B, and we say that aO = b. ) We sometimes say that a - b. g. 1 that if a0 = b, bo = a and if bo = a, a0 = b. Such a mapping exists only when every element of B is the image of one and only one element of A. We thus need two conditions on 0.

Such a product exists only when the object space of each is the same as the image space of the one preceding (or possibly contains the image space as a subgroup, although this case is better avoided, as will be discussed later). It is very convenient to show the various homomorphisms diagrammatically as in figure 4. 0 0 V, G-H->K-M Fig. 4 The inverse of a homomorphism We have seen that for a mapping to have an inverse it must be both onto and 1-1: thus in the homomorphism case the mapping must be both epimorphic and monomorphic-in other words it must be an isomorphism.

### An Introduction to Abstract Algebra (Vol II) by F.M. Hall

by Thomas

4.4