Lagrange Theorem

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.

#Algebra #LagrangeTheorem #KeyTechniques #Olympiad

Note by Calvin Lin
7 years, 2 months ago

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Really useful notes Aditya

Usama Khidir - 6 years, 9 months ago
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