Motivation

As I often forget how to prove the isomorphism theorems in group theory, here is one (in all painstaking details) for the record.

Theorem statement

Given a homomorphism \(\Phi: G \rightarrow H\), then

  • (1) \(\text{ker}(\Phi)\trianglelefteq G\)
  • (2) \(G/\text{ker}(\Phi) \equiv \text{im}(\Phi)\)

Proof of (1)

We will follow a 3-step process:

  • 1) Determine what is \(\text{ker}(\Phi)\) as a set.
  • 2) Show \(\text{ker}(\Phi)\) is a group.
  • 3) Show \(\text{ker}(\Phi)\) is a normal subgroup.

For 1)

$$\text{ker}(\Phi)=\{g\in G \Phi(g) = e_H\}\(, where\)e_H\(is an identity element in\)H$$.

For 2)

Check the group axioms:

  • Identity:
  • Inverse:
  • Associativity:
  • Closure: