K are homomorphisms, so is the mapping 00: G - K. If g and g' are any two elements of G, and if gO = h, g'9 = h', ho = k, h'¢ = k', we know that (gg') 0 = hh' and (hh') 0 = kk' by the homomorphism property. Hence (gg') e0 = (hh') 0 = kk'. But k = gOg and k' = g'90 and so (gg') 00 = (gOc) (g'60), showing that 60 is a homomorphism. We may similarly prove that the product of any number of homomorphisms is a homomorphism. 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).
The image GO is a subgroup of H. Let any two elements of GO be gO and g'O (they must be of this form for some g and g' in G, since otherwise they would not be in the image GO). 3 Then (gO) (g'0)-1 = (gg'-1) 0 and so is the image of gg'-1, and hence is in GO. 1. 1. Changing the image and object spaces Suppose we are given a homomorphism 0: G - H. Then the image space H may be extended at will without affecting the homomorphism. For example, a homomorphism into the real numbers may equally well be considered as one into the com- plex numbers.