Group Homomorphisms

You have met a whole gallery of groups — the integers under addition, the clocks \mathbb{Z}_n, the units modulo a prime, the symmetry groups D_n. Each has its own set and its own operation. The natural next question is: how do you compare two groups? What is the right kind of map between them — not just any function of sets, but one that respects the multiplication?

That map is a group homomorphism, and it is the single most important idea in all of group theory. Think of it as a translation dictionary between two groups that is careful to preserve grammar: if you combine two words on one side and then translate, you get the same thing as translating each word first and then combining on the other side. The whole subject — quotient groups, the isomorphism theorems, representation theory — is built on maps that keep the structure intact.

The definition

Let (G, \cdot) and (H, *) be groups. A function \varphi : G \to H is a homomorphism if for all a, b \in G,

\varphi(a \cdot b) = \varphi(a) * \varphi(b).

Read the two sides carefully, because they live in different worlds. On the left, you first combine a and b using G's operation, then send the result across to H. On the right, you send a and b across first, then combine their images using H's operation. A homomorphism is exactly a map for which those two routes always agree — "the picture of the product is the product of the pictures."

Most of the time we drop the dots and just write \varphi(ab) = \varphi(a)\varphi(b), trusting you to remember that the multiplication on the left is happening in G and the one on the right in H.

Two free consequences

You do not have to add "and \varphi sends the identity to the identity" and "and \varphi respects inverses" to the definition — both follow automatically from the one condition above.

The identity maps to the identity: \varphi(e_G) = e_H.
Inverses map to inverses: \varphi(a^{-1}) = \varphi(a)^{-1} for every a \in G.

For the first, note \varphi(e_G) = \varphi(e_G \cdot e_G) = \varphi(e_G)\,\varphi(e_G). Now cancel one copy of \varphi(e_G) (multiply both sides by its inverse in H) to get e_H = \varphi(e_G). For the second, \varphi(a)\,\varphi(a^{-1}) = \varphi(a \cdot a^{-1}) = \varphi(e_G) = e_H, so \varphi(a^{-1}) is precisely the thing that undoes \varphi(a) in H — that is, \varphi(a)^{-1}. Both facts drop out of the single defining equation, which is a good sign that we picked the right definition.

The kernel and the image

Every homomorphism carries two subgroups that capture almost everything about it. The kernel is the set of elements that get crushed down to the identity of H:

\ker\varphi = \{\, g \in G : \varphi(g) = e_H \,\} \subseteq G.

The image is the set of values \varphi actually hits:

\operatorname{im}\varphi = \{\, \varphi(g) : g \in G \,\} \subseteq H.

\ker\varphi is a subgroup of the domain G (in fact a normal one).
\operatorname{im}\varphi is a subgroup of the codomain H.

Why is the kernel a subgroup? It contains e_G (since \varphi(e_G) = e_H); it is closed, because if \varphi(a) = \varphi(b) = e_H then \varphi(ab) = e_H e_H = e_H; and it contains inverses, because \varphi(a^{-1}) = \varphi(a)^{-1} = e_H^{-1} = e_H. The same three checks, run on the image, show it too is a subgroup. Keep the sides straight: the kernel lives upstream in G, the image downstream in H.

Injectivity is controlled entirely by the kernel

Here is the fact that makes the kernel so useful. To check whether a homomorphism is one-to-one, you do not have to compare every pair of inputs — you only have to look at what maps to the identity.

A homomorphism \varphi is injective (one-to-one) if and only if \ker\varphi = \{e_G\}.

The reason is short. Suppose \varphi(a) = \varphi(b). Multiply by \varphi(b)^{-1} to get \varphi(a)\varphi(b)^{-1} = e_H, i.e. \varphi(ab^{-1}) = e_H, so ab^{-1} \in \ker\varphi. If the kernel is only \{e_G\}, then ab^{-1} = e_G, forcing a = b — injective. Conversely if \varphi is injective, nothing but e_G can map to e_H = \varphi(e_G), so the kernel is trivial. The size of the kernel is a one-number measure of "how far from injective" a homomorphism is.

A picture: reduction modulo 3

Let \varphi : \mathbb{Z}_6 \to \mathbb{Z}_3 be reduction mod 3, i.e. \varphi(x) = x \bmod 3. This is a homomorphism, because (a+b) \bmod 3 = (a \bmod 3) + (b \bmod 3) — reducing respects addition. Step through the diagram. The two elements 0 and 3 both map to 0, so \ker\varphi = \{0, 3\} (a subgroup of \mathbb{Z}_6 of size 2); meanwhile every element of \mathbb{Z}_3 gets hit, so \operatorname{im}\varphi = \mathbb{Z}_3 and \varphi is surjective. Notice the arithmetic: |\mathbb{Z}_6| = |\ker\varphi| \cdot |\operatorname{im}\varphi| = 2 \cdot 3 — a first glimpse of the first isomorphism theorem.

Worked examples — spot the kernel and image

Homomorphisms are everywhere once you learn to look. In each case, the trick is to ask two questions: what collapses to the identity (the kernel), and what gets reached (the image)?

Determinant. \det : GL_n(\mathbb{R}) \to \mathbb{R}^{\times}. Since \det(AB) = \det(A)\det(B) it is a homomorphism from invertible matrices (under multiplication) to nonzero reals (under multiplication). Its kernel is \{A : \det A = 1\} = SL_n(\mathbb{R}); its image is all of \mathbb{R}^{\times}.
Sign of a permutation. \operatorname{sgn} : S_n \to \{+1, -1\}. The sign is +1 for even permutations and -1 for odd, and \operatorname{sgn}(\sigma\tau) = \operatorname{sgn}(\sigma)\operatorname{sgn}(\tau). Its kernel is the alternating group A_n (the even permutations); its image is \{\pm 1\}.
Reduction mod n. \varphi : \mathbb{Z} \to \mathbb{Z}_n, k \mapsto k \bmod n. Kernel = n\mathbb{Z} (the multiples of n); image = \mathbb{Z}_n, so it is surjective but very far from injective.
Squaring. \varphi : \mathbb{C}^{\times} \to \mathbb{C}^{\times}, z \mapsto z^2. Since (zw)^2 = z^2 w^2 it is a homomorphism; its kernel is \{+1, -1\} (the square roots of 1) and its image is all of \mathbb{C}^{\times}.

Isomorphism: the same group in different clothes

A homomorphism that is also a bijection is called an isomorphism. If one exists between G and H we write G \cong H and say the groups are isomorphic. An isomorphism is a perfect relabelling: it pairs the elements up one-for-one and matches the multiplication tables exactly, so G and H are the same abstract group wearing different costumes. Anything you prove about one is instantly true of the other.

A small vocabulary grows around this, all built from the Greek morphe ("form"):

Monomorphism — an injective homomorphism (trivial kernel).
Epimorphism — a surjective homomorphism (image is all of H).
Isomorphism — bijective: both mono and epi at once.
Automorphism — an isomorphism from a group to itself, \varphi : G \to G (a symmetry of the group's own structure).

Two traps catch beginners. First: a homomorphism is not required to be a bijection. Reduction \mathbb{Z} \to \mathbb{Z}_n is surjective but wildly non-injective (infinitely many integers share each residue); the inclusion \mathbb{Z} \hookrightarrow \mathbb{R} under addition is injective but far from surjective. Injective, surjective, both, or neither — all four are allowed. Only the "both" case earns the name isomorphism.

Second — the sneakier one: the defining law is \varphi(ab) = \varphi(a)\varphi(b), where each side uses that group's own operation. It does not say \varphi(a+b) = \varphi(a) + \varphi(b) unless both groups happen to be additive. When the operations differ, the equation genuinely mixes them. Take the exponential \exp : (\mathbb{R}, +) \to (\mathbb{R}^{\times}, \times): the homomorphism property is \exp(a + b) = \exp(a)\cdot\exp(b) — a plus on the left turns into a times on the right. Writing \exp(a+b) = \exp(a) + \exp(b) would be flatly false. Always ask which operation lives on which side before you apply the rule.

Here is an isomorphism that feels like it shouldn't exist. The additive group of all real numbers, (\mathbb{R}, +), is isomorphic to the multiplicative group of positive reals, (\mathbb{R}^{+}, \times):

\exp : (\mathbb{R}, +) \;\xrightarrow{\;\cong\;}\; (\mathbb{R}^{+}, \times), \qquad x \mapsto e^{x}.

It is a homomorphism because e^{x+y} = e^{x}e^{y}; it is injective (the exponential never repeats a value, so the kernel is just \{0\}); and it is surjective onto the positive reals, with inverse \ln. So "the reals under addition" and "the positive reals under multiplication" are, as groups, the very same object — two costumes on one skeleton. This is not a curiosity: it is exactly why a slide rule turns multiplication into the physical act of adding lengths, and why logarithms were history's first great computational shortcut. An isomorphism lets you do your arithmetic in whichever costume is easier.