If $H$ is a normal subgroup of $G$, then the mapping $f: G \to G/H$ defined as $f(g) = [g]$ is an epimorphism (since all elements of $G/H$ are left cosets of some element in $G$).
The kernel of a group homomorphism is the set of elements in the domain that map to the identity element of the codomain. Hence, the kernel $K$ of a homomorphism $f: G \to G’$ is defined as
$K$ is a normal subgroup of $G$. We can show this in two steps:
(i) $K$ is a subgroup, by the one-step subgroup test. $\forall k_{1}, k_{2} \in K$, $f(k_{1} \cdot k_{2}^{-1}) = f(k_{1}) \ast f(k_{2}^{-1}) = e_{G’} \ast (e_{G’})^{-1} = e_{G’} \Rightarrow k_{1} \cdot k_{2}^{-1} \in K$.
(ii) $K$ is a normal subgroup, i.e. $g \cdot K \cdot g^{-1} \subseteq K$, $\forall g \in G$. Let $K’$ be the set of images of $g \cdot K \cdot g^{-1}$ under $f$. Now, for all $k \in K$, $f(g \cdot k \cdot g^{-1}) = f(g) \ast f(k) \ast f(g^{-1}) = f(g) \ast e_{G’} \ast [f(g)]^{-1} = e_{G’}$. This shows that $f(g \cdot K \cdot g^{-1})$ is simply the singleton set consisting of $e_{G’}$, and consequently $g \cdot K \cdot g^{-1} \subseteq K$.
We will prove Langrange’s Theorem and three of its corollaries.
The order of a finite group $G$ is divisible by the order of its subgroup $H$.
Proof. Let $n$ be the order of $G$ and $m$ the order of H. We know that all left cosets of $H$ have $m$ elements, and form a partition of $G$. Since $G$ is finite, the number of these cosets is also finite; call it $k$. Hence $n = mk$ and therefore, $m$ divides $n$, QED.
The order of $G$ is divisible by the index $k$ [number of distinct cosets] of $H$.
Proof. It has been shown that $n = mk$; from this it follows directly that $k$ divides $n$.
The order of $G$ is divisble by the orders of all $g \in G$.
Proof. Let the order of $g \in G$ be $\gamma$. Consider the set
| Clearly, this set is closed and finite; hence it is a subgroup of $G$. However, its order is $\gamma$, so by Langrange’s theorem, $\gamma | n$. |
If $n = |G|$, then $g^n = e$, $\forall g \in G$.
Proof. Let $n = p\gamma$, where $p$ is an integer. Now $g^n = g^{p\gamma} = (g^{\gamma})^p = e^p = e$, QED.