📂Abstract Algebra

Proof of the First Isomorphism Theorem

Theorem¹

Let $G,G'$ be groups.

• First Isomorphism Theorem: If a homomorphism $\phi : G \to G'$ exists, then $$ G / \ker ( \phi ) \simeq \phi (G) $$
• Second Isomorphism Theorem: If $H \le G$ and $N \triangleleft G$, then $$ (HN) / N \simeq H / (H \cap N) $$
• Third Isomorphism Theorem: If $H , K \triangleleft G$ and $K \leq H$, then $$ G/H \simeq (G/K) / (H/K) $$

The isomorphism theorem refers to the three independent theorems above, proved by the algebraist Emmy Noether.

---

• $\ker$ is the kernel.
• $N \triangleleft G$ means that $N$ is a normal subgroup of $G$.

Explanation

The First Isomorphism Theorem means that there exists an isomorphism $\color{red} {\mu}$ corresponding to the red part in the diagram above. This suggests that, in a group, one can discard the parts unnecessary for $\phi$ and keep only a structure that takes the kernel as a kind of ‘unit’.

Proof

Let $K : = \ker ( \phi )$, and define $\mu : G / K \to \phi (G)$ by $\mu (xK) = \phi ( x)$. It suffices to show that this $\mu$ is an isomorphism.

---

Part 1. $\mu$ is a function.

For $x,y \in G$ and the identity element $e'$ of $G'$,

$$ \begin{align*} & xK = yK \\ \iff & x^{-1} y \in K \\ \iff & \phi ( x^{-1} y ) = e' \\ \iff & \phi ( x^{-1} ) \phi ( y ) = e' \\ \iff & \phi ( x ) ^{-1} \phi ( y ) = e' \\ \iff & \phi ( x ) = \phi ( y ) \end{align*} $$ In other words, $xK = yK \implies \phi ( x ) = \phi ( y )$, so $\mu$ is a function.

---

Part 2. $\mu$ is injective.

Tracing the process of Part 1 backwards gives $\phi ( x ) = \phi ( y ) \implies xK = yK$, so $\mu$ is injective.

---

Part 3. $\mu$ is surjective.

Since $\mu ( G / K ) = \left\{ \mu (xK) \ | \ x \in G \right\} = \left\{ \phi (x) \ | \ x \in G \right\} = \phi (G)$, $\mu$ is surjective.

---

Part 4. $\mu$ is a homomorphism.

For $x,y \in G$, $$ \mu (xKyK) = \mu (xyK) = \phi (xy) = \phi (x) \phi (y) = \mu (xK) \mu (yK) $$ so $\mu$ is a homomorphism.

■

Generalization

Meanwhile, a theorem extending the First Isomorphism Theorem to rings is known. The proof is almost identical; the difference is that, unlike groups, two operations—addition and multiplication—are considered.

Fundamental Theorem of Homomorphisms: For rings $R$, $r '$, if a homomorphism $\phi : R \to r '$ exists, then $R / \ker ( \phi ) \simeq \phi (R)$