Lagrange Theorem

This note has been used to help create the Lagrange's Theorem wiki

In this post, we will build up towards proving Lagrange's Theorem, which can be viewed as an extension of Euler's Theorem.

Let GG be a group with operation * and consider an arbitrary subset HGH \subseteq G . In general, HH is not a group under *. In fact, * may not even be a well-defined function on H×HHH \times H \rightarrow H, since its image may lie outside HH. Hence, we make the following definition.

A subgroup of GG is a subset HGH \subseteq G such that:
(a) eHe \in H ,
(b) hHh1H h \in H \Rightarrow h^{-1} \in H ,
(c) h,hHhhHh, h' \in H \Rightarrow hh' \in H .

Note that any such HH becomes a group in its own right, with the operation * "inherited" from G.G.

For any group G,G, GG itself and {e}\{e\} are subgroups of G.G. Most groups have other, non-trivial subgroups. The following are some examples of subgroups, that correspond to the example list of groups from the post Group Theory.

1) Z\mathbb{Z} , the set of integers
2) R+ \mathbb{R}^+ , the set of positive real numbers.
3) Sn1 S_{n-1} , the set of functions which also satisfy f(n)=n f(n) = n .
4) mZn m \mathbb{Z}_n , the subset of multiples of mm. Note that this is strictly smaller than Zn\mathbb{Z}_n if gcd(m,n)1.\gcd(m,n)\neq 1 .
5) (Zn×)2(\mathbb{Z}_n ^\times )^2 , the set of squares modulo nn.

A general example of a subgroup is obtained by picking an element gGg \in G and taking all powers of gg. This is known as the subgroup generated by gg, and is denoted by:

<g>:={,g2,g1,e,g,g2,}. < g > := \{ \ldots, g^{-2}, g^{-1}, e, g, g^2, \ldots \}.

Proposition. The order of the subgroup <g> < g > is the smallest positive mm for which gm=eg^m = e . If such an mm does not exist, then the order is infinite.

As such, we define the order of element gg to be the smallest positive mm for which gm=e g^m = e , and write o(g)=m o(g) = m .

Let HH be a subgroup of the group GG, and let gGg \in G . The set gH={gh:hH} g H = \{ gh : h \in H \} is called a (left) coset of HH in G.G. . In other words, the coset gHgH is the image (range) of the function ϕg:HG \phi_g : H \rightarrow G where ϕg(h)=gh \phi_g (h) = gh . Because this function is one-to-one (by the Cancellation Property!) gH=H.\lvert gH \rvert = \lvert H \rvert. Note also that g=gegH.g=ge\in gH.

The most important property of the cosets is that g1Hg_1H and g2Hg_2H either coincide or do not intersect (see Worked Example 2 below). This implies the following classical theorem.

Lagrange's Theorem. If GG is a finite group and H H is a subgroup of GG, then H \lvert H \rvert divides G \lvert G \rvert .

As an immediate corollary, we get that if GG is a finite group and gGg\in G , then o(g) o(g) divides GG. In particular, gG=e g^{ \lvert G \rvert} = e for all gG g \in G .

Worked Examples

1. Prove the proposition.

Solution: Suppose mm is finite. We claim that <g>={e,g,g2,,gm1} < g > = \{e, g, g^2, \ldots, g^{m-1} \} and that all these mm elements are distinct. For the first statement, note that for any nZn \in \mathbb{Z}, we can write n=qm+rn = qm+r where 0rm1 0 \leq r \leq m-1 . Hence gn=(gm)q(gr)=e(gr)=grg^n = (g^m)^q(g^r) = e(g^r) = g^r. To prove the second statement, suppose gi=gjg^i = g^j for 0ijm1 0 \leq i \leq j \leq m-1 , then gji=eg^{j-i} = e, which contradicts the minimality of mm.


2. Let g,gG g, g' \in G and HG H \subseteq G be a subgroup.

(a) If g1gH g^{-1} g' \in H , then gH=gH gH = g'H;

(b) If g1g∉Hg^{-1} g' \not \in H , then gHgH= gH \cap g'H = \emptyset.

Solution: For the first statement, any element of gHg'H can be written as ghg'h, for some hHh\in H . But gh=g(g1g)hgH g' h = g(g^{-1} g') h \in gH since g1gHg^{-1} g' \in H . This tells us that gHgH g'H \subseteq gH .
Conversely, any element of gH gH can be written as gh,hHgh, h \in H and gh=g(g1g)h gh = g'(g'^{-1}g)h. But g1g=(g1g)1 g'^{-1}g = (g^{-1}g')^{-1} lies in HH since HH is a subgroup of GG. Hence the result follows and gH=gHgH = g′H.
For the second statement, suppose xgHgH x \in gH \cap g'H . Then it can be written as x=gh=gh x = gh = g'h' for some h,hH h, h' \in H . Thus g1g=hh1Hg^{-1}g' = hh'^{-1} \in H which contradicts the fact that g1g∉H g^{-1}g' \not \in H .


3. Prove Lagrange's Theorem

Solution: The above shows that the cosets partition the entire group GG into mutually disjoint subsets, all of which have H \lvert H \rvert elements. Hence, Lagrange's Theorem follows.

Note by Calvin Lin
7 years, 4 months ago

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Really useful notes Aditya

Usama Khidir - 6 years, 10 months ago

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