**Lemma**. Let be a commutative ring and let be a group. We define the map by

Then is an onto ring homomorphism and is a basis for the free -module

*Proof*. Obviously is well-defined, additive and onto. Now, let Then

So is a ring homomorphism. Therefore is an ideal of and hence an -module. Now, if and only if if and only if

Thus as an -module, is generated by the set Then, obviously, the set still generates To show that is a basis for as an -module, we suppose that Then

But for all and so for all

**Definition**. The ring homomorphism as defined in the lemma, is called the **augmentation map** and is called the **augmentation ideal** of

**Hurewicz Theorem**. Let be a group with the commutator subgroup Let be the augmentation ideal of and consider as an additive group. Then

*Proof*. Define the map by Clearly is well-defined because for all Also, since we have

Thus is a group homomorphism. So and therefore, since is an abelian group, we have Thus and hence we have a group homomorphism defined by

Now, to show that is an isomorphism, we will find an inverse for it. Define the map by Note that since for all the map is a well-defined group homomorphism. Now,

So because, by the lemma, the set generates the additive group Hence the map defined by

is a well-defined group homomorphism. It is now easy to see that both and are identity maps.

Thanks!

Thank you very much…