We have seen that the (left or right) cosets of a subgroup form a partition on the group. If the partitions formed by the left and right cosets are identical, the subgroup is called a normal subgroup.
Equivalently, $H$ is normal if $H \cdot g = g \cdot H$, $\forall g \in G$. It is not immediately clear that this is true, since it is not necessary that each element forms an equal left and right coset.
To prove this, we need to show that if $g_{1} \cdot H = H \cdot g_{2}$, then $g_{1} \cdot H = H \cdot g_{1}$. First, we note that $g_{1} \in g_{1} \cdot H = H \cdot g_{2} \Rightarrow g_{1} \in H \cdot g_{2}$. However, this means that $H \cdot g_{1} \cap H \cdot g_{2} \neq \phi$, so $H \cdot g_{1} = H \cdot g_{2}$, which is equal to $g_{1} \cdot H$. This completes the proof.
Another way to state the condition is $g^{-1} \cdot H \cdot g \subseteq H$, $\forall g \in G$.
To prove that $x \cdot H = H \cdot x$, $\forall x \in G$ from this, first we let $g = x$ and premultiply by $x$. Hence, $H \cdot x \subseteq x \cdot H$. Then, we let $g = x^{-1}$ and postmultiply by $x$. Therefore, $x \cdot H \subseteq H \cdot x$. These two equations prove that $x \cdot H = H \cdot x$.
Let $H$ be a normal subgroup of a group $\langle G, \cdot \rangle$, and let us denote the left coset $gH \in G$ as $[g]$. Let the set of left cosets be $G/H$.
We will define a binary operation $\circ$ on $G/H$ as $[g_{1}] \circ [g_{2}] = [g_{1} \cdot g_{2}]$, $\forall g_{1}, g_{2} \in G$.
$\langle G/H, \circ \rangle$ forms a group, called the quotient or factor group of $G$ relative to $H$. The identity element of this group is $[e] = H$.
A homomorphism (or morphism) between semigroups $\langle S, \cdot \rangle$ and $\langle T, \ast \rangle$ is a function $\theta: S \to T$, such that $\theta (s_{1} \cdot s_{2}) = \theta (s_{1}) \ast \theta (s_{2})$, $\forall s_{1}, s_{2} \in S$.
If $\langle S, \cdot \rangle$ and $\langle T, \ast \rangle$ are monoids, then for $\theta$ to be a homomorphism, it must additionally satisfy $\theta(e_{S}) = e_{T}$.
If $\langle S, \cdot \rangle$ and $\langle T, \ast \rangle$ are groups, then there are two more conditions on $\theta$: $\theta (e_{S}) = e_{T}$, and $\theta (s^{-1}) = [\theta(s)]^{-1}$, $\forall s \in S$. These conditions, however, are redundant and can be derived from the first as follows:
Let $\theta$ be a homomorphism between semigroups $\langle S, \cdot \rangle$ and $\langle T, \ast \rangle$, which are also groups. Now, $\theta(e_{S}) = \theta (e_{S} \cdot e_{S}) = \theta (e_{S}) \ast \theta(e_{S})$, i.e., $\theta(e_{S})$ is idempotent in $T$. However, since $T$ is a group, its only idempotent is $e_{T}$; hence $\theta(e_{S}) = e_{T}$.
Let $\theta$ be a homomorphism between semigroups $\langle S, \cdot \rangle$ and $\langle T, \ast \rangle$, which are also groups. Then $\theta (e_{S}) = \theta (s \cdot s^{-1}) = \theta (s) \ast \theta(s^{-1}) = e_{T}$. Similarly, $\theta (s^{-1}) \ast \theta (s) = e_{T}$. This shows that $\theta (s^{-1}) = [\theta(s)]^{-1}$, $\forall s \in S$.
Let $g$ be a homomorphism from a structure $\langle X, \cdot \rangle$ to $\langle Y, \ast \rangle$.
If $g: X \to Y$ is surjective, it is called an epimorphism.
If $g: X \to Y$ is injective, it is called an monomorphism.
If $g: X \to Y$ is bijective, it is called an isomorphism.
If $g: X \to Y$ is bijective and $X = Y$, it is called an automorphism.