Group Homomorphisms and the Isomorphism Theorems

⚡ Key Takeaways

A group homomorphism is a structure-preserving map $\varphi: G \to H$ satisfying $\varphi(ab) = \varphi(a)\varphi(b)$ . The First Isomorphism Theorem states that every homomorphism factors as $G/\ker\varphi \cong \operatorname{im}\varphi$ , identifying quotient groups with images. These results, developed by Noether, Dedekind, and Jordan (19th–early 20th c.), are the central organizing tool of group theory.

💡 The Core Insight

A homomorphism $\varphi: G \to H$ is completely controlled by two pieces of data: its kernel $K = \ker\varphi$ (a normal subgroup of $G$ ) and its image $\operatorname{im}\varphi$ (a subgroup of $H$ ). The fibers of $\varphi$ are precisely the cosets of $K$ , so the quotient $G/K$ — the set of cosets with induced multiplication — is in canonical bijection with the image, and that bijection is itself an isomorphism. To know a homomorphism is to know which normal subgroup it kills.

DefinitionGroup Homomorphism

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

1

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

Derivation hint: This single axiom is the definition; everything else (identity, inverses) is derived from it. Here $a \cdot b$ is multiplication in $G$ and $\varphi(a) * \varphi(b)$ is multiplication in $H$ . We write both as juxtaposition when context is clear. Two immediate consequences:

2

\quad \varphi(e_G) = e_H, \qquad \varphi(a^{-1}) = \varphi(a)^{-1}.

Derivation hint: Apply [1] to $e_G\,e_G = e_G$ and cancel; apply [1] to $a\,a^{-1}=e_G$ and use the first part.

DefinitionKernel and Image

For a homomorphism $\varphi: G \to H$ , define the kernel and image:

3

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

Derivation hint: The kernel is the preimage of the trivial subgroup $\{e_H\}$ ; the image is the set-theoretic range. The kernel is a normal subgroup of $G$ (written $K \trianglelefteq G$ ), and the image is a subgroup of $H$ .

DefinitionQuotient Group

If $N \trianglelefteq G$ , the set of left cosets $G/N = \{ gN : g \in G\}$ is a group under

4

\quad (aN)(bN) = (ab)N,

Derivation hint: Well-definedness requires normality: if $aN=a'N$ and $bN=b'N$ , normality $N a = a N$ lets you re-bracket $a'b' \in abN$ . with identity $eN = N$ and inverse $(gN)^{-1} = g^{-1}N$ . The map $\pi: G \to G/N$ , $g \mapsto gN$ , is a surjective homomorphism called the canonical projection, with $\ker\pi = N$ .

TheoremFirst Isomorphism Theorem

Let $\varphi: G \to H$ be a group homomorphism with kernel $K = \ker\varphi$ . Then $K \trianglelefteq G$ and the induced map

5

\quad \bar\varphi: G/K \longrightarrow \operatorname{im}\varphi, \qquad \bar\varphi(gK) = \varphi(g)

Derivation hint: Define $\bar\varphi$ on cosets; well-defined because $gK = g'K \iff g^{-1}g' \in K \iff \varphi(g)=\varphi(g')$ . is a group isomorphism. Equivalently, $\varphi$ factors as $\varphi = \iota \circ \bar\varphi \circ \pi$ , where $\pi$ is projection and $\iota$ the inclusion of $\operatorname{im}\varphi$ into $H$ :

6

\quad G/\ker\varphi \;\cong\; \operatorname{im}\varphi.

Derivation hint: This is the statement that $\bar\varphi$ is a bijective homomorphism; see the proof below.

TheoremSecond (Diamond) Isomorphism Theorem

Let $S \le G$ be a subgroup and $N \trianglelefteq G$ a normal subgroup. Then $SN = \{sn : s\in S, n \in N\}$ is a subgroup, $N \trianglelefteq SN$ , $S \cap N \trianglelefteq S$ , and

7

\quad S/(S \cap N) \;\cong\; SN/N.

Derivation hint: Apply the First Theorem to the composite $S \hookrightarrow SN \twoheadrightarrow SN/N$ ; its kernel is $S \cap N$ , its image is $SN/N$ .

TheoremThird (Quotient) Isomorphism Theorem

Let $N \trianglelefteq G$ and $K \trianglelefteq G$ with $N \le K$ . Then $K/N \trianglelefteq G/N$ and

8

\quad (G/N)\big/(K/N) \;\cong\; G/K.

Derivation hint: The map $G/N \to G/K$ , $gN \mapsto gK$ , is a well-defined surjective homomorphism with kernel $K/N$ ; apply the First Theorem.

TheoremFourth (Correspondence/Lattice) Theorem

Let $N \trianglelefteq G$ with projection $\pi: G \to G/N$ . There is an inclusion-preserving bijection between subgroups of $G$ containing $N$ and subgroups of $G/N$ , given by $S \mapsto S/N$ . Under it, normal subgroups correspond to normal subgroups, and indices are preserved.

ProofFirst Isomorphism Theorem

We prove each claim about $\varphi: G \to H$ with $K = \ker\varphi$ .

Step 1 — $K$ is a subgroup. If $a,b \in K$ then $\varphi(ab^{-1}) = \varphi(a)\varphi(b)^{-1} = e_H e_H^{-1} = e_H$ by [1] and [2], so $ab^{-1} \in K$ . Also $e_G \in K$ by [2]. Hence $K \le G$ .

Step 2 — $K$ is normal. For $g \in G$ and $k \in K$ :

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

so $gkg^{-1} \in K$ . Thus $gKg^{-1} \subseteq K$ for all $g$ , i.e. $K \trianglelefteq G$ .

Step 3 — Fibers are cosets. For $g, g' \in G$ :

\varphi(g) = \varphi(g') \iff \varphi(g^{-1}g') = e_H \iff g^{-1}g' \in K \iff gK = g'K.

So the level sets of $\varphi$ are exactly the cosets of $K$ .

Step 4 — $\bar\varphi$ is well-defined. Define $\bar\varphi(gK) = \varphi(g)$ . By Step 3, if $gK = g'K$ then $\varphi(g) = \varphi(g')$ , so the value does not depend on the coset representative.

Step 5 — $\bar\varphi$ is a homomorphism. Using [4] and [1]:

\bar\varphi\big((aK)(bK)\big) = \bar\varphi(abK) = \varphi(ab) = \varphi(a)\varphi(b) = \bar\varphi(aK)\bar\varphi(bK).

Step 6 — $\bar\varphi$ is injective. If $\bar\varphi(gK) = e_H$ then $\varphi(g) = e_H$ , so $g \in K$ , hence $gK = K$ , the identity of $G/K$ . The kernel of $\bar\varphi$ is trivial, so it is injective.

Step 7 — $\bar\varphi$ is surjective onto $\operatorname{im}\varphi$ . Every element of $\operatorname{im}\varphi$ has the form $\varphi(g) = \bar\varphi(gK)$ . By Steps 5–7, $\bar\varphi: G/K \to \operatorname{im}\varphi$ is a bijective homomorphism, i.e. an isomorphism, proving [6]. $\blacksquare$

∎

ProofThird Isomorphism Theorem (illustrating reuse)

Define $\psi: G/N \to G/K$ by $\psi(gN) = gK$ . Since $N \le K$ , if $gN = g'N$ then $g^{-1}g' \in N \subseteq K$ , so $gK = g'K$ ; thus $\psi$ is well-defined. It is a homomorphism: $\psi(aN\,bN) = \psi(abN) = abK = (aK)(bK)$ . It is surjective since every $gK$ is hit. Its kernel is

\ker\psi = \{ gN : gK = K\} = \{ gN : g \in K \} = K/N.

By the First Isomorphism Theorem applied to $\psi$ , $(G/N)/(K/N) \cong G/K$ , which is [8]. $\blacksquare$

∎

mathematician · Emmy Noether | 1882–1935

Emmy Noether crystallized the modern abstract viewpoint in which the Isomorphism Theorems are stated. Working in Göttingen, she reformulated algebra around homomorphisms, ideals, modules, and chain conditions rather than explicit element computations. Her 1927 paper Abstrakter Aufbau der Idealtheorie and her influence on van der Waerden's Moderne Algebra (1930–31) made the "First Isomorphism Theorem" framing standard. Noether's broader program — treating algebraic structures via their structure-preserving maps — is the conceptual ancestor of category theory. She is equally celebrated in physics for Noether's theorem linking symmetries to conservation laws.

🧠 Intuition

Think of a homomorphism as a lossy compression of a group. The kernel is exactly the information that gets thrown away: everything in the same coset of $K$ becomes indistinguishable after applying $\varphi$ . The quotient $G/K$ is the "compressed file" — and the First Isomorphism Theorem says this compressed file is a perfect, lossless copy of the image. Nothing is wasted: the only collapse that happens is the unavoidable one forced by the kernel. A second picture: $\varphi$ partitions $G$ into the fibers $\varphi^{-1}(h)$ . These fibers all have the same size (they are cosets of $K$ ), so $\varphi$ is like a flattening that stacks each coset onto one point. The third theorem then says quotients can be taken in stages — "dividing by $K$ all at once" equals "dividing by $N$ first, then by the leftover $K/N$ " — just like cancelling fractions.

Figure

The First Isomorphism Theorem as a factorization diagram

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ExampleDeterminant:

GL_2(\mathbb{R}) \to \mathbb{R}^\times

Given: The determinant map $\det: GL_2(\mathbb{R}) \to \mathbb{R}^\times$ (nonzero reals under multiplication).

Step 1 — Homomorphism. $\det(AB) = \det(A)\det(B)$ , so [1] holds.

Step 2 — Image. For any $r \neq 0$ , the matrix $\operatorname{diag}(r,1)$ has determinant $r$ , so $\operatorname{im}\det = \mathbb{R}^\times$ (surjective).

Step 3 — Kernel. $\ker\det = \{A : \det A = 1\} = SL_2(\mathbb{R})$ , the special linear group.

Step 4 — Apply [6]. The First Isomorphism Theorem gives:

\boxed{\,GL_2(\mathbb{R})\big/SL_2(\mathbb{R}) \;\cong\; \mathbb{R}^\times\,}

Interpretation: Collapsing all volume-preserving (det $=1$ ) transformations identifies a matrix only by its scaling factor on area. The quotient remembers exactly the determinant and nothing else.

ExampleReduction mod

n

:

\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}

, with a Third-Theorem computation

Given: $\varphi: \mathbb{Z} \to \mathbb{Z}/12\mathbb{Z}$ , $\varphi(k) = k \bmod 12$ , and the subgroups $N = 12\mathbb{Z}$ , $K = 4\mathbb{Z}$ with $N \le K \le \mathbb{Z}$ .

Step 1 — First Theorem. $\varphi$ is surjective with $\ker\varphi = 12\mathbb{Z}$ , so $\mathbb{Z}/12\mathbb{Z} \cong \mathbb{Z}/12\mathbb{Z}$ (the canonical case).

Step 2 — Set up the Third Theorem. Take $G = \mathbb{Z}$ , $N = 12\mathbb{Z}$ , $K = 4\mathbb{Z}$ . Then $K/N = 4\mathbb{Z}/12\mathbb{Z}$ is the subgroup $\{0, 4, 8\} \subseteq \mathbb{Z}/12\mathbb{Z}$ , of order $3$ .

Step 3 — Apply [8].

(\mathbb{Z}/12\mathbb{Z})\big/(4\mathbb{Z}/12\mathbb{Z}) \;\cong\; \mathbb{Z}/4\mathbb{Z}.

Step 4 — Order check. $|\mathbb{Z}/12\mathbb{Z}| = 12$ , $|4\mathbb{Z}/12\mathbb{Z}| = 3$ , and $12/3 = 4$ . Consistent.

\boxed{\,(\mathbb{Z}/12\mathbb{Z})\big/\{0,4,8\} \;\cong\; \mathbb{Z}/4\mathbb{Z}\,}

Interpretation: "Reducing mod 12, then mod 4" equals "reducing mod 4" — cancellation of the modulus, exactly the staged-quotient idea.

⚠️ Common mistake

Not every subgroup is a kernel. Only normal subgroups arise as kernels of homomorphisms, and only normal subgroups admit a well-defined quotient group via [4]. If $N$ is not normal, the coset multiplication $(aN)(bN) = abN$ is not well-defined, so $G/N$ is at best a set of cosets, not a group.

$\operatorname{im}\varphi$ need not be normal in $H$ . The image is always a subgroup of $H$ , but it need not be normal. Normality belongs to the kernel inside $G$ , not the image inside $H$ .

$G/K \cong \operatorname{im}\varphi$ , not $G/K \cong H$ . The quotient matches the image, which may be a proper subgroup of the codomain. Confusing image with codomain is the most common error.

Counterexample

A non-normal subgroup with no quotient group. Let $G = S_3$ and $S = \{e, (1\,2)\}$ , a subgroup of order 2. It is not normal: $(1\,3)(1\,2)(1\,3)^{-1} = (2\,3) \notin S$ . Try to multiply cosets. The left cosets are $S$ , $(1\,3)S = \{(1\,3),(1\,2\,3)\}$ , $(2\,3)S = \{(2\,3),(1\,3\,2)\}$ . Attempt $(1\,3)S \cdot (1\,3)S$ via representatives:

Using $(1\,3)\cdot(1\,3) = e$ : product coset $= S$ .
Using the other representatives $(1\,2\,3)\cdot(1\,2\,3) = (1\,3\,2)$ : product coset $= (2\,3)S$ .

Two representatives give different answers ( $S \neq (2\,3)S$ ), so coset multiplication is ill-defined. Hence $S_3/S$ is not a group, confirming that normality is essential and that $S$ cannot be a kernel.

connection

The Isomorphism Theorems sit at the hub of algebra. They are the group-theoretic special case of theorems holding for Rings and Ring Homomorphisms (with ideals replacing normal subgroups) and Modules over a Ring (with submodules). The kernel–image dichotomy generalizes to the notion of exact sequences: $1 \to K \to G \to \operatorname{im}\varphi \to 1$ is a short exact sequence, the starting point of Homological Algebra. The Correspondence Theorem underlies the Lattice of Subgroups and is used heavily in Galois Theory, where the Galois correspondence is its field-theoretic analogue. All of this is unified by Category Theory, where homomorphisms are morphisms, quotients are coequalizers, and the First Isomorphism Theorem becomes the existence of an image factorization.

🔬 Where this appears

⏳ History

- 1830–1832 — Évariste Galois introduces normal subgroups (his "diviseurs propres") implicitly while studying solvability of equations; the seed of the quotient idea.

- 1854 — Arthur Cayley gives an abstract definition of a group, enabling maps between groups to be considered.

- 1869–1870 — Camille Jordan develops permutation-group theory in his Traité des substitutions, proving early forms of composition-series results (the Jordan–Hölder theorem) that depend on quotients.

- 1889 — Otto Hölder completes and corrects Jordan's results, firmly establishing quotient groups

G/N

as legitimate objects and proving the Jordan–Hölder theorem.

- 1895–1896 — Heinrich Weber systematizes abstract group theory in his Lehrbuch der Algebra, standardizing homomorphism terminology.

- 1920s — Emmy Noether reframes algebra around homomorphisms and quotients; the Isomorphism Theorems acquire their modern statement and proof via the kernel–image factorization.

- 1927 — Noether's Abstrakter Aufbau der Idealtheorie articulates the structural viewpoint for rings/modules, with the group case as prototype.

- 1930–1931 — B. L. van der Waerden publishes Moderne Algebra, the textbook that disseminated Noether's and Artin's formulations worldwide, fixing the "First/Second/Third Isomorphism Theorem" naming.

- 1945 — Eilenberg & Mac Lane found category theory, recasting the theorems as universal properties of quotients and images.

open-problem

The Isomorphism Theorems are theorems (fully proved), but the structures they describe leave deep open territory:

Classifying all groups up to isomorphism is hopeless in general. Even with the theorems, the number of finite groups of order $2^n$ grows super-polynomially; counting groups of a given order is a hard, partly open enumeration problem.
The extension problem. Knowing $N \trianglelefteq G$ and $G/N$ does not determine $G$ : there may be many non-isomorphic groups with the same normal subgroup and quotient (e.g. $\mathbb{Z}/4$ vs. $\mathbb{Z}/2 \times \mathbb{Z}/2$ , both extensions of $\mathbb{Z}/2$ by $\mathbb{Z}/2$ ). Classifying extensions is governed by group cohomology $H^2(G/N, N)$ , an active area.
Infinite and topological refinements. For topological or Lie groups, $G/K$ must be given a quotient topology and the "isomorphism" must be a homeomorphism; the Open Mapping Theorem (for locally compact groups) is needed for the topological First Isomorphism Theorem to hold, and counterexamples exist without completeness hypotheses.

Group Homomorphisms and the Isomorphism Theorems

Formal Statement

Why It Holds

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also