The Lagrange theorem states:
If $G$ is a finite group, and $H$ a subgroup of $G$, then the order of $H$ will divide the order of $G$. More precisely, $|G| = |H| \cdot (\text{number of left cosets of }H)$:
$$|G| = |H| \cdot (G:H)$$
The proof I have in my notes says:
$G$ consists of $\{G:H\}$ cosets, each of them consists of $|H|$ elements, the cosets are disjoint.
That's it. How does this prove the theorem?
$\endgroup$5 Answers
$\begingroup$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 $n$.
$\endgroup$ 0 $\begingroup$Maybe it helps to add that the union of all the left cosets of H in G is again G. So G is the disjoint union of [G:H] sets that all have the same number of elements, namely $|H|$. Thus $|G| = |H|[G:H]$
$\endgroup$ $\begingroup$One simple approach is to show that the (left) cosets are distinct. Once we have that, the rest follows fairly obviously.
Step 1: We are given that $H$ is a group (a subgroup of $G$). In particular for every $h\in H$ we also have $h^{-1}\in H$. So for some element $j\in G, j \not \in H$ we have that $jh \not \in H$, since otherwise $jhh^{-1}=j\in H$.
Step 2: Form the coset $jH$ from $\bigcup_{h\in H} jh$, and note $|jH| = |H|$. Of course since $e\in H$, $j\in jH$. Now consider the coset $kH$ with $k\in jH$. We must have $k=jh_0$ so also $j=kh_0^{-1}$. Then any $jh_i = kh_0^{-1}h_i$ and of course $h_0^{-1}h_i \in H$ so $kH = jH$.
Step 3: Thus cosets are distinct (since any common member implies that the cosets are identical), all are of size $|H|$, and every member of $G$ is in a coset, so $|H|$ divides $|G|$
$\endgroup$ $\begingroup$[ Here is one way we could proceed ]
Let $ G $ be a finite group, and $ H \subseteq G $ a subgroup.
Consider the map $ f : G \longrightarrow f(G) $ sending $ g \mapsto gH $, where $ gH := \{ gh : h \in H \} $. In trying to understand fibres of $ f $, we find $ f^{-1} ( \{ gH \} ) = gH $.
[ For a surjective map $ f : X \longrightarrow Y $, the sets $ f^{-1} ( \{y\} ) := \{ x \in X : f(x) = y \} $ where $ y \in Y $ are called "fibres" of $ f $. These fibres partition the set $ X $ ]
Also each fibre has the same size as $ H $ (because $ H \longrightarrow gH $ sending $ h \mapsto gh $ is a bijection), and hence $ |H| $ divides $ |G| $.
$\endgroup$ $\begingroup$Given a group $G$ and $H\le G$:
- $x\sim y \stackrel{(def.)}{\iff} xy^{-1}\in H$ is an equivalence relation on $G$;
- $[x]_\sim = Hx$;
- the map $h\mapsto hx$ is a bijection from $H$ to $Hx$.
If $G$ is finite, from 1 to 3 follows that $|H|\mid |G|$.
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