Just as in the case of abelian groups, each subgroup gives rise to a coset decomposition of or, more precisely, to *two* decompositions, into *left cosets* i.e.

and *right cosets,* i.e.

It is proved by the same argument as in the case of abelian groups that one indeed has decompositions of in this way.

Unlike in the case of abelian groups, these two decompositions need not be the same: an example is given by

The following is an important and elementary implication of the fact that all the cosets of have the same cardianlity:

Let and Then divides

An extremely important case (and, in fact, the only case) when these two decompositions are the same is when where is a group homomorphism. (Just as in the case of abelian groups, one can show that and )

We need to show that for any and one has for some We compute Hence

A subgroup satisfying for all is called

normal(notation: or when )

Often it is more convenient to work with the equivalent condition for normality, namely that for all

**Examples of normal subgroups.**

- The subgroup generated by in is normal.
- defined by is normal.
- of
*index*2 (i.e. the one that has exactly 2 left cosets in ) is normal.

The 1st is easy to establish directly (it also follows from the 3rd).The 2nd follows from the identity

The 3rd follows from the equality and an observation that so

Cosets play an important role in constructing “easier” groups from “complicated” ones, by the quotient group construction. The difference with the case of abelian groups is that only normal subgroups can be kernels of homomorphism.

Let Then is a group with operation and a group homomorphism with the kernel

We need to use to establish that the multiplication in is well-defined:

Then is a group with the identity element as associativity in follows from the associativity in and

It can be readily checked that is a homomorphism, and that Indeed, Also, for and so On the other hand if then and so

**Matrix groups**

Subgroups of the group, with respect to matrix multiplication, of invertible matrices with entries in a field provide a rich playground for various examples of normal subgroups and quotient groups. To this end, we will work with *block matrices*, i.e. we will partition our matrices into rectangular blocks of appropriate size, so that this block decomposition is preserved under the matrix multiplication. The simplest case is the following partition into 4 blocks (i.e. *block matrices*).

Given an matrix and positive integer we can view as where is a matrix, a matrix (and certainly and being of size ).

Let us fix and

First, we need to convince ourselves (an easy exercise in matrix multiplication) that the following multiplication law works for two block matrices and :

It should of course look very familiar, as this is exactly how two matrices are multiplied.

When we see that the shape is preserved, i.e. More precisely

We see that the blocks with indices 11 and 22 behave as if “nothing around them matters”, as if they sit in their own groups and (that is, assuming the matrices we talk about are invertible). We can formalise this observation as follows.

Let be a subgroup of block matrices, with diagonal blocks of sizes and and such that for all Then

- the maps

and

are group homomorphisms.- for

The reader is encouraged to provide a proof.

More generally, this result can be interpreted in terms of the actions of on the dimensional vectorspace over

**Exercise.** Prove that the intersection of two normal subgroups of a group is a normal subgroup. Based on this, describe and derive that is a normal subgroup of

Note that (where or ) cannot be obtained by simply setting the remaining off-diagonal block of to 0, and the remaining diagonal block to the identity matrix. For instance, let be generated by the matrix where Then is a cyclic group of order 4, and it satisfies the conditions of the above lemma with It is easy to see that is an isomorphism, and is a homomorphism with the kernel generated by

Let us denote and

As by the usual argument involving decomposition of into the cosets of computing for for prime, can be accomplished by computing and separately. In turn, so we can simplify the task of computing too.

To illustrate this principle, let us compute the order of

First we derive the following:

where is prime.

Indeed, we can fill in the 1st row of an matrix in way (everything goes, except all zeros). Fixing this 1st row vector we can again fill in the 2nd row with entries of any vector which is linearly independent (to get an invertible matrix) of i.e. lies outside of the subspace spanned by As we get ways to choose Now we fix as well, and fill in the 3rd row with entries of any vector which is linearly independent of and i.e. lies outside of the subspace spanned by and As we get ways to choose

So when filling the -th row, we have choices, giving us the formula above.

Now, we claim that Indeed, consists of such that whereas we can fill the entries of without any strings attached – the result will be invertible. Thus we have

**Excercise.** Compute for and as in (1).

(Hint: use the fact that is a group homomorphism.)

2 January, 2011 at 22:01 |

[…] Cosets and normal subgroups March 2009 4 […]

14 August, 2012 at 17:20 |

i want previous years questions of iit jam about normal subgroup.

14 August, 2012 at 19:00 |

Hmm, I don’t have an idea what you are talking about…