is the map $gH\rightarrow H$ which sends an element $h'\in gH$ to $g^{-1}h'$. 
One can adapt this easily to the case of right cosets.
