Homo(Iso)morphism
Definition 1.14 – Image and Kernel of a Homomorphism
Consider the homomorphism $\mu: G \to G'$. All elements from $G$ map to some, but not all elements of $G'$.
- The image $\cal{I}_{\mu}$ is defined as the set of elements in $G'$ to which the elements of $G$ map.
- The kernel $\cal{K}_{\mu}$ is defined as the set of elements in $G$ which map to the identity of $G'$.
Theorem 1.7
The image $\cal{I}_{\mu}$ of a map $\mu: G \to G'$ is a subgroup of $G'$.
Proof
Consider groups $G$ and $G'$ with $\lvert G \rvert = n$ and $\lvert G' \rvert = m$, and the homomorphism $\mu: G \to G'$.
The mapping means that, $\forall g_{a},g_{b},g_{c} \in G$, $g_{a} g_{b} = g_{c} \to \mu(g_{a}) \mu(g_{b}) = \mu(g_{c})$. Note that $\mu(g_{a}), \mu(g_{b}), \mu(g_{c}) \in \cal{I}_{\mu}$. Closure is thus shown.
Therefore, considering theorem 1.6, $\cal{I}_{\mu}$ is indeed a subgroup of $G'$.
Theorem 1.8
The kernel $\cal{K}_{\mu}$ of a map $\mu: G \to G'$ is an invariant subgroup of $G$.
Proof
The elements of $\cal{K}_{\mu}$ will always map to the identity $e' \in G'$. $\forall g_{a},g_{b} \in \cal{K}_{\mu}$, $\mu(g_{a}) = \mu(g_{b}) = e'$. Consequently, $\mu(g_{a}) \mu(g_{b}) = e'e' = e'$. Thus, closure is verified, and so, by theorem 1.6, $\cal{K}_{\mu}$ is a subgroup.
To show that the subgroup is invariant, we must show that $\cal{K}_{\mu}$ is the union of equivalence classes.
Consider the element $g_{a} \in G$.
$$ \forall g \in G, \quad \mu(g^{-1} g_{a} g) = \mu(g^{-1})e' \mu(g) = \mu(g^{-1})\mu(g) = \mu(g^{-1} g) = e' \implies g^{-1} g_{a} g \in \cal{K}_{\mu} $$Therefore, all the elements of an equivalence class in $G$ are in $\cal{K}_{\mu}$, meaning $\cal{K}_{\mu}$ is an invariant subgroup.
Theorem 1.9 – Isomorphism Theorem
Proof
Since there exists an invariant subgroup of $G$, $\cal{K}_{\mu}$, a quotient group can be constructed: $G / \cal{K}_{\mu}$.
The identity of $G / \cal{K}_{\mu}$ is $\cal{K}_{\mu}$, the image of which is the identity of $G'$. The remaining elements of the quotient group are cosets of $\cal{K}_{\mu}$, and the image of the elements in each coset is an element of $\cal{I}_{\mu}$. Therefore, $G / \cal{K}_{\mu}$ and $\cal{I}_{\mu}$ are isomorphs.