The isomorphism theorems: three theorems, one idea

The isomorphism theorems form the structural backbone of group theory. Through concrete examples -- the determinant, the sign map, the exponential function -- we see that all three rest on a single principle: collapse the kernel and you get the image.

Folio Official

March 1, 2026

In the previous article we introduced the kernel and image of a homomorphism $φ : G \to H$ and proved that the kernel is always a normal subgroup. A question was left open: what, concretely, is the quotient group $G / ker φ$ ?

The answer is startlingly simple: $G / ker φ ≅ Im φ$ . Collapse the kernel and what remains is the image. This is the First Isomorphism Theorem, and it is the single most important structural result in group theory. The Second and Third Isomorphism Theorems, despite their billing as independent results in many textbooks, are both applications of the First.

1 The First Isomorphism Theorem

Theorem 1 (First Isomorphism Theorem).

Let

φ : G \to H

be a group homomorphism. Then

G / ker φ ≅ Im φ .

To see why this is almost forced, consider the following diagram. Write $K = ker φ$ , let $π : G ↠ G / K$ denote the natural projection $g \mapsto g K$ , and let $\overset{φ}{ˉ} : G / K \to Im φ$ be the map induced by $φ$ , namely $\overset{φ}{ˉ} (g K) = φ (g)$ :

Commutativity of the diagram means $φ = \overset{φ}{ˉ} \circ π$ : whether you go across the top or down-and-diagonal, you arrive at the same element of $H$ .

Proof.

We verify four properties of

\overset{φ}{ˉ} : G / K \to Im φ

Well-defined. If

g K = g^{'} K

, then

g^{- 1} g^{'} \in K

, so

φ (g^{- 1} g^{'}) = e_{H}

, and hence

φ (g) = φ (g^{'})

Homomorphism.

\overset{φ}{ˉ} (g K \cdot g^{'} K) = \overset{φ}{ˉ} (g g^{'} K) = φ (g g^{'}) = φ (g) φ (g^{'}) = \overset{φ}{ˉ} (g K) \overset{φ}{ˉ} (g^{'} K) .

Injective. If

\overset{φ}{ˉ} (g K) = e_{H}

, then

φ (g) = e_{H}

, so

g \in K

, and

g K = K

is the identity in

G / K

Surjective. Every element of

Im φ

has the form

φ (g) = \overset{φ}{ˉ} (g K)

. □

2 Putting the theorem to work

The real power of the First Isomorphism Theorem lies in its ability to identify unknown quotient groups. A quotient $G / N$ is, by definition, a set of cosets equipped with a multiplication. But as an abstract object, it can be hard to visualize. The strategy is this:

Find a homomorphism $φ : G \to H$ whose kernel is $N$ .
Then $G / N = G / ker φ ≅ Im φ$ , and the quotient is identified.

Let us see this strategy in action.

Example 1: The determinant. The homomorphism $det : G L_{n} (R) \to R^{*}$ has kernel $S L_{n} (R)$ and is surjective. The theorem gives

G L_{n} (R) / S L_{n} (R) ≅ R^{*} .

The left side is, by definition, the collection of equivalence classes of invertible matrices grouped by determinant. The right side tells us that this abstract object is nothing more than the multiplicative group of nonzero reals. The determinant strips away everything about a matrix except the single number that measures how it scales volume.

Example 2: The sign map. The surjective homomorphism $sgn : S_{n} \to {+ 1, - 1}$ has kernel $A_{n}$ . The theorem gives

S_{n} / A_{n} ≅ {+ 1, - 1} ≅ Z /2 Z .

The full complexity of

S_{n}

, with its

n!

elements, collapses to a single binary distinction: even or odd.

Example 3: The exponential. The map $exp : (R, +) \to (R_{> 0}, \times)$ has trivial kernel and is surjective, so

(R, +) ≅ (R_{> 0}, \times) .

Additive reals and positive multiplicative reals are isomorphic as groups. The exponential function and the logarithm are the bridge between two arithmetic worlds, and this isomorphism is the group-theoretic distillation of that relationship.

Example 4: Reduction mod $n$ . The natural projection $π : Z \to Z / n Z$ has kernel $n Z$ and is surjective. The theorem gives $Z / n Z ≅ Z / n Z$ – a tautology, but an instructive one. The left side is a quotient of the abstract group $Z$ by its subgroup $n Z$ ; the right side is the concrete set ${0, 1, \dots, n - 1}$ with modular arithmetic. The theorem confirms that these two descriptions agree.

Here is a less obvious application. Define $φ : Z \times Z \to Z$ by $φ (a, b) = a - b$ .

$ker φ = {(a, a) ∣ a \in Z}$ (the diagonal subgroup).
$Im φ = Z$ (surjective).

The First Isomorphism Theorem yields $(Z \times Z) / {(a, a)} ≅ Z$ . Without the theorem, identifying this quotient would require considerably more work.

3 The Second Isomorphism Theorem

Theorem 2 (Second Isomorphism Theorem).

Let

H \leq G

be a subgroup and

N ⊴ G

a normal subgroup. Then

H N / N ≅ H / (H \cap N) .

Here $H N = {hn ∣ h \in H, n \in N}$ . The left side forms $H N$ (which is a subgroup because $N$ is normal) and quotients out $N$ . The right side takes $H$ and quotients out the part of $H$ that overlaps with $N$ . The theorem asserts these produce the same group.

The intuition is this: viewing elements of $H$ through a lens that blurs everything in $N$ is the same as blurring only the part of $N$ that $H$ actually intersects.

Proof.

Define

φ : H \to H N / N

φ (h) = h N

Homomorphism.

φ (h h^{'}) = h h^{'} N = (h N) (h^{'} N) = φ (h) φ (h^{'})

Surjective. A typical element of

H N / N

has the form

(hn) N = h N

(since

n N = N

), which is

φ (h)

Kernel.

φ (h) = N

if and only if

h \in N

, i.e.,

h \in H \cap N

. So

ker φ = H \cap N

By the First Isomorphism Theorem,

H / (H \cap N) ≅ H N / N

. □

Example 3.

Take

G = Z

H = 4 Z

N = 6 Z

. Then

$H N = 4 Z + 6 Z = 2 Z$ (multiples of $g cd (4, 6) = 2$ ).
$H \cap N = 4 Z \cap 6 Z = 12 Z$ (multiples of $lcm (4, 6) = 12$ ).

The theorem gives

2 Z /6 Z ≅ 4 Z /12 Z

. The left side has three cosets:

{0, 6, 12, \dots}

{2, 8, 14, \dots}

{4, 10, 16, \dots}

. The right side also has three cosets:

{0, 12, 24, \dots}

{4, 16, 28, \dots}

{8, 20, 32, \dots}

. They look different, but both are cyclic of order

3

, isomorphic to

Z /3 Z

4 The Third Isomorphism Theorem

Theorem 4 (Third Isomorphism Theorem).

Let

N \subseteq H

both be normal subgroups of

G

. Then

(G / N) / (H / N) ≅ G / H .

In words: dividing in two stages yields the same result as dividing all at once. First quotient by $N$ to get $G / N$ ; then quotient the result by the image of $H$ (which is $H / N$ ). The outcome is the same as if you had simply formed $G / H$ directly.

Proof.

Define

φ : G / N \to G / H

φ (g N) = g H

Well-defined. If

g N = g^{'} N

, then

g^{- 1} g^{'} \in N \subseteq H

, so

g H = g^{'} H

Surjective. Every coset

g H

is the image of

g N

Kernel.

φ (g N) = H

if and only if

g \in H

, i.e.,

g N \in H / N

. So

ker φ = H / N

By the First Isomorphism Theorem,

(G / N) / (H / N) ≅ G / H

. □

Example 5.

Take

G = Z

N = 6 Z

H = 2 Z

(note

6 Z \subseteq 2 Z

). The left side:

Z /6 Z = {0, 1, 2, 3, 4, 5}

modulo

2 Z /6 Z = {0, 2, 4}

produces two cosets,

{0, 2, 4}

and

{1, 3, 5}

. The right side:

Z /2 Z = {0, 1}

. Manifestly isomorphic.

Example 6.

Take

G = Z

N = 12 Z

H = 4 Z

(with

12 Z \subseteq 4 Z

). The left side:

Z /12 Z = {0, 1, \dots, 11}

modulo

4 Z /12 Z = {0, 4, 8}

gives four cosets:

{0, 4, 8}

{1, 5, 9}

{2, 6, 10}

{3, 7, 11}

. The right side:

Z /4 Z = {0, 1, 2, 3}

. Once again, they agree.

5 Three theorems, one idea

Step back and compare the three proofs. Every one follows the same template:

Construct a homomorphism $φ$ from the appropriate domain to the desired target.
Verify that $φ$ is surjective.
Identify $ker φ$ .
Apply the First Isomorphism Theorem.

The First Isomorphism Theorem is the engine; the Second and Third are instances obtained by feeding in cleverly chosen homomorphisms. Textbooks sometimes present the three theorems side by side as though they were results of comparable depth and independence. They are not. The First is the genuine theorem; the Second and Third are its corollaries, distinguished only by the particular homomorphisms used in their proofs. The underlying principle throughout is the same: collapse the kernel, and what remains is the image.

6 Takeaway

The isomorphism theorems are tools for unmasking quotient groups – for determining what a quotient "really is" in concrete terms. The First Isomorphism Theorem provides the fundamental mechanism: $G / ker φ ≅ Im φ$ . The Second relates $H N / N$ to $H / (H \cap N)$ by constructing the right homomorphism and reading off its kernel. The Third tells us that quotienting in two stages is the same as quotienting all at once. Behind all three lies a single refrain: build a homomorphism, identify its kernel, and let the First Isomorphism Theorem do the rest.

Group Theory Algebra Between the Lines

Folio Official

Mathematics "between the lines" — exploring the intuition textbooks leave out, written in LaTeX on Folio.

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The isomorphism theorems: three theorems, one idea

Folio Official

March 1, 2026

1 The First Isomorphism Theorem

Theorem 1 (First Isomorphism Theorem).

Let

φ : G \to H

be a group homomorphism. Then

G / ker φ ≅ Im φ .

Commutativity of the diagram means $φ = \overset{φ}{ˉ} \circ π$ : whether you go across the top or down-and-diagonal, you arrive at the same element of $H$ .

Proof.

We verify four properties of

\overset{φ}{ˉ} : G / K \to Im φ

Well-defined. If

g K = g^{'} K

, then

g^{- 1} g^{'} \in K

, so

φ (g^{- 1} g^{'}) = e_{H}

, and hence

φ (g) = φ (g^{'})

Homomorphism.

\overset{φ}{ˉ} (g K \cdot g^{'} K) = \overset{φ}{ˉ} (g g^{'} K) = φ (g g^{'}) = φ (g) φ (g^{'}) = \overset{φ}{ˉ} (g K) \overset{φ}{ˉ} (g^{'} K) .

Injective. If

\overset{φ}{ˉ} (g K) = e_{H}

, then

φ (g) = e_{H}

, so

g \in K

, and

g K = K

is the identity in

G / K

Surjective. Every element of

Im φ

has the form

φ (g) = \overset{φ}{ˉ} (g K)

. □

2 Putting the theorem to work

Find a homomorphism $φ : G \to H$ whose kernel is $N$ .
Then $G / N = G / ker φ ≅ Im φ$ , and the quotient is identified.

Let us see this strategy in action.

Example 1: The determinant. The homomorphism $det : G L_{n} (R) \to R^{*}$ has kernel $S L_{n} (R)$ and is surjective. The theorem gives

G L_{n} (R) / S L_{n} (R) ≅ R^{*} .

Example 2: The sign map. The surjective homomorphism $sgn : S_{n} \to {+ 1, - 1}$ has kernel $A_{n}$ . The theorem gives

S_{n} / A_{n} ≅ {+ 1, - 1} ≅ Z /2 Z .

The full complexity of

S_{n}

, with its

n!

elements, collapses to a single binary distinction: even or odd.

Example 3: The exponential. The map $exp : (R, +) \to (R_{> 0}, \times)$ has trivial kernel and is surjective, so

(R, +) ≅ (R_{> 0}, \times) .

Here is a less obvious application. Define $φ : Z \times Z \to Z$ by $φ (a, b) = a - b$ .

$ker φ = {(a, a) ∣ a \in Z}$ (the diagonal subgroup).
$Im φ = Z$ (surjective).

The First Isomorphism Theorem yields $(Z \times Z) / {(a, a)} ≅ Z$ . Without the theorem, identifying this quotient would require considerably more work.

3 The Second Isomorphism Theorem

Theorem 2 (Second Isomorphism Theorem).

Let

H \leq G

be a subgroup and

N ⊴ G

a normal subgroup. Then

H N / N ≅ H / (H \cap N) .

The intuition is this: viewing elements of $H$ through a lens that blurs everything in $N$ is the same as blurring only the part of $N$ that $H$ actually intersects.

Proof.

Define

φ : H \to H N / N

φ (h) = h N

Homomorphism.

φ (h h^{'}) = h h^{'} N = (h N) (h^{'} N) = φ (h) φ (h^{'})

Surjective. A typical element of

H N / N

has the form

(hn) N = h N

(since

n N = N

), which is

φ (h)

Kernel.

φ (h) = N

if and only if

h \in N

, i.e.,

h \in H \cap N

. So

ker φ = H \cap N

By the First Isomorphism Theorem,

H / (H \cap N) ≅ H N / N

. □

Example 3.

Take

G = Z

H = 4 Z

N = 6 Z

. Then

$H N = 4 Z + 6 Z = 2 Z$ (multiples of $g cd (4, 6) = 2$ ).
$H \cap N = 4 Z \cap 6 Z = 12 Z$ (multiples of $lcm (4, 6) = 12$ ).

The theorem gives

2 Z /6 Z ≅ 4 Z /12 Z

. The left side has three cosets:

{0, 6, 12, \dots}

{2, 8, 14, \dots}

{4, 10, 16, \dots}

. The right side also has three cosets:

{0, 12, 24, \dots}

{4, 16, 28, \dots}

{8, 20, 32, \dots}

. They look different, but both are cyclic of order

3

, isomorphic to

Z /3 Z

4 The Third Isomorphism Theorem

Theorem 4 (Third Isomorphism Theorem).

Let

N \subseteq H

both be normal subgroups of

G

. Then

(G / N) / (H / N) ≅ G / H .

Proof.

Define

φ : G / N \to G / H

φ (g N) = g H

Well-defined. If

g N = g^{'} N

, then

g^{- 1} g^{'} \in N \subseteq H

, so

g H = g^{'} H

Surjective. Every coset

g H

is the image of

g N

Kernel.

φ (g N) = H

if and only if

g \in H

, i.e.,

g N \in H / N

. So

ker φ = H / N

By the First Isomorphism Theorem,

(G / N) / (H / N) ≅ G / H

. □

Example 5.

Take

G = Z

N = 6 Z

H = 2 Z

(note

6 Z \subseteq 2 Z

). The left side:

Z /6 Z = {0, 1, 2, 3, 4, 5}

modulo

2 Z /6 Z = {0, 2, 4}

produces two cosets,

{0, 2, 4}

and

{1, 3, 5}

. The right side:

Z /2 Z = {0, 1}

. Manifestly isomorphic.

Example 6.

Take

G = Z

N = 12 Z

H = 4 Z

(with

12 Z \subseteq 4 Z

). The left side:

Z /12 Z = {0, 1, \dots, 11}

modulo

4 Z /12 Z = {0, 4, 8}

gives four cosets:

{0, 4, 8}

{1, 5, 9}

{2, 6, 10}

{3, 7, 11}

. The right side:

Z /4 Z = {0, 1, 2, 3}

. Once again, they agree.

5 Three theorems, one idea

Step back and compare the three proofs. Every one follows the same template:

Construct a homomorphism $φ$ from the appropriate domain to the desired target.
Verify that $φ$ is surjective.
Identify $ker φ$ .
Apply the First Isomorphism Theorem.

6 Takeaway

Group Theory Algebra Between the Lines

Folio Official

Mathematics "between the lines" — exploring the intuition textbooks leave out, written in LaTeX on Folio.

1 followers·107 articles

The isomorphism theorems: three theorems, one idea

1 The First Isomorphism Theorem

2 Putting the theorem to work

3 The Second Isomorphism Theorem

4 The Third Isomorphism Theorem

5 Three theorems, one idea

6 Takeaway

Share your expertise with the world

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The isomorphism theorems: three theorems, one idea

1 The First Isomorphism Theorem

2 Putting the theorem to work

3 The Second Isomorphism Theorem

4 The Third Isomorphism Theorem

5 Three theorems, one idea

6 Takeaway

Share your expertise with the world

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