Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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, $|G| = |H| \cdot (\mathrm{number\, of\, left\, cosets\, of\, H})$:

$$|G| = |H| \cdot (G:H)$$

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.

Thats it. How does this prove the theorem?

share|improve this question

2 Answers 2

up vote 6 down vote accepted

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 $n$.

share|improve this answer

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 $|G| = |H|[G:H]$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.