You've got most of the pieces you need: we also use that the union of all left cosets of H in G IS G. Now we just put the pieces together:
Let $H\le G$, $|G| = n$, and let $|H| = m$.
Since every coset (left or right) of a subgroup $H\le G$ has the same number of elements as $H$, we know that every coset of $H$ also has $m$ elements. Let $r$ be the number of cells in the partition of G into left cosets of $H$ (because the union of the left cosets of $H$ in $G$ is $G$, and these cosets are disjoint, they partition $G$).
Then $n = rm$: i.e., $|G| = (G:H)\cdot |H|$, so $m = |H|$ is indeed a divisor of $n = |G|$.
To elaborate on how relates to Lagrange's theorem:
$$\text{For}\;\; H\le G, \;G\;\text{ finite}: |G| = |H| \cdot (G:H) \implies \dfrac{|G|}{|H|} = (G:H) = r,\;\; r\in \mathbb{N}.$$
So $|H|$ divides $|G|$, since the index is the number of left cosets of $H$ in $G$ and is hence an integer, say $ (G:H) = r \ge 1$.
Alternatively, let $(G:H) = r,\;\; r \ge 1\in \mathbb{N}$. Then $|G| = r\cdot|H|$. That is, $|G|$ is an positive integer multiple of $|H|$, so $|H|$ divides $|G|$, for $G$ of finite order $n$.
Let $H\subset G$ be a group of $p$ elements. $H$ is both left and right coset and it is easy to see that then $G\backslash H$ [just in case: elements of $G$ that are not in $H$] is also both left and right coset. If you take $g\in G\backslash H$, then all the elements of the kind $gg_1$, where $g_1\in G$ are different. Because for $g_1\in H$ we already get all $p$ elements of $G\backslash H$ as $gg_1$, it follows that $g^2\in H$. Now there are two cases:
- For all $g\in G\backslash H$ we have $g^2=e$. Then there are $p$ elements of the order $2$, just like in dihedral group.
- For some $g\in G\backslash H$ we have $g^2\in H$ but $g^2\ne e$. Therefore, $g^2$ is of the order $p$, and $g$ - of the order $2p$ - abelian case.
Best Answer
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 $H$.
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 $P\neq G$.