Group Homomorphisms

In this lesson you’ll learn what a homomorphism is, how to compute a kernel and an image, why the kernel is always normal, and how the first isomorphism theorem ties every homomorphism back to a quotient group.

A homomorphism is a map $\varphi: G \to H$ between groups satisfying

$$\varphi(ab) = \varphi(a)\varphi(b) \qquad \text{for all } a, b \in G$$

Read it carefully: the product on the left happens in $G$, the product on the right happens in $H$. The condition says combining first and then mapping gives the same result as mapping first and then combining.

That single equation forces everything else.

$$\varphi(e_G) = e_H, \qquad \varphi(a^{-1}) = \varphi(a)^{-1}, \qquad \varphi(a^n) = \varphi(a)^n$$

The first: $\varphi(e) = \varphi(ee) = \varphi(e)\varphi(e)$, and cancelling one copy gives $\varphi(e) = e_H$. The second follows by applying $\varphi$ to $aa^{-1} = e$.

So you never need to require that identities and inverses are respected. They come free, which is a nice illustration of how much a well-chosen definition does.

Two subsets carry all the information.

$$\ker \varphi = \{g \in G : \varphi(g) = e_H\}, \qquad \operatorname{im} \varphi = \{\varphi(g) : g \in G\}$$

The kernel is everything crushed to the identity. The image is what actually gets hit.

$\ker \varphi \trianglelefteq G$. It’s a subgroup: if $\varphi(a) = \varphi(b) = e$ then $\varphi(ab^{-1}) = \varphi(a)\varphi(b)^{-1} = e$. And it’s normal: for $k \in \ker\varphi$ and any $g$,

$$\varphi(gkg^{-1}) = \varphi(g)\varphi(k)\varphi(g)^{-1} = \varphi(g) e_H \varphi(g)^{-1} = e_H$$

$\operatorname{im}\varphi \le H$, by a similar one-line check. But it is generally not normal in $H$.

That asymmetry is worth noting: kernels are always normal, images generally aren’t. In fact every normal subgroup is the kernel of some homomorphism - the projection $G \to G/N$ - so “normal subgroup” and “kernel” describe the same class of objects.

$$\varphi \text{ is injective} \iff \ker \varphi = \{e_G\}$$

If the kernel is trivial and $\varphi(a) = \varphi(b)$, then $\varphi(ab^{-1}) = e$, so $ab^{-1} \in \ker\varphi = \{e\}$, giving $a = b$. Conversely if $k \ne e$ is in the kernel, then $k$ and $e$ have the same image, so $\varphi$ isn’t injective.

The kernel also tells you the exact amount of collapsing: every fibre has size $|\ker \varphi|$. All the preimages are cosets of the kernel, which is why they came out equal-sized in the figure.

$$G / \ker \varphi \;\cong\; \operatorname{im} \varphi$$

This is the central theorem of the lesson, and arguably of the whole group-theory half of the section. In words: quotienting out exactly what a map forgets leaves precisely what it remembers.

The proof is short. Define $\Phi: G/\ker\varphi \to \operatorname{im}\varphi$ by $\Phi(g \ker\varphi) = \varphi(g)$. It’s well-defined because $g\ker\varphi = g'\ker\varphi$ implies $g^{-1}g' \in \ker\varphi$, hence $\varphi(g) = \varphi(g')$. It’s a homomorphism because $\varphi$ is. It’s surjective onto the image by construction, and injective because $\Phi(g\ker\varphi) = e$ forces $g \in \ker\varphi$, the identity coset. ∎

The counting corollary is used constantly:

$$|G| = |\ker\varphi| \cdot |\operatorname{im}\varphi|$$

• Monomorphism - injective. Trivial kernel. Embeds $G$ inside $H$.
• Epimorphism - surjective. The image is all of $H$.
• Isomorphism - both. The groups are structurally identical.
• Endomorphism - a homomorphism from $G$ to itself.
• Automorphism - an isomorphism from $G$ to itself. These form a group $\operatorname{Aut}(G)$ under composition.

Example 1: Is $\varphi: \mathbb{Z} \to \mathbb{Z}$ with $\varphi(x) = 3x$ a homomorphism?

Solution. Both groups are additive, so check $\varphi(a+b) = \varphi(a) + \varphi(b)$:

$$\varphi(a+b) = 3(a+b) = 3a + 3b = \varphi(a) + \varphi(b) \ \checkmark$$

Yes. $\ker\varphi = \{x : 3x = 0\} = \{0\}$, so it’s injective. $\operatorname{im}\varphi = 3\mathbb{Z}$, so it’s not surjective.

By the first isomorphism theorem, $\mathbb{Z}/\{0\} \cong 3\mathbb{Z}$, which just says $\mathbb{Z} \cong 3\mathbb{Z}$: an infinite group can be isomorphic to a proper subgroup of itself.

Example 2: Is $\varphi(x) = x^2$ a homomorphism on $\mathbb{Z}$?

Solution. Need $\varphi(a+b) = \varphi(a) + \varphi(b)$, that is $(a+b)^2 = a^2 + b^2$.

Take $a = b = 1$: left side $4$, right side $2$. No.

But on the multiplicative group of nonzero reals, $\varphi(x) = x^2$ satisfies $(ab)^2 = a^2b^2$ and is a homomorphism. The same formula can be a homomorphism for one operation and not another, which is why naming the operation matters.

Example 3: Kernel and image of reduction mod $n$.

Let $\varphi: \mathbb{Z} \to \mathbb{Z}_6$ be $\varphi(x) = x \bmod 6$.

Solution. Homomorphism: $(a+b) \bmod 6 = (a \bmod 6) + (b \bmod 6)$ in $\mathbb{Z}_6$. ✓

$$\ker\varphi = 6\mathbb{Z}, \qquad \operatorname{im}\varphi = \mathbb{Z}_6$$

Surjective, not injective. The first isomorphism theorem gives

$$\mathbb{Z}/6\mathbb{Z} \cong \mathbb{Z}_6$$

which is the formal justification of the notation $\mathbb{Z}/n\mathbb{Z}$ for $\mathbb{Z}_n$.

Example 4: The determinant.

Show $\det: GL_2(\mathbb{R}) \to \mathbb{R}^*$ is a homomorphism and find its kernel.

Solution. $\det(AB) = \det(A)\det(B)$ is the multiplicative property of determinants, which is exactly the homomorphism condition from matrix multiplication to real multiplication. ✓

$$\ker(\det) = \{A : \det A = 1\} = SL_2(\mathbb{R})$$

the special linear group. Since kernels are normal, $SL_2(\mathbb{R}) \trianglelefteq GL_2(\mathbb{R})$ for free - a genuinely useful fact obtained without any conjugation computation.

The map is surjective (scale a diagonal matrix), so

$$GL_2(\mathbb{R})/SL_2(\mathbb{R}) \cong \mathbb{R}^*$$

Example 5: The sign homomorphism.

Solution. Define $\operatorname{sign}: S_n \to \{1, -1\}$ sending even permutations to $1$ and odd to $-1$. Parities add, so signs multiply, and it’s a homomorphism.

$$\ker(\operatorname{sign}) = A_n$$

So $A_n$ is normal in $S_n$, again for free. The map is surjective for $n \ge 2$, so

$$S_n / A_n \cong \mathbb{Z}_2, \qquad |A_n| = \frac{n!}{2}$$

Both of those facts were asserted in earlier lessons; here they drop out of one homomorphism.

Example 6: Counting with the kernel.

$\varphi: \mathbb{Z}_{20} \to \mathbb{Z}_{20}$ is given by $\varphi(x) = 8x$. Find the kernel, the image, and check the counting formula.

Solution. $\ker\varphi = \{x : 8x \equiv 0 \pmod{20}\}$. Since $\gcd(8,20) = 4$, this needs $2x \equiv 0 \pmod 5$, so $x \equiv 0 \pmod 5$:

$$\ker\varphi = \{0, 5, 10, 15\}, \qquad |\ker\varphi| = 4$$

The image is the multiples of $\gcd(8,20) = 4$:

$$\operatorname{im}\varphi = \{0,4,8,12,16\}, \qquad |\operatorname{im}\varphi| = 5$$

Check: $4 \times 5 = 20 = |\mathbb{Z}_{20}|$ ✓. And $\mathbb{Z}_{20}/\ker\varphi \cong \mathbb{Z}_5$, matching the image.

Example 7: A map that fails the definition.

Is $\varphi: \mathbb{Z}_6 \to \mathbb{Z}_6$ with $\varphi(x) = x + 1$ a homomorphism?

Solution. Homomorphisms must send the identity to the identity, and $\varphi(0) = 1 \ne 0$.

No. Check directly: $\varphi(1 + 2) = \varphi(3) = 4$, but $\varphi(1) + \varphi(2) = 2 + 3 = 5$.

This is a common trap: shifting looks structure-like but isn’t. A translation is a bijection, and it is a symmetry of the set, but it does not respect the operation.

Hash functions and checksums. A checksum that satisfies $h(x + y) = h(x) + h(y)$ is a homomorphism, and its kernel is exactly the set of undetectable errors. Designing a good checksum means choosing a kernel that avoids the error patterns you expect, which is why CRC polynomials are chosen so carefully.

Homomorphic encryption. An encryption scheme is homomorphic when $E(a) \cdot E(b) = E(a + b)$, letting a server compute on encrypted data without decrypting it. Fully homomorphic encryption, first constructed in 2009, is the extreme version, and it is a straight application of this definition to real privacy problems.

Fourier and Laplace transforms. The Fourier transform converts convolution into pointwise multiplication, which is a homomorphism between two group structures on functions. This is precisely why signal processing and differential-equation solving get easier in the transform domain.

Logarithms. $\log(xy) = \log x + \log y$ says the logarithm is a homomorphism from the positive reals under multiplication to the reals under addition. It is an isomorphism, and slide rules were physical devices for exploiting it.

Physics. A representation of a symmetry group is a homomorphism from the group into a group of matrices. Particle physics classifies particles by which representation they belong to, so this definition is the organizing principle of the Standard Model.

Compiler optimization. A program transformation is safe when it commutes with evaluation, which is a homomorphism condition. Reasoning about which optimizations preserve semantics is reasoning about structure-preserving maps.