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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