10.3 Lagrange’s theorem
Let G=D_{2n} be the dihedral group of order 2n, and consider the orders of elements. Every reflection has order 2, which divides |G|. The rotation a has order n, which divides |G|. Every other rotation a^i has order n/\gcd(n,i), which divides |G|. In this section we’ll show that this is a general phenomenon for finite groups. In fact, we’ll prove a more general fact about orders of subgroups (of course, this includes the case of the order of an element x, since \operatorname{ord}(x) is equal to the order of the cyclic subgroup \langle x\rangle generated by x).
The theorem in question is named after Joseph-Louis Lagrange, an Italian mathematician ( 1736 - 1813 ), who proved a special case of the theorem in 1770 (long before abstract group theory existed).
Let G be a finite group, and H\leq G a subgroup. Then |H| divides |G|.
The idea of the proof is to partition the set G into subsets, each with the same number of elements as H, so that |G| is just the number of these subsets times |H|.
The set of left cosets partition G.
For any group G and subgroup H \leq G, \{xH:x\in G\} is a partition of the set G.
Clearly property 1 is satisfied: G = \cup_{x\in G} xH because any x=xe\in xH. So we just need to check property 2, that for x,y\in G we have xH\cap yH\neq\emptyset if and only if xH=yH.
Suppose xH\cap yH\neq\emptyset, and choose g\in xH\cap yH. Then g=xa=yb for some a,b\in H, so that x=yba^{-1}. If h\in H then xh=y(ba^{-1}h)\in yH since ba^{-1}h\in H. So every element of xH is in yH. Similarly every element of yH is in xH, so that xH=yH.□
Given that we have a partition on the set G, we can construct an equivalence relation on the set G and say x\sim_H y if and only if there exists z such that x,y \in zH. With a bit of work, we can show that this is the same as saying x, y\in G are equivalent (with respect to/modulo H) if and only if y^{-1}x \in H [you can check this is an equivalence relation].
Compare this with the (infinite) group G=(\mathbb{Z},+) and (infinite) subgroup H = (n\mathbb{Z},+). We partitioned \mathbb{Z} using n\mathbb{Z} and defined an equivalence relation a\equiv b \pmod{n} if and only if a-b\in n\mathbb{Z}.Next we verify that each left coset has the same cardinality.
Let H\leq G and x\in G. Then there is a bijection \alpha:H\to xH, so that |xH|=|H|.
Define \alpha by \alpha(h)=xh. Then \alpha is surjective, since by definition every element of xH is of the form xh=\alpha(h) for some h\in H. But also \alpha is injective, since if h,h'\in H then \alpha(h)=\alpha(h')\Rightarrow xh=xh'\Rightarrow h=h'.
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Everything so far works for possibly infinite H,G, but now for finite G (and hence H) we can put everything together to prove Lagrange’s Theorem.
Suppose that k is the number of distinct left cosets xH. By the two previous lemmas, the cosets partition G and each coset contains |H| elements. So the number of elements of G is k|H|.
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Let G=D_6 again, but let H=\langle b\rangle=\{e,b\} be the cyclic subgroup generated by b. Then
eH=\{e,b\}=bH;
aH=\{a,ab\}=abH;
a^2H=\{a^2,a^2b\}=a^2bH,
You may have seen the word index (Latin for “pointer”) in the context of summation (e.g., \sum_{i=1}^3 i^2, where i is the index). In group theory, there are many time where one might want to do \sum_{i=1}^{|G:H|}. Here, the index is telling us how many terms are in our sum.
Let G be a finite group with |G|=n. Then for any x\in G, the order of x divides n, and so x^n=e.
By Lagrange’s Theorem, the order of the cyclic subgroup \langle x\rangle divides n. But the order of this subgroup is just \operatorname{ord}(x), so \operatorname{ord}(x) divides n, and so x^n=e.
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