First Isomorphism Theorem

Statement

Let $f : G \to H$ be a group homomorphism. Then:

More precisely, the map $\bar{f} : G/\ker(f) \to H$ defined by $\bar{f}(g\ker f) = f(g)$ is a well-defined injective group homomorphism whose image equals $\operatorname{im}(f)$, giving the isomorphism.

Visualization

Take the canonical surjection $f : \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$, $f(k) = k \bmod n$.

Kernel / Image Diagram (n = 6)

ℤ
│   f
│──────────────→  ℤ/6ℤ = {0,1,2,3,4,5}
│
ker(f) = 6ℤ = {..., -12, -6, 0, 6, 12, ...}
im(f)  = ℤ/6ℤ (all of the target; f is surjective)

Cosets of 6ℤ in ℤ:
  0 + 6ℤ = {..., -12, -6,  0,  6, 12, ...}  →  [0]
  1 + 6ℤ = {..., -11, -5,  1,  7, 13, ...}  →  [1]
  2 + 6ℤ = {..., -10, -4,  2,  8, 14, ...}  →  [2]
  3 + 6ℤ = {...,  -9, -3,  3,  9, 15, ...}  →  [3]
  4 + 6ℤ = {...,  -8, -2,  4, 10, 16, ...}  →  [4]
  5 + 6ℤ = {...,  -7, -1,  5, 11, 17, ...}  →  [5]

Six cosets, one for each element of $\mathbb{Z}/6\mathbb{Z}$. The First Isomorphism Theorem says the bijection between cosets and target elements is actually a group isomorphism:

Another example: $f : \mathbb{Z} \to \mathbb{Z}$ given by $f(k) = 2k$. Then $\ker(f) = \{0\}$ and $\operatorname{im}(f) = 2\mathbb{Z}$, so $\mathbb{Z}/\{0\} \cong 2\mathbb{Z}$, i.e.\ $\mathbb{Z} \cong 2\mathbb{Z}$ — both are infinite cyclic.

Proof Sketch

1. Well-defined: if $g\ker f = g'\ker f$ then $g^{-1}g' \in \ker f$, so $f(g^{-1}g') = e$, so $f(g) = f(g')$.
2. Homomorphism: $\bar{f}(g\ker f \cdot h\ker f) = \bar{f}(gh\ker f) = f(gh) = f(g)f(h) = \bar{f}(g\ker f)\bar{f}(h\ker f)$.
3. Injective: $\bar{f}(g\ker f) = e \implies f(g) = e \implies g \in \ker f \implies g\ker f = \ker f$ (the identity coset).
4. Surjective onto image: by definition of $\operatorname{im}(f)$.

Connections

The First Isomorphism Theorem is the engine of exact sequences, appearing throughout the Fundamental Theorem of Galois TheoryFundamental Theorem of Galois Theory$$\{\text{intermediate fields}\} \longleftrightarrow \{\text{subgroups of } \text{Gal}(K/F)\}$$Bijection between intermediate fields and subgroups of the Galois groupRead more → (quotients of Galois groups correspond to sub-extensions). It quantifies Lagrange's TheoremLagrange's Theorem$$|H| \;\text{divides}\; |G|$$Subgroup order divides group orderRead more →: $|G/\ker f| = |\operatorname{im}(f)|$ gives $|\ker f| \cdot |\operatorname{im}(f)| = |G|$. In the proof of the Impossibility of the Quintic FormulaImpossibility of the Quintic Formula$$S_5 \text{ is not solvable} \implies \text{no radical formula for degree } \geq 5$$No general algebraic formula exists for polynomials of degree 5 or higherRead more →, surjective maps from $\text{Gal}(K/\mathbb{Q})$ onto quotients detect composition factors. Cayley's TheoremCayley's Theorem$$G \hookrightarrow S_{|G|}$$Every group embeds into a symmetric groupRead more → provides the embedding $G \hookrightarrow \text{Sym}(G)$ whose injectivity is a special case of this theorem ($\ker = 1$).

/-- **First Isomorphism Theorem**: G / ker(f) ≃* range(f) for any group homomorphism f.
    Mathlib: `QuotientGroup.quotientKerEquivRange`
    in `Mathlib.GroupTheory.QuotientGroup.Basic`. -/
noncomputable def first_iso {G H : Type*} [Group G] [Group H] (f : G →* H) :
    G ⧸ f.ker ≃* f.range :=
  QuotientGroup.quotientKerEquivRange f