Group homomorphisms Let G be the
cyclic group on 6 elements \{0,1,2,3,4,5\} with
modular addition, H be the cyclic on 2 elements \{0,1\} with modular addition, and f the homomorphism that maps each element g \in G to the element g modulo 2 in H. Then \ker f = \{0,2,4\}, since all these elements are mapped to 0 \in H. The quotient group G / \ker{f} has two elements: \{0,2,4\} and \{1,3,5\}, and is isomorphic to H. Given a
isomorphism \varphi: G \to H, one has \ker \varphi = 1. Then the function f: \mathbb{R} \to S^1 sending x \mapsto e^{2\pi ix}=\cos(2\pi x)+i\sin(2\pi x) is a homomorphism with the integers being the kernel. The first isomorphism theorem then implies that \mathbb{R}/\mathbb{Z} \cong S^1. The
symmetric group on n elements, S_n, has a surjective homomorphism \epsilon: S_n \to \mathbb{Z}_2 that takes each
permutation to the parity of the number of transpositions whose product is that permutation. The
alternating group A_n = \ker \epsilon is the kernel of this homomorphism, consisting of the even permutations. The alternating group is a non-abelian
simple group for n \geq 5. The
determinant of n \times n
invertible matrices of the
real numbers \mathbb{R}, whose set is denoted GL(n, \mathbb{R}) and called the
general linear group of n \times n matrices of \mathbb{R}, is a homomorphism onto the multiplication group \mathbb{R}^\times (consisting of all non-zero real numbers), and the kernel of the determinant is called the
special linear group SL(n,\mathbb{R}) of n \times n matrices of \mathbb{R}. These are the matrices whose determinant is precisely 1. Given a group G and an element, the mapping x \mapsto gxg^{-1} is an
automorphism - an isomorphism whose domain and image are the same group. This gives a homomorphism from G to its automorphism group \text{Aut}(G), mapping each g to its respective
inner automorphism as described, and the kernel of this homomorphism is the
center Z(G) of G, consisting of g \in G where for every x \in G, we have gxg^{-1}=x, or equivalently gx=xg. More generally, for every
normal subgroup H of G (i.e. groups closed under conjugation), this conjugation map is also an automorphism on H, giving another homomorphism G to \text{Aut}(H), with the kernel being the
centralizer C_G(H) of H in G, being the set of g \in G where for every h \in H, we have ghg^{-1}=h.
Ring homomorphisms Consider the mapping \varphi : \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} where the later ring is the integers modulo 2 and the map sends each number to its
parity; 0 for even numbers, and 1 for odd numbers. This mapping turns out to be a homomorphism, and since the additive identity of the later ring is 0, the kernel is precisely the even numbers. Let \varphi: \mathbb{Q}[x] \to \mathbb{Q} be defined as \varphi(p(x))=p(0). This mapping, which happens to be a homomorphism, sends each polynomial to its constant term. It maps a polynomial to zero
if and only if said polynomial's constant term is 0.
Linear maps Let \varphi: \mathbb{C}^3 \to \mathbb{C} be defined as \varphi(x,y,z) = x+2y+3z, then the kernel of \varphi (that is, the null space) will be the set of points (x,y,z) \in \mathbb{C}^3 such that x+2y+3z=0, and this set is a subspace of \mathbb{C}^3 (the same is true for every kernel of a linear map). If D represents the
derivative operator on real
polynomials, then the kernel of D will consist of the polynomials with deterivative equal to 0, that is the
constant functions. Consider the mapping (Tp)(x)=x^2p(x), where p is a polynomial with real coefficients. Then T is a linear map whose kernel is precisely 0, since 0 is the only polynomial to satisfy x^2p(x) = 0 for all x \in \mathbb{R}. == Quotient algebras ==