TOPIC 2 / 3

Homomorphisms, Kernels & Images

Useful preparation: Groups, Subgroups & Cosets

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

Homomorphisms preserve operations while possibly merging inputs. The kernel records what disappears; every nonempty fiber is one translated copy of that kernel.

Group homomorphism: For groups G and H, a map $\displaystyle \phi : G \to H$ is a homomorphism when $\displaystyle \phi\, \left(ab\right) = \phi\, \left(a\right)\, \phi\, \left(b\right)$ for all a,b in G. In additive groups, replace products by sums.
Kernel and image: The kernel is the set of inputs sent to the identity of H. The image is the set of all attained outputs; it need not equal the entire codomain.
Fiber and isomorphism: The fiber over y is the set of inputs mapped to y. A bijective homomorphism is an isomorphism and identifies the two group structures.

Conditions before calculation

Check that a proposed rule is well-defined on the declared elements. For residue classes, different integer representatives must produce the same output class.
The finite counting identity requires a finite domain group and a genuine homomorphism. An arbitrary function need not have fibers of equal size.
The quotient G/ker $\displaystyle \phi$ is a group because the kernel is normal. The first isomorphism theorem identifies this quotient with the image, not automatically with the entire codomain.

Why the method works

1. Find what collisions mean

Two elements have the same output exactly when their relative difference belongs to the kernel. This follows from preservation of inverses and multiplication.

\varphi(a)=\varphi(b)\iff\varphi(b^{-1}a)=e_H\iff b^{-1}a\in\ker\varphi

2. Describe a whole fiber

If $\displaystyle \phi\, \left(a\right) = y$ , then multiplying a by any kernel element leaves its output unchanged. Conversely every input giving y differs from a by a kernel element; the fiber is a ker $\displaystyle \phi$ .

3. Count and interpret

Kernel cosets partition G and pair with image elements. For finite G, their equal sizes give the counting formula. A trivial kernel means no two inputs collide.

|G|=|\ker\varphi|\,|\operatorname{im}\varphi|,\qquad G/\ker\varphi\cong\operatorname{im}\varphi

Worked examples

EXAMPLE 1

Let $\displaystyle \phi$ : $\mathbb{Z}/18\mathbb{Z}$ → $\mathbb{Z}/18\mathbb{Z}$ send [x] to [6x]. Find its kernel and image.

Changing x by 18 changes 6x by 108, a multiple of 18. Thus the rule is well-defined; distributivity proves it preserves addition.
The kernel condition is 18 dividing 6x, equivalent to 3 dividing x.
$\ker\varphi=\{0,3,6,9,12,15\}$
The attained outputs are 0,6,12. Each has six preimages, consistent with the total 18.
$\operatorname{im}\varphi=\{0,6,12\},\qquad18=6\cdot3$

Conclusion: The map is neither injective nor surjective onto its declared codomain.

EXAMPLE 2

Let ψ:(Z,+)→(Z,+) be ψ $\displaystyle \left(n\right) = 4\, n$ . Is it an isomorphism?

For any integers m,n, multiplication distributes over addition.
$\psi(m+n)=4m+4n=\psi(m)+\psi(n)$
The only integer sent to 0 is 0, so the kernel is trivial and the map is injective.
The integer 1 has no integer preimage; the image is 4Z. Restricting the codomain to 4Z makes the map bijective.

Conclusion: It is not an isomorphism onto Z, but it is an isomorphism from Z to 4Z.

Pitfalls & counterexamples

Tempting claim: A trivial kernel guarantees surjectivity.

Correction: It guarantees injectivity. Surjectivity must be tested against the specified codomain.

Test the claim: The map n↦4n on Z has trivial kernel but misses 1, 2, and 3.

Check your understanding

A homomorphism has a domain group of order 42 and a kernel of order 7. How many elements are in its image?

Write a reason before opening the explanation. This reflection is not graded or saved.

Compare your reasoning

The image indexes the six equal kernel cosets, since $\displaystyle \frac{42}{7} = 6$ .

Connect this idea

Kernels group indistinguishable inputs. Linear maps have analogous kernels and images, with dimensions replacing finite group sizes.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.