would exist such that and . Thus . f is a group homomorphism, so inverses and group operations are preserved, giving ; in other words, , and ker f would not be the singleton. Conversely, distinct elements of the kernel violate injectivity directly: if there would exist an element , then , thus f would not be injective. is a subgroup of G and further it is a normal subgroup. Thus, there is a corresponding quotient group . This is isomorphic to f(G), the image of G under f (which is a subgroup of H also), by the first isomorphism theorem for groups. In the special case of abelian groups, there is no deviation from the previous section. Example Let G be the cyclic group on 6 elements with modular addition, H be the cyclic on 2 elements with modular addition, and f the homomorphism that maps each element g in G to the element g modulo 2 in H. Then , since all these elements are mapped to 0H. The quotient group has two elements: and . It is indeed isomorphic to H. Ring homomorphisms Let R and S be rings (assumed unital) and let f be a ring homomorphism from R to S. If 0S is the zero element of S, then the kernel of f is its kernel as linear map over the integers, or, equivalently, as additive groups. It is the preimage of the zero ideal {0S}, which is, the subset of R consisting of all those elements of R that are mapped by f to the element 0S. The kernel is usually denoted (or a variation). In symbols: Since a ring homomorphism preserves zero elements, the zero element 0R of R must belong to the kernel. The homomorphism f is injective if and only if its kernel is only the singleton set {0R}. This is always the case if R is a field, and S is not the zero ring. Since ker f contains the multiplicative identity only when S is the zero ring, it turns out that the kernel is generally not a subring of R. The kernel is a subrng, and, more precisely, a two-sided ideal of R. Thus, it makes sense to speak of the quotient ring R/(ker f). The first isomorphism theorem for rings states that this quotient ring is naturally isomorphic to the image of f (which is a subring