Question

In the notes we proved that if φ:G→H is a homomorphism and H is cyclic, then G must be as well.

Show that the converse is false. That is, find groups G and H, where G is cyclic, but H is not and a homomorphism φ:G→H.

Answer 1

To show that the converse of the statement is false, one example of groups G and H where G is cyclic but H is not, and a homomorphism φ:G→H exists, is G = Z and H = D_4. The homomorphism can be defined by assigning even integers in G to the identity element in H, and odd integers in G to any non-identity element in H.

Step-by-step explanation:

To show that the converse of the statement is false, we need to find groups G and H such that G is cyclic, but H is not, and there exists a homomorphism φ:G→H. One example of such groups is G = Z (the integers under addition) and H = D_4 (the dihedral group of order 8). G is cyclic because it consists of all integer multiples of a single element, such as 1. H is not cyclic because it does not have a single generator that can generate all its elements. The homomorphism φ:G→H can be defined by assigning even integers in G to the identity element in H, and odd integers in G to any non-identity element in H.