PlanetMath: calculus of subgroup orders

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calculus of subgroup orders

(Application)

Recall that for any group and any subgroup of we can define left and/or right cosets of in . As the cardinality of left cosets equals the cardinality of right cosets, the index, denoted is well-defined as this cardinality.

The theorem of Lagrange is the first of many basic theorems on the calculus of indices of subgroups and it can be stated as follows:

$\displaystyle \vert G\vert=[G:H]\vert H\vert.$

(1)

When

is a finite group then we can rewrite this in the familiar form (as actually proved by LaGrange)

$\displaystyle \vert G\vert/\vert H\vert=[G:H].$

(2)

and so conclude the familiar statement: “The order of every subgroup divides the order of the group.”

As the proof of (2) can be written with bijections instead of specific integer values, the proof of (1) is immediate from the usual proof of the Lagrange's theorem.

The first corollary to the theorem states that given $K\leq H\leq G$ then

$\displaystyle [G:K]=[G:H][H:K].$

(3)

So when

is finite we have

$\displaystyle \frac{[G:K]}{[H:K]}=[G:H].$

(4)

This should be contrasted with the third isomorphism theorem which claims if

and

are normal in

then

$\displaystyle \frac{G/K}{H/K}\cong G/H.$

(5)

Remark 1 It is preferrable to express the various equations relating indices of subgroups with multiplications. This is to allow for infinite indices, as we can multiply cardinal numbers, but we may not always be able to make sense of cardinal number division.

When only finite groups are considered so that division is allowed, expressing the theorems as quotients is often easier to understand.

Next suppose and are any two subgroups of then we define

$\displaystyle HK:=\{ hk:h\in H, k\in K\}$

(sometimes called the complex of

and

.) Caution: it is not always true that

is a subgroup of

. It is true if either

is a normal subgroup of

and occassionally it is true even without

being normal - for example when

so called permutable subgroups.

If and are finite subgroups then we can express this as:

$\displaystyle \vert HK\vert/\vert K\vert=[H:H\cap K]$

(6)

Once again if we have normality, say

is normal in

, then this is mimicks second isomorphism theorem:

$\displaystyle HK/K\cong H/H\cap K.$

(7)

Notice that if is a subset of the subgroup $\langle H,K\rangle\leq G$ . So it is possible to state (6) with all subgroups using inequalities such as

$\displaystyle [H:H\cap K]\leq [G:K]$

(8)

even when all the groups are infinite. Furthermore, if

is a subgroup of

then we can write

$\displaystyle [H:H\cap K]=[HK:K]$

(9)

to apply even for infinite groups. This equation is often called the parallelogram law for groups because it can be described with with the following picture:

$\displaystyle \begin{xy}<5mm,0mm>:<0mm,10mm>:: (0,5) *+{G} =''G''; (0,4) *+{HK}... ...; ''H''; ''HiK'' **@{-}; ''K''; ''HiK'' **@{-}; ''HiK''; ''1'' **@{-}; \end{xy}$

Note the diagram is a Hasse diagram of the lattice of subgroups of

. We further inforce a policy of drawing edges of the same length if the index of the corresponding subgroups are equal. Thus (9) is simply a proof that the picture is accurate: opposite sides of a parallelogram are congruent.

The most common use of index calculus is for subgroups of finite index in . This allows one to solve for indices from given assumptions. It is also quite common to prove certain configurations of subgroups are impossible as the indices are relatively prime.