Intuition behind why order of a subgroup divides order of a group. I'm trying to get some intuition on why order of a subgroup divides order of a group. I know Lagrange's Theorem and its proof (every element of a group is an element of some coest, cosets are disjoint and each one has the same number of elements as subgroup), but the proof is more like magic than any intuition.
So I've been trying to prove this fact (divisibility of order) without Lagrange theorem. Let's say we take a group $H$ and try to extend it, by taking $g \notin H$ and considering group $G$ generated by set $H \cup {g}$. Such a group would have to contain $g^{-1}$ and every word over alphabet $\{g, g^{-1}, H\}$. Is it possible to prove intuitively that such construct is of size $|H|\cdot n$ for some $n \in \mathbb{N}$? This would essentially prove that any group $G$ containing $H$ has order divisible by $|H$|, because we would be able to take $g \in G$ such that $g \notin H$, extend $H$ by it just like above and continue this process until $H$ is extended into $G$.
 A: Here's a (partial) response to the question in the title.
It's easy to develop that intuition for finite cyclic groups. Division with remainder implies that the order of every element is divides the order of the generator, which is the order of the group. Then it's only a short step to see that any subgroup of a cyclic group is cyclic.
I don't know whether the argument in the body of the question is easy to complete.
A: Let $H$ be a subgroup of the finite group $G$. For each $g\in G$ the set $gH$ has exactly $|H|$ elements. It is easily seen that two such sets   with a common element are in fact equal. Therefore $G$ is partitioned into finitely many subsets, each of size $|H|$.
A: As far as I know, the only known proof of Lagrange's theorem is the standard one.
That is, cosets of $H$ partition $G$ and each coset has the same size; thus $|G|= |H|[G:H]$.
I believe the following is how Jordan proved Lagrange's theorem, which to some people might be easier to understand. But the proof really is the same old argument.

Proof: List the elements of $H$: $h_1,h_2,h_3, \ldots$. 
Let $p$ be some element not in this list. Then $ph_1, ph_2, ph_3, \ldots$ are all distinct and none of them lie in $H$, as you can easily prove.
Next let $q$ be some element not among
  $$h_1,h_2,h_3,\ldots$$
$$ph_1,ph_2,ph_3,\ldots$$
You can show that $qh_1, qh_2, qh_3, \ldots$ are all distinct and none of them are among the elements above.
Continuing in this manner, we eventually exhaust all elements of $G$ (since $G$ is finite):
$$h_1,h_2,h_3,\ldots$$
$$ph_1,ph_2,ph_3,\ldots$$
$$qh_1,qh_2,qh_3,\ldots$$
$$rh_1,rh_2,rh_3,\ldots$$
$$\vdots$$
So we conclude that the order of $H$ divides the order of $G$.

Note that in the proof we count the elements of $G$ by first listing the elements of $H$, then the elements of $pH$, $qH$, $rH$, $\ldots$
So it is really pretty much the same as the usual proof.
