First isomorphism theorem

The first isomorphism theorem is a fundamental result in abstract algebra that describes the relationship between a homomorphism, its kernel, and its image. The theorem appears uniformly across algebraic structures such as groups, rings, and modules, and serves as a prototype for many structural results in algebra. Specifically, given a homeomorphism, the quotient of its domain by its kernel is isomorphic to its image.

Group theory

Statement

Let $G$ and $H$ be groups and $f : g \to H$ a group homomorphism. Then,

The kernel of $f$ , $\ker f ⊴ G$ is a normal subgroup of $G$ .
The image of $f$ , $im f < G$ is a subgroup of $H$ .
$G / \ker f ≅ im f$ .

Proof

Proof of 1

By definition, $\ker f = {g \in G ∣ f (g) = e_{H}}$ where $e_{H}$ is the identity of $H$ . $\ker f$ is a subgroup of $G$ because:

Identity: Since $f$ is a homomorphism, $f (e_{G}) = e_{H}$ . Therefore $e_{G} \in \ker f$ , implying $\ker f$ is non-empty and has an identity.
Closure: Let $a, b \in \ker f$ , then

$f (a b) = f (a) f (b) = e_{H} e_{H} = e_{H} .$ Thus $a b \in \ker f$ .

Inverses: Let $a \in \ker f$ , then $f (a^{- 1}) = f (a)^{- 1} = e_{H}^{- 1} = e_{H} .$

Thus $a^{- 1} \in \ker f$ .

Therefore, $\ker f$ is a subgroup of $G$ .

Let $g \in G$ and $k \in \ker f$ , then $f (g k g^{- 1}) = f (g) f (k) f (g^{- 1}) = f (g) e_{H} f (g^{- 1}) = f (g) f (g^{- 1}) = f (g) f (g)^{- 1} = e_{H},$ thus $g k g^{- 1} \in \ker f$ .

Therefore, $\ker f ⊴ G$ is a normal subgroup.

□

Proof of 2

By definition, $im f = {h \in H ∣ \exists g \in G : f (g) = h}$ . $im f$ is a subgroup of $H$ because:

Identity: Since $f$ is a homomorphism, $f (e_{G}) = e_{H}$ . Therefore $e_{H} \in im f$ , implying $im f$ is non-empty and has an identity.
Closure: Let $h_{1}, h_{2} \in im f$ , then by definition, there exists $g_{1}, g_{2} \in G$ such that $f (g_{1}) = h_{1}$ and $f (g_{2}) = h_{2}$ . Thus

$h_{1} h_{2} = f (g_{1}) f (g_{2}) = f (g_{1} g_{2}) .$ Thus $h_{1} h_{2} \in im f$ .

Inverses: Let $h \in im f$ , and $g \in G$ such that $f (g) = h$ . Then $h^{- 1} = {(f (g))}^{- 1} = f (g^{- 1}) .$

Thus $h^{- 1} \in im f$ .

Therefore, $im f < H$ is a subgroup.

□