Let be a homomorphism of abelian groups and (we denoted operations in both groups by the same symbol – these are different operations, but no confusion will arise; you will always see from the context in which group we work; same for 0s in these groups).

The

imageof is the set

thekernelof is the setIt is easy to check (check this!) using the definition of homomorphism that sets just defined are subgroups: and

Note that image and kernel are concepts parallel to the concepts of the image and kernel of a linear map between vectorspaces; in particular such linear maps provide homomorphisms between additive groups of the corresponding vectorspaces.

**Cosets and quotient, a.k.a. factor, groups .**

Let Given the set is called the

cosetof in

For any the following holds: either or (Indeed, if then we have the former, as is closed under addition, otherwise the latter – check this.) Thus we have apartitionof into cosets of

A fancy way of saying the latter is to say that being in the same coset of defines an *equivalence relation* on

As an example, consider and a positive integer Then the cosets of in are the sets of integers of the form with Then and are in the same coset of iff i.e. when divides Recalling the procedure of division with reminder, is easy to see that the cosets are where

The set of cosets of in has a natural structure of group defined on it.

The

quotient grouphas the set of cosets of in as the set of elements; the addition is defined by

In a way, we can identify the operation with the in

The former is the same as the latter taken “modulo “.

So we still need to check that this way one ideed has an abelian group. Commutativity and associativity in is immediate from the associativity and commutativity in Then, (note that is a coset of in , just any other coset), and is

Continuing the above example we have isomorphic to This is in fact not suprising, and is implied by the following general theorem.

Let be a homorphism of abelian groups. Then we can identify with certain quotient group of Namely Specifically, the latter isomorphism is given by

The proof of this is left as an easy exercise (that you should do!). The (rather unfortunate) name (1st…) comes from the fact that there are more similar results in group theory (sometimes also the already mentioned facts that and are subgroups is included as its part).

It follows that every is the kernel of the homomorphism What distinguishes such homomorphisms from the general ones is that they are *surjective*, i.e. the target of the mapping equals its image.

(Thus there is a 1-1 correspondence between the surjective homomorphisms from and subgroups of .)

A question to think about: what are the groups in the cases when and

Armed with the 1st isomorpshism theorem, we can easily give a proof of our claim about cyclic groups. Namely, let be a cyclic group. Consider a homomorphism defined by

Trivially, is surjective. Let the minimal positive integer such that provided that such an integer exists. We claim that if does not exist then is an isomorphism, and otherwise Of course, if for a negative , then too, so there is no loss in generality in assuming if it exists. Thus in the case of no such we indeed have an isomorphism. In the case when such an exists it is obvious that (check this!). Thus, by the 1st isomorphism theorem, completing the proof of the claim.

Sometimes, although not always, a subgroup isomorphic to is readily found inside This is for instance always the case when is the additive group of a finite-dimensional vectorspace, and the additive group of a subspace. Then as is known from linear algebra.

(A fancy way to say the latter is to say that is a *split extension* of )

You are encouraged to show that this is indeed not true in general (Hint: consider and ) (Such extensions are called *non-split*.)

10 February, 2009 at 16:36 |

a typo on the paragraph before 1st isomorphism theorem. it should be B/H instead of H/B.

10 February, 2009 at 17:40 |

thanks for pointing this out.

10 February, 2009 at 16:38 |

Shouldn’t A be isomorphic to Z/mZ instead of Z/Z_m in the second last paragraph?

10 February, 2009 at 17:42 |

and for this one, too. Indeed, $ latex \mathbb{Z}_m$ isn’t even a subgroup of

25 February, 2009 at 11:02 |

Under Cosets and Quotients, given b in B, the set b+H is called the coset of H, not of b, right?

25 February, 2009 at 11:10 |

b+H is

acoset of H (notthecoset, as there could be more than one!) with a representative b.In view of this, b+H is the coset of b.