Homomorphisms and Quotient Groups - Key Proof

We present a complete proof of the First Isomorphism Theorem, the fundamental result connecting homomorphisms and quotient groups.

TheoremFirst Isomorphism Theorem (Restatement)

Let $\phi: G \to H$ be a group homomorphism. Then: $G/\ker(\phi) \cong \text{im}(\phi)$

ProofProof of First Isomorphism Theorem

Let $K = \ker(\phi)$ and $I = \text{im}(\phi)$ . We construct an isomorphism $\overline{\phi}: G/K \to I$ .

Step 1: Define the map

For each coset $gK \in G/K$ , define: $\overline{\phi}(gK) = \phi(g)$

We must verify this is well-defined: if $gK = g'K$ , does $\phi(g) = \phi(g')$ ?

If $gK = g'K$ , then $g^{-1}g' \in K = \ker(\phi)$ , so: $\phi(g^{-1}g') = e_H$ $\phi(g)^{-1}\phi(g') = e_H$ $\phi(g') = \phi(g)$

Thus $\overline{\phi}$ is well-defined.

Step 2: Verify $\overline{\phi}$ is a homomorphism

For cosets $gK, g'K \in G/K$ : $\overline{\phi}((gK)(g'K)) = \overline{\phi}((gg')K) = \phi(gg')$ $= \phi(g)\phi(g') = \overline{\phi}(gK)\overline{\phi}(g'K)$

So $\overline{\phi}$ preserves the group operation.

Step 3: Prove injectivity

Suppose $\overline{\phi}(gK) = e_H$ . Then $\phi(g) = e_H$ , so $g \in K = \ker(\phi)$ . Therefore $gK = K$ (the identity coset). This shows $\ker(\overline{\phi}) = \{K\}$ , so $\overline{\phi}$ is injective.

Step 4: Prove surjectivity

Take any $h \in I = \text{im}(\phi)$ . By definition, $h = \phi(g)$ for some $g \in G$ . Then: $\overline{\phi}(gK) = \phi(g) = h$

So every element of $I$ is in the image of $\overline{\phi}$ , making it surjective.

Conclusion: $\overline{\phi}$ is a bijective homomorphism, hence an isomorphism.

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Remark

The key insight is that elements of $K$ contribute nothing to the image under $\phi$ —they all map to $e_H$ . The quotient $G/K$ eliminates this redundancy, allowing a one-to-one correspondence with $I$ . Each coset $gK$ represents all elements mapping to the same image $\phi(g)$ .

ExampleApplication: Cyclic Groups

Define $\phi: \mathbb{Z} \to \mathbb{Z}_n$ by $\phi(k) = k \bmod n$ . This is a surjective homomorphism with:

$\ker(\phi) = n\mathbb{Z} = \{nk : k \in \mathbb{Z}\}$
$\text{im}(\phi) = \mathbb{Z}_n$

By the First Isomorphism Theorem: $\mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}_n$

This confirms that the quotient construction for $\mathbb{Z}$ by $n\mathbb{Z}$ yields the familiar cyclic group of order $n$ .

The proof technique—showing well-definedness, then verifying the homomorphism property and bijectivity—is standard for quotient constructions. Similar arguments prove the other isomorphism theorems.