Elements of odd order generate a proper subgroup of a group I am stuck with the following question:
Given is a group $G$ with a subgroup $H$ of index $2$, so $\left [ G:H \right ]=2$. I have to show that the elements of odd order of $G$ generate a proper subgroup.
What i know is that $H$ as a subgroup of index $2$ has only $2$ left cosets (and also right too). So i know that this subgroup is normal in $G$. What can i do with the order? Can anybody help me with this exercise, please?
Thank you in advance!
 A: The elements of odd order generate a subgroup, let's call it $P$, because
$1 \in G$ has order $1$, which is odd. So $1 \in P$.
Thus $P$ contains some elements and since we are generating a group, we automatically generate the inverses and all products, so $P$ will be a subgroup. 

We still need to show that $P\neq G$ for $P$ to be a proper subgroup.
We know that there is an $H \leq G$ with $[G:H]=2$, so $H\neq G$. 
We will show that $P$ lies in $H$. 

Theorem 2. 
  If $G$ is a finite group and $N \triangleleft G$ then any element of $G$ with order relatively prime to $[G:N]$ lies in $N$.  In
  particular, if $N$ has index $2$ then all elements of $ G$ with odd
  order lie in $N$.
Proof:   Let $g$ be an element of $G$ with order $m$, which is relatively prime to $[G:N]$. The equation $g^m=e$ gives
  $(gN)^m=N \in G/N$.   Also $(gN)^{[G:N]}=N$, as $[G:N]$ is the order
  of $G/N$.
So the order of $gN \in G/N$ divides $m$ and $[G:N]$.
These numbers are relatively prime, so $gN=N$, which means $g \in N$.

Now we know that all odd elements lie in $H$. Since $H$ is a subgroup, it is closed, so the whole subgroup generated by odd elements must lie in $H$. Thus $P\neq G$.
