The First Group Isomorphism Theorem

# The First Group Isomorphism Theorem

Recall from The Kernel of a Group Homomorphism is a Normal Subgroup of the Domain page that if $G$ and $H$ are homomorphic groups with homomorphism $\phi : G \to H$, then $\ker (\phi)$ is a subgroup of $G$, and more precisely, $\ker (\phi)$ is a normal subgroup of $G$.

Hence, we can consider the quotient group $G / \ker (\phi)$. As we will prove below, this quotient group is isomorphic to $\phi (G)$.

Thereom 1 (The First Group Isomorphism Theorem): Let $G$ and $H$ be groups and let $\phi : G \to H$ be a group homomorphism from $G$ to $H$. Then $G / \ker (\phi) \cong \phi (G)$. |

*The theorem above is sometimes called "The Fundamental Theorem of Group Homomorphisms".*

**Proof:**Let $K = \ker (\phi)$ and let $\psi : G / K \to \phi (G)$ be defined for all $aK \in G / K$ by:

\begin{align} \quad \psi (a K) = \phi (a) \end{align}

- We note that the function $\psi$ is well-defined, for if $aK = bK$ then for some $k \in K$ we must have that $a = bk$ and so:

\begin{align} \quad \phi (a) = \psi (aK) = \psi (bkK) = \psi (bK) = \phi (b) \end{align}

- Now let $aK, bK \in G/K$. Then:

\begin{align} \quad \psi (aK * bK) = \psi ((a * b)K) = \phi(a * b) = \phi (a) * \phi(b) = \psi (aK) * \psi (bK) \end{align}

- We now show that $\psi$ is bijective.

- Let $aK, bK \in G/K$ and suppose that $\psi (aK) = \psi (bK)$. Then:

\begin{align} \quad \phi(a) &= \phi (b) \\ \quad \phi (a) * [\phi (b)]^{-1} &= e \\ \quad \phi(a) * \phi(b^{-1}) &= e \\ \quad \phi (ab^{-1}) &= e \end{align}

- So $ab^{-1} \in \ker (\phi) = K$. Hence $aK = bK$. This shows that $\psi$ is injective.

- Now let $a \in \phi (G)$. Then $aK \in G / K$ is such that $\psi (aK) = a$. So $\psi$ is surjective.

- Therefore $\psi : G / K \to \phi(G)$ is an isomorphism, and hence:

\begin{align} \quad G / \ker (\phi) \cong \phi (G) \quad \blacksquare \end{align}