An Illustrated Proof of the First Group Isomorphism Theorem

# An Illustrated Proof of the First Group Isomorphism Theorem

Recall from The First Group Isomorphism Theorem page that The First Group Isomorphism Theorem (also known as The Fundamental Theorem of Group Homomorphisms) states that if $G$ and $H$ are groups and if $\phi : G \to H$ is a group homomorphism from $G$ to $H$ then:

(1)
\begin{align} \quad G / \ker (\phi) \cong \phi(G) \end{align}

We will now describe this very important result intuitively - using pictures, but first, we need the following proposition:

 Proposition 1: Let $G$ and $H$ be groups and let $\phi : G \to H$ be a homomorphism from $G$ to $H$. Let $g \in G$. Then $x \in g \ker(\phi)$ if and only if $\phi(x) = \phi(g)$.
• Proof: $\Rightarrow$ Let $x \in g \ker (\phi)$. Then $x = gh$ for some $h \in \ker (\phi)$. Since $\phi$ is a homomorphism, we have that:
(2)
\begin{align} \quad \phi(x) = \phi(gh) = \phi(g) \phi(h) = \phi(g) e_H = \phi(g) \end{align}
• $\Leftarrow$ Let $\phi(x) = \phi(g)$. Observe that $x = gg^{-1}x$. Let $h = g^{-1}x$. Then:
(3)
\begin{align} \quad \phi(h) = \phi(g^{-1}x) = \phi(g^{-1}) \phi(x) = \phi(g)^{-1} \phi(x) = \phi(g^{-1}) \phi(g) = e_H \end{align}
• So $h \in \ker (\phi)$. Thus $x = gh \in g\ker(\phi)$. $\blacksquare$

Let $G$ and $H$ be groups and let $\phi : G \to H$ be a group homomorphism of $G$ to $H$. Recall that $\ker (\phi)$ is always a normal subgroup of $G$, and so we can construct the quotient $G / \ker (\phi)$ without any problems - that is, $G / \ker (\phi)$ is the group of all left cosets of $\ker (\phi)$ in $G$.

Now, for each $x \in G$ we have that either $x \in \ker (\phi)$ or $x \not \in \ker (\phi)$.

By Proposition 1, we see that $x \in \ker (\phi)$ ($= e_G \ker (\phi)$) if and only if $\phi(x) = e_H$ ($= \phi(e_G)$). Furthermore, Proposition 1 tells us that $x \in g \ker (\phi)$ if and only if $\phi(x) = \phi(g)$.

Now note again that $G / \ker(\phi)$ is the set of all left cosets of $\ker (\phi)$ in $G$. That is:

(4)
\begin{align} \quad G / \ker (\phi) = \{ g \ker (\phi) : g \in G \} \end{align}

So the cosets in the quotient $G / \ker (\phi)$ are really sets of the form $\{ \phi^{-1}(h) : h \in H \}$. Now suppose that we represent the group $G$ as a circle, and represent elements of $G$ as dots:

Recall that the cosets of $\ker (\phi)$ in $G$ partition $G$. This is evident as we see that each coset of $\ker(\phi)$ is the preimage of a singleton set in $\phi(G) \subseteq H$. If $x, y \in g \ker(\phi)$ then $\phi(x) = \phi(y) = \phi(g) \in \phi(G)$. So we can colour each element of $G$ such that two elements of $G$ are the same colour if and only if they are mapped to the same element in $\phi(G)$:

From the following diagram, it is clear now that the cosets of $\ker(\phi)$, i.e., $G / \ker(\phi)$ are in bijection with $\phi(G)$ (the range of $\phi$):

Furthermore, the map $\psi : G / \ker(\phi) \to \phi(G)$ defined for all $g \ker(\phi) \in G / \ker(\phi)$ by $\psi(g \ker(\phi)) = \phi(G)$ is that bijection - and is also a homomorphism, i.e., $G / \ker(\phi)$ is isomorphic to $\phi(G)$.