First Isomorphism Theorem (for groups)

Field: Algebra · Level: First two years of university · Last updated: 2026-09-14

In one sentence: if $\varphi$ is a structure-preserving map from a group $G$ to a group $H$ , then $G$ "divided by" the part that $\varphi$ collapses (the kernel) has exactly the same structure as the part of $H$ that $\varphi$ actually reaches (the image). $G / \ker\varphi \;\cong\; \im\varphi$

In this chapter

What the theorem says
Preparation: the words we need
Three examples first
Statement
Why it is true: the intuition
Proof
Where it is used
Common misconceptions
Exercises
References

What the theorem says

A map $\varphi$ may send several elements of $G$ to the same place. The kernel $\ker\varphi$ is what decides which elements share a destination. The first isomorphism theorem says:

If you look at $G$ while treating elements that differ only by the kernel as "the same", what you see is indistinguishable, as a group, from the collection of destinations (the image).

In other words, the information that $\varphi$ loses and the information that $\varphi$ keeps separate cleanly.

This is the basic tool for understanding an abstractly defined quotient group $G/N$ as a concrete, familiar group.

Preparation: the words we need

Each of these notions gets its own chapter. Here we recall only what this chapter needs.

Group (coming soon)
Group homomorphism (coming soon)
Normal subgroup (coming soon)
Quotient group (coming soon)

Definition (group)

A set $G$ with a "multiplication" (a binary operation $a \cdot b$ ) is a group if (1) the operation is associative, $(ab)c = a(bc)$ ; (2) there is an identity $e$ with $ea = ae = a$ ; and (3) every $a$ has an inverse $a^{-1}$ with $aa^{-1} = a^{-1}a = e$ . Examples: the integers $\Z$ under addition (identity $0$ , inverse of $n$ is $-n$ ); the nonzero reals $\R^\times$ under multiplication.

Definition (homomorphism)

A map $\varphi$ from a group $G$ to a group $H$ is a homomorphism if for all $a, b \in G$ $\varphi(ab) = \varphi(a)\,\varphi(b).$ "Multiply first and then map, or map first and then multiply — same answer." A homomorphism preserves the structure of the operation. It follows automatically that $\varphi(e) = e$ and $\varphi(a^{-1}) = \varphi(a)^{-1}$ .

Definition (kernel and image)

For a homomorphism $\varphi : G \to H$ , $\ker\varphi = \set{\,g \in G \mid \varphi(g) = e\,}, \qquad \im\varphi = \set{\,\varphi(g) \mid g \in G\,}$ are the kernel and the image of $\varphi$ . The kernel is the set of elements sent to the identity; the image is the set of values that actually occur. The kernel is a subgroup of $G$ , the image a subgroup of $H$ .

Definition (normal subgroup and quotient group)

A subgroup $N$ of $G$ is normal if $g n g^{-1} \in N$ for every $g \in G$ and $n \in N$ . Given a normal subgroup $N$ , declare $g$ and $g'$ to be "the same" when $g^{-1}g' \in N$ . The set of resulting bundles is written $G/N$ , and the bundle (coset) containing $g$ is written $gN$ . When $N$ is normal, the rule $(gN)(g'N) = (gg')N$ is a well-defined operation and makes $G/N$ a group, the quotient group.

Quotient groups look very abstract at first. The value of the first isomorphism theorem is that it translates the abstract $G/N$ into a group you already know.

Three examples first

Before stating the theorem, let us see the pattern "divide by the kernel, get the image" in three cases.

Example 1: the sign of a real number

From the multiplicative group $\R^\times$ of nonzero reals to the group $\set{+1, -1}$ (under multiplication), define $\varphi(x) = \begin{cases} +1 & (x > 0) \\ -1 & (x < 0). \end{cases}$ Since "positive times positive is positive, positive times negative is negative, negative times negative is positive", we have $\varphi(xy) = \varphi(x)\varphi(y)$ : a homomorphism. The kernel is the set of $x$ with $\varphi(x) = +1$ , i.e. the positive reals $\R_{>0}$ . The image is all of $\set{+1,-1}$ . The theorem says $\R^\times / \R_{>0} \cong \set{+1, -1}$ : "ignore magnitude (identify numbers that differ by a positive factor) and all that remains is the sign." Exactly what intuition suggests.

Example 2: wrapping the line around a circle

From the additive group $\R$ to the multiplicative group $S^1 = \set{z \in \C \mid |z| = 1}$ of complex numbers of absolute value $1$ , take $\varphi(t) = e^{2\pi i t}.$ The exponential law $e^{2\pi i(s+t)} = e^{2\pi i s} e^{2\pi i t}$ says precisely that addition is sent to multiplication: a homomorphism. $\varphi(t) = 1$ exactly when $t$ is an integer, so the kernel is $\Z$ . The image is the whole circle $S^1$ . The theorem: $\R / \Z \cong S^1$ . "Roll up the real line, identifying points that differ by an integer, and you get a circle." It is the same idea as a clock face, where 13 o'clock and 1 o'clock are the same.

Example 3: the determinant

The invertible $n \times n$ real matrices $GL_n(\R)$ form a group under matrix multiplication, and the determinant $\det : GL_n(\R) \to \R^\times$ satisfies $\det(AB) = \det A \det B$ , so it is a homomorphism. The kernel is the set of matrices with determinant $1$ , written $SL_n(\R)$ . The image is all of $\R^\times$ (adjust one diagonal entry to get any nonzero value). The theorem: $GL_n(\R)/SL_n(\R) \cong \R^\times$ . "Ignore the determinant-one matrices, and what remains of an invertible matrix is a single number: its determinant."

In every example the kernel is "what $\varphi$ cannot tell apart", the image is "what $\varphi$ does tell apart", and dividing by the former leaves the latter.

Statement

Theorem (First Isomorphism Theorem)

Let $\varphi : G \to H$ be a homomorphism of groups. Then

$\ker\varphi$ is a normal subgroup of $G$ ;
the rule $gN \mapsto \varphi(g)$ (with $N = \ker\varphi$ ) defines a group isomorphism from the quotient $G/\ker\varphi$ onto the image $\im\varphi$ .

Hence $G / \ker\varphi \;\cong\; \im\varphi .$ In particular, if $\varphi$ is surjective (its image is all of $H$ ), then $G/\ker\varphi \cong H$ .

"Isomorphic", written $\cong$ , means that a bijective homomorphism exists: after renaming elements, the two groups are literally the same.

Why it is true: the intuition

Think of $\varphi$ as a light that projects the elements of $G$ onto $H$ . Some elements cast the same shadow: those $g, g'$ with $\varphi(g) = \varphi(g')$ . In that case $\varphi(g^{-1}g') = \varphi(g)^{-1}\varphi(g') = e,$ so $g^{-1}g'$ lies in the kernel. Conversely, if $g^{-1}g' \in \ker\varphi$ then $\varphi(g) = \varphi(g')$ . So:

"same shadow" $\iff$ "differ by an element of the kernel"

This one line is the heart of the theorem. The quotient $G/\ker\varphi$ is exactly "the world in which elements differing by the kernel are identified", so it matches the world of shadows, the image, one for one. And because $\varphi$ preserves products, the match respects the group structure.

Proof

Write $N = \ker\varphi$ . The proof has four steps; each begins by saying what it is going to show.

Step 1: the kernel is a normal subgroup

To show: if $g \in G$ and $n \in N$ then $g n g^{-1} \in N$ .

Because $\varphi$ is a homomorphism, $\varphi(g n g^{-1}) = \varphi(g)\,\varphi(n)\,\varphi(g)^{-1} = \varphi(g)\, e\, \varphi(g)^{-1} = e .$ So $g n g^{-1} \in N$ , and the quotient group $G/N$ exists.

Step 2: the map $\psi : G/N \to H$ is well defined

To show: setting $\psi(gN) = \varphi(g)$ does not depend on the representative $g$ chosen for the coset.

Suppose $gN = g'N$ . That means $g^{-1}g' \in N$ , so $\varphi(g^{-1}g') = e$ , i.e. $\varphi(g)^{-1}\varphi(g') = e$ , hence $\varphi(g) = \varphi(g')$ . Changing the representative does not change the value.

Step 3: $\psi$ is a homomorphism

To show: $\psi\big((gN)(g'N)\big) = \psi(gN)\,\psi(g'N)$ .

By the definition of the quotient operation and the homomorphism property of $\varphi$ , $\psi\big((gN)(g'N)\big) = \psi(gg'N) = \varphi(gg') = \varphi(g)\varphi(g') = \psi(gN)\,\psi(g'N).$

Step 4: $\psi$ is injective and its image is $\im\varphi$

Injective: if $\psi(gN) = \psi(g'N)$ then $\varphi(g) = \varphi(g')$ , so $\varphi(g^{-1}g') = e$ , so $g^{-1}g' \in N$ , i.e. $gN = g'N$ . Distinct cosets go to distinct places.

Image: the values of $\psi$ are exactly the elements of the form $\varphi(g)$ , so the image of $\psi$ is $\im\varphi$ . Therefore $\psi$ , with its codomain restricted to $\im\varphi$ , is a bijective homomorphism $G/N \to \im\varphi$ : an isomorphism.

Looking back, all we used was " $\varphi$ preserves products" and the definition of the kernel. That this alone forces the abstract quotient to coincide with the image is what makes the theorem beautiful.

Where it is used

Corollary (counting)

If $G$ is finite, isomorphic groups have the same number of elements and $G/N$ has $|G|/|N|$ elements, so $|G| = |\ker\varphi| \cdot |\im\varphi| .$ "Size of $G$ = what was collapsed × what remains."

Identifying quotient groups. To understand $G/N$ , find one homomorphism $\varphi$ whose kernel is $N$ ; then $G/N$ is the familiar group $\im\varphi$ . Example 2, $\R/\Z \cong S^1$ , is the model case.
Counting even permutations. The sign map $\operatorname{sgn} : S_n \to \set{+1,-1}$ on the permutations of $n$ objects is a homomorphism with kernel the even permutations $A_n$ and (for $n \ge 2$ ) image $\set{+1,-1}$ . The corollary gives $|A_n| = n!/2$ : even and odd permutations are exactly half and half, with no counting.
A template for later theorems. The second and third isomorphism theorems, the isomorphism theorems for rings and modules, and the rank–nullity theorem of linear algebra ( $\dim V = \dim\ker f + \dim\im f$ ) all have the same shape as this theorem.

Common misconceptions

Misconception 1: “ $G/\ker\varphi$ is a subgroup of $G$ ”

No. The elements of $G/\ker\varphi$ are not elements of $G$ but bundles (cosets) of elements of $G$ . The theorem does not speak of a subgroup; it says that $G$ , regrouped, is isomorphic to the image.

Misconception 2: “ $\cong$ means $=$ ”

Isomorphic means "same structure as a group", not equal as sets. In Example 1 the elements of $\R^\times/\R_{>0}$ are the two sets "all positive reals" and "all negative reals", not the numbers $+1$ and $-1$ themselves.

Misconception 3: “the image is always all of $H$ ”

The image is a subgroup of $H$ but need not be all of it. For $\varphi : \Z \to \Z$ , $n \mapsto 2n$ , the image is the even integers. One may write $G/\ker\varphi \cong H$ only when $\varphi$ is surjective.

Misconception 4: “you can form a quotient by any subgroup”

$G/N$ is a group only when $N$ is normal. Kernels are always normal (Step 1), so there is no problem here; conversely every normal subgroup is the kernel of some homomorphism (the natural projection $G \to G/N$ has kernel $N$ ).

Exercises

Exercise 1. Let $\varphi : \Z \to \Z$ be $\varphi(n) = 2n$ (additive groups). Check that $\varphi$ is a homomorphism, find its kernel and image, and state what the theorem says.

Solution

$\varphi(m+n) = 2(m+n) = 2m + 2n = \varphi(m) + \varphi(n)$ , so $\varphi$ is a homomorphism. $2n = 0$ only for $n = 0$ , so $\ker\varphi = \set{0}$ . The image is the even integers $2\Z$ . The theorem says $\Z/\set{0} \cong 2\Z$ : the integers and the even integers are isomorphic groups (via $n \leftrightarrow 2n$ ), a phenomenon possible only for infinite sets.

Exercise 2. Let $\varphi : \Z/12\Z \to \Z/4\Z$ send "the remainder mod $12$ " to "its remainder mod $4$ " (re-reading a 12-hour clock with a 4-hour period). Explain why this is well defined, find the kernel and the image, and verify $|G| = |\ker\varphi|\cdot|\im\varphi|$ .

Solution

Since $4$ divides $12$ , numbers differing by a multiple of $12$ have the same remainder mod $4$ , so the map is well defined. The kernel consists of the remainders divisible by $4$ : $\set{0, 4, 8}$ (three elements). The image is all of $\set{0,1,2,3}$ (four elements). Indeed $12 = 3 \times 4$ . The theorem gives $(\Z/12\Z)/\set{0,4,8} \cong \Z/4\Z$ .

Exercise 3. For $n \ge 2$ , use the sign homomorphism $\operatorname{sgn} : S_n \to \set{+1,-1}$ from the previous section to show that the number of even permutations is $n!/2$ .

Solution

$\operatorname{sgn}$ is a homomorphism (the sign of a composition is the product of the signs) with kernel $A_n$ . For $n \ge 2$ a transposition exists and has sign $-1$ , so the image is all of $\set{+1,-1}$ . The corollary gives $n! = |A_n| \cdot 2$ , hence $|A_n| = n!/2$ .

References

D. S. Dummit, R. M. Foote, Abstract Algebra, 3rd ed., Wiley, 2004. §3.3 "The Isomorphism Theorems".
M. Artin, Algebra, 2nd ed., Pearson, 2011. Chapter 2.
I. N. Herstein, Topics in Algebra, 2nd ed., Wiley, 1975. §2.7.
Mathlib (the Lean 4 mathematics library): QuotientGroup.quotientKerEquivRange.

Formal verification in Lean✓ verified 2026-09-14 · leanprover/lean4:v4.34.0-rc2

Below is this theorem written in the Lean 4 proof assistant with the Mathlib library. The computer checks every step of the proof mechanically.

lean/MathThemodel/FirstIsomorphism.lean

import Mathlib

/-!
# 準同型定理(群の第一同型定理) / First Isomorphism Theorem for groups

ページ: https://math.themodel.be/first-isomorphism-theorem/ja/ ,
https://math.themodel.be/first-isomorphism-theorem/en/

主張: 群の準同型 `φ : G →* H` に対して `G ⧸ φ.ker ≃* φ.range`。
ページ本文の証明(第1段〜第4段)に対応する補題を先に置き、最後に Mathlib の定理で同型を与える。
-/

namespace MathThemodel

variable {G H : Type*} [Group G] [Group H] (φ : G →* H)

/-- 補題1: 準同型の核 ker φ は G の正規部分群である。
    Lemma 1: the kernel of a homomorphism is a normal subgroup. -/
theorem ker_normal : φ.ker.Normal := inferInstance

/-- 補題2: 核 ker φ による剰余類の上で φ は矛盾なく定まり(well-defined)、
    誘導される写像 G / ker φ → H は [g] ↦ φ(g) で与えられる。
    Lemma 2: φ descends to the quotient, sending the coset [g] to φ(g). -/
theorem induced_map_apply (g : G) :
    QuotientGroup.kerLift φ (g : G ⧸ φ.ker) = φ g :=
  QuotientGroup.kerLift_mk φ g

/-- 補題3: 誘導された写像 G / ker φ → H は単射である。
    Lemma 3: the induced map is injective.
    証明: [a], [b] の像が等しいなら φ(a⁻¹ b) = φ(a)⁻¹ φ(b) = 1、
          すなわち a⁻¹ b ∈ ker φ、これは [a] = [b] を意味する。 -/
theorem induced_map_injective : Function.Injective (QuotientGroup.kerLift φ) := by
  intro a b h
  induction a using QuotientGroup.induction_on with
  | H a =>
    induction b using QuotientGroup.induction_on with
    | H b =>
      rw [QuotientGroup.kerLift_mk, QuotientGroup.kerLift_mk] at h
      rw [QuotientGroup.eq]
      exact φ.mem_ker.mpr (by rw [map_mul, map_inv, h, inv_mul_cancel])

/-- 準同型定理: G / ker φ ≅ im φ(群同型)。
    First Isomorphism Theorem: G / ker φ ≅ im φ. -/
noncomputable def firstIsomorphism : G ⧸ φ.ker ≃* φ.range :=
  QuotientGroup.quotientKerEquivRange φ

/-- 同型は [g] ↦ φ(g) で与えられる。
    The isomorphism is induced by φ itself. -/
theorem firstIsomorphism_apply (g : G) :
    ((firstIsomorphism φ) (g : G ⧸ φ.ker) : H) = φ g := rfl

/-- 系: φ が全射なら G / ker φ ≅ H。
    Corollary: if φ is surjective then G / ker φ ≅ H. -/
noncomputable def firstIsomorphismOfSurjective (hφ : Function.Surjective φ) :
    G ⧸ φ.ker ≃* H :=
  QuotientGroup.quotientKerEquivOfSurjective φ hφ

end MathThemodel