Kernels and Images in Homomorphisms and Isomorphisms

students, imagine two different algebra systems talking to each other 📘🔁. A homomorphism is a rule that translates the operation from one structure to another without breaking the algebraic pattern. In this lesson, you will learn two of the most important ideas connected to a homomorphism: the kernel and the image.

Learning goals:

Explain what the kernel and image of a homomorphism mean.
Identify kernels and images in examples.
Use kernels and images to understand whether a map is one-to-one or onto.
Connect these ideas to isomorphisms and the First Isomorphism Theorem.
Apply algebra reasoning to solve problems involving homomorphisms.

These ideas matter because they tell us what a homomorphism “forgets” and what it “keeps.” That makes them central to understanding when two algebraic structures are essentially the same. 🌟

What a Homomorphism Does

A homomorphism is a function between algebraic structures that preserves the operation. For example, if $f$ is a homomorphism between groups, then for all elements $a$ and $b$ in the first group,

$$f(ab)=f(a)f(b).$$

If we are working with additive groups, the rule becomes

$$f(a+b)=f(a)+f(b).$$

This preservation property is what makes homomorphisms so useful. They let us study one structure by mapping it into another. Think of it like translating a sentence from one language to another while keeping the meaning. The exact words may change, but the structure of the message stays intact.

Once we have a homomorphism, two questions become especially important:

Which elements get sent to the identity element?
Which elements appear as outputs?

These questions lead directly to the kernel and image.

The Kernel: What Gets Sent to the Identity

The kernel of a homomorphism is the set of elements in the domain that map to the identity element in the codomain.

If $f:G\to H$ is a group homomorphism, then the kernel is

$$\ker(f)=\{g\in G\mid f(g)=e_H\}.$$

Here, $e_H$ means the identity element in $H$.

If the groups are written additively, the identity is $0$, so the kernel is

$$\ker(f)=\{g\in G\mid f(g)=0\}.$$

Why the kernel matters

The kernel tells us exactly which inputs become invisible after applying the homomorphism. If an element is in the kernel, then the map sends it to the identity, so it disappears in the output. In that sense, the kernel measures what the homomorphism collapses.

A key fact in abstract algebra is that the kernel of a homomorphism is always a subgroup of the domain. In group theory, it is even a normal subgroup. That is one reason kernels are so important: they are not random sets, but structured pieces of the original algebraic object.

Example 1: Modulo 3 map

Consider the function $f:\mathbb{Z}\to\mathbb{Z}_3$ defined by

$$f(n)=n \bmod 3.$$

This is a homomorphism from the integers under addition to the integers mod $3$ under addition.

Which integers map to $0$ in $\mathbb{Z}_3$? Exactly those divisible by $3$.

$$\ker(f)=\{\dots,-6,-3,0,3,6,9,\dots\}=3\mathbb{Z}.$$

This means all multiples of $3$ vanish under the map. The kernel captures that pattern perfectly.

Example 2: Squaring map in a group

Let $f:\mathbb{R}^\times\to\mathbb{R}^\times$ be defined by

$$f(x)=x^2,$$

where $\mathbb{R}^\times$ is the group of nonzero real numbers under multiplication.

This is a homomorphism because

$$f(xy)=(xy)^2=x^2y^2=f(x)f(y).$$

To find the kernel, solve

$$x^2=1.$$

The solutions are

$$x=1 \quad \text{or} \quad x=-1.$$

Thus,

$$\ker(f)=\{-1,1\}.$$

These are exactly the elements that become the identity $1$ after squaring.

The Image: What the Map Actually Reaches

The image of a homomorphism is the set of all outputs produced by the map.

If $f:G\to H$, then the image is

$$\operatorname{im}(f)=\{f(g)\mid g\in G\}.$$

The image is the part of the codomain that is actually hit by the function.

Why the image matters

The image tells us how much of the target structure is covered. If the image is the whole codomain, then the homomorphism is surjective or onto. If the image is smaller than the codomain, then some elements of the codomain are never reached.

The image of a homomorphism is always a subgroup of the codomain. So again, we get structure, not just a set of outputs.

Example 3: Modulo 3 map again

For $f:\mathbb{Z}\to\mathbb{Z}_3$ given by $f(n)=n \bmod 3$, the image is

$$\operatorname{im}(f)=\mathbb{Z}_3.$$

Every residue class $0$, $1$, and $2$ appears as an output. So this map is surjective.

Example 4: Inclusion of even integers

Consider the map $i:2\mathbb{Z}\to\mathbb{Z}$ defined by

$$i(n)=n.$$

This is a homomorphism from the even integers to the integers. Its image is

$$\operatorname{im}(i)=2\mathbb{Z}.$$

The codomain is $\mathbb{Z}$, but the map only reaches the even integers. So this homomorphism is not surjective.

Kernel and Image Together

The kernel and image give two complementary views of a homomorphism:

The kernel tells us what is lost.
The image tells us what is reached.

Together, they describe how the map behaves.

A homomorphism with a small kernel is closer to being one-to-one. A homomorphism with a large image is closer to being onto. That is why these two objects are so useful when studying isomorphisms.

One-to-one and the kernel

A homomorphism is injective if different inputs always give different outputs. For group homomorphisms, there is a very important result:

$$f \text{ is injective } \iff \ker(f)=\{e_G\}.$$

In other words, a homomorphism is one-to-one exactly when the only element that goes to the identity is the identity itself.

Example 5: Injective map

Let $f:\mathbb{Z}\to\mathbb{Z}$ be defined by

$$f(n)=2n.$$

This is a homomorphism under addition.

$$f(n)=0,$$

then

$$2n=0,$$

$$n=0.$$

Thus,

$$\ker(f)=\{0\}.$$

So $f$ is injective.

Onto and the image

A homomorphism is surjective if its image equals the codomain:

$$f \text{ is surjective } \iff \operatorname{im}(f)=H.$$

This means every element of the codomain has at least one preimage in the domain.

Isomorphisms: When the Structure is the Same

An isomorphism is a homomorphism that is both injective and surjective. It preserves operation structure perfectly and matches every element of one algebraic structure with exactly one element of another.

For a group homomorphism $f:G\to H$ to be an isomorphism, both of these must be true:

$$\ker(f)=\{e_G\}$$

and

$$\operatorname{im}(f)=H.$$

So kernels and images help us check whether a map is an isomorphism.

Example 6: A simple isomorphism

Define $f:\mathbb{Z}\to 2\mathbb{Z}$ by

$$f(n)=2n.$$

This map is a homomorphism.

Its kernel is $\{0\}$, so it is injective.
Its image is $2\mathbb{Z}$, which is the whole codomain.

Therefore, $f$ is an isomorphism from $\mathbb{Z}$ onto $2\mathbb{Z}$.

This shows something important: two structures can be isomorphic even if they look different at first. The integers and the even integers are the same as groups under addition, even though one is a subset of the other. 🔍

The Big Idea Behind the First Isomorphism Theorem

The First Isomorphism Theorem connects kernels, images, and quotient structures.

For a group homomorphism $f:G\to H$,

$$G/\ker(f) \cong \operatorname{im}(f).$$

This means that if we collapse the kernel of $f$ down to the identity, the remaining structure is isomorphic to the image.

Why is this important? Because it tells us that every homomorphism factors through a quotient group. In simple terms, the kernel describes exactly what must be “compressed” so that the map becomes injective.

Intuitive meaning

If two elements of $G$ differ by something in the kernel, then $f$ sends them to the same output. So the quotient $G/\ker(f)$ treats those elements as equivalent. After we identify them, the map becomes one-to-one on the quotient.

This theorem is one of the strongest reasons kernels matter. They are not just a definition; they reveal the hidden structure behind a homomorphism.

Conclusion

students, kernels and images are two of the most important tools for understanding homomorphisms. The kernel shows which elements disappear under the map, while the image shows which elements are reached. The kernel helps determine whether the map is injective, and the image helps determine whether it is surjective. Together, they help identify isomorphisms and lead to the First Isomorphism Theorem, which explains how every homomorphism is built from a quotient and an image.

If you remember just one idea, remember this: a homomorphism keeps the algebraic operation, but the kernel records what gets collapsed and the image records what survives. That is the heart of the lesson. ✅

Study Notes

A homomorphism preserves the operation, such as $f(ab)=f(a)f(b)$ or $f(a+b)=f(a)+f(b)$.
The kernel is the set of elements mapped to the identity: $\ker(f)=\{g\in G\mid f(g)=e_H\}$.
The image is the set of all outputs: $\operatorname{im}(f)=\{f(g)\mid g\in G\}$.
For group homomorphisms, $\ker(f)$ is a subgroup of the domain, and often a normal subgroup.
The image of a homomorphism is always a subgroup of the codomain.
A homomorphism is injective if and only if $\ker(f)=\{e_G\}$.
A homomorphism is surjective if and only if $\operatorname{im}(f)=H$.
An isomorphism is both injective and surjective.
The First Isomorphism Theorem says $G/\ker(f)\cong \operatorname{im}(f)$.
Kernels show what gets collapsed; images show what is reached.