Homomorphisms
When considering sets, functions are allowed to do anything, as we can map any element in domain to any element in codomain. However, as groups have some additional structure on top of sets, we are more interested in functions that respect/preserve the group structure, and they are called homomorphisms.
The definition of functions should be revised first.
Definition. Let $(G, \ast)$ and $(H, \times)$ be two groups. A function $f$ is a group homomorphism iff
\[(\forall a, b \in G) \, f(a \ast b) = f(a) \times f(b)\]
Properties
Property. Homomorphism has identity preserved, i.e.
\[f(e_G) = e_H\]where $e_G$ and $e_H$ are the identity of $G$ and $H$ repsectively.
Proof.
\[\begin{align*} f(e_G) &= f(e_G \ast e_G) = f(e_G) \times f(e_G) \\ f(e_G)^{-1} \times f(e_G) &= (f(e_G)^{-1} \times f(e_G)) \times f(e_G) \\ e_H &= f(e_G) \end{align*}\]
Property. Homomorphism has inverse preserved, i.e.
\[f(a^{-1}) = f(a)^{-1}\]Proof.
\[\begin{gather} f(e_G) = f(a \ast a^{-1}) = f(a) \times f(a^{-1}) = e_H \\ f(e_G) = f(a^{-1} \ast a) = f(a^{-1}) \times f(a) = e_H \end{gather}\]Hence, $f(a^{-1})$ is the inverse of $f(a)$ in $H$, i.e. $f(a^{-1}) = f(a)^{-1}$.
Property. $\text{ord}(f(a)) \mid \text{ord}(a)$.
Proof.
Let $G$ be a finite group and $f: G \to H$ be a homomorphism. For $a \in G$, let $\text{ord}(a) = k$, we have
\[f(e_G) = f(a^k) = [f(a)]^k = e_H\]Hence, $\text{ord}(f(a))$ is finite, say is $l$. Let $k = ml + r$, where $0 \le r < l$, we have
\[[f(a)]^k = [f(a)]^{ml}[f(a)]^{r} = [f(a)]^{r} = e_H\]which implies $r = 0$ and therefore $k = ml$ and $\text{ord}(f(a)) \mid \text{ord}(a)$.
Property. The composite of two homomorphisms is a homomorphism.
Proof.
Let $f: G_1 \to G_2$ and $g: G_2 \to G_3$ be homomorphisms, then for $a, b \in G_1$,
\[\begin{align*} (g \circ f)(a \ast_1 b) = g(f(a \ast_1 b)) &= g(f(a) \ast_2 f(b)) \\ &= g(f(a)) \ast_3 g(f(b)) \\ &= (g \circ f)(a) \ast_3 (g \circ f)(b) \end{align*}\]
Image and Kernel
Definition. The image of a homomorphism $f: G \to H$ is defined by
\[\text{im}\, f = f(G) = \set{f(g) \in H : g \in G}\]which is same as the codomain of a function.
Definition. The kernel of a homomorphism $f: G \to H$ is defined by
\[\ker f = f^{-1}(\set{e_H}) = \set{g \in G : f(g) = e_H}\]which is the set of elements in $G$ that are mapped to identity of $H$.
Theorem. Image is a subgroup of $H$, i.e. $\text{im}\, f \le H$.
Proof.
As $f(e_G) = e_H$, $e_H \in \text{im}\, f$. Also, for all $h_1, h_2 \in \text{im}\, f$, there exists $g_1, g_2 \in G$ with $h_i = f(g_i)$. Hence,
\[f(g_1 \ast g_2^{-1}) = f(g_1) \times f(g_2^{-1}) = h_1 \times h_2^{-1} \in \text{im}\, f\]By super-efficient subgroup criterion, $\text{im}\, f \le H$.
Theorem. Kernel is a subgroup of $G$, i.e. $\ker f \le G$.
Proof.
$e_G \in \ker f$ as homomorphism preserves identity. For all $g_1, g_2 \in \ker f$, i.e. $f(g_1) = f(g_2) = e_H$, we have
\[f(g_1 \ast g_2^{-1}) = f(g_1) \times f(g_2^{-1}) = e_H \times e_H = e_H\]Hence, $g_1 \ast g_2^{-1} \in \ker f$. By super-efficient subgroup criterion, $\ker f \le G$.
Injectivity via Kernel
Theorem. A homomorphism $f: G \to H$ is
injective iff $\ker f = \set{e_G}$
surjective iff $\text{im}\, f = H$
Proof.
$(\Leftarrow)$ If $\ker f = \set{e_G}$, given $a, b \in G$ such that $f(a) = f(b)$,
\[f(e_G) = f(a \ast a^{-1}) = f(a) \times f(a^{-1}) = f(b) \times f(a^{-1}) = f(b \ast a^{-1}) = e_H\]Hence, $b \ast a^{-1} = e_G$ and $b = a$.
$(\Rightarrow)$ Conversely, if $f$ is injective, for $k \in \ker f$, $f(k) = e_H = f(e_G) \implies k = e_G$.
Isomorphisms
Definition. Isomorphisms are bijective homomorphisms. Two groups $G$ and $H$ are isomorphic if there exists an isomorphism between them, denoted as $G \cong H$.
Theorem. Inverse of isomorphism $f^{-1}: H \to G$ is an isomorphism.
Proof.
As $f$ is bijective, $f^{-1}: H \to G$ exists and is bijective, and
\[f^{-1}(h_1 \times h_2) = f^{-1}(f(g_1 \ast g_2)) = g_1 \ast g_2 = f^{-1}(h_1) \ast f^{-1}(h_2)\]$f^{-1}$ is a bijective homomorphism and therefore an isomorphism.
Isomorphic groups are considered to be “the same”, except the elements and operations are being relabeled.
Endomorphisms and Automorphisms
Definition. Endomorphisms are homomorphisms that has the same domain as codomain, meaning it is a mapping to itself.
Definition. Automorphisms are isomorphisms that are also endomorphic.
Automorphism can be thought as a symmetry of the group as it is a bijection to itself preserving the structure. Hence, a set of automorphisms is also a group as it is the set of symmetries of something.
Examples
Exponential Map
In Power Series form:
\[\exp(x) = 1 + x + {x^2 \over 2!} + {x^3 \over 3!} + ...\]which has the property
\[\exp(a + b) = \exp(a)\exp(b)\]It is a homomorphism from $(\mathbb{R}, +)$ to $(\mathbb{R}^{\ast}, \times)$, but not an isomorphism as it is not surjective. However, it is an isomorphism between $(\mathbb{R}, +)$ and $(\mathbb{R}_{> 0}, \times)$, and the inverse is the logarithmic map $\log(x)$. $\text{im} \exp = \mathbb{R}_{> 0}$ and $\ker \exp = \set{0}$.
Determinants
In linear algebra, using 2-by-2 matrices as example, we have
\[\det(A) = ad - bc\]which has the property
\[\det(AB) = \det(A) \det(B)\]It is a homomorphism from $(\text{GL}_n(\mathbb{R}), \times)$ to $(\mathbb{R}^{\ast}, \times)$. $\ker \det = \text{SL}_n(\mathbb{R})$.
Integers modulo m
In number theory, using $(\mathbb{Z}/5\mathbb{Z})^\times$ as example, it possesses a primitive root of unity $g$ where
\[(\mathbb{Z}/5\mathbb{Z})^\times = \set{ g^n : n \in \mathbb{Z}/4\mathbb{Z} }\]meaning every elements can be represented by some powers of $g$. We also have
\[g^{a + b} = (g^a)(g^b)\]Therefore, it is a homomorphism from $(\mathbb{Z}/4\mathbb{Z}, +)$ to $((\mathbb{Z}/5\mathbb{Z})^\times, \times)$.
On the other hand, let $f(a) = 2a$, then $f$ is an automorphism of $(\mathbb{Z}/5\mathbb{Z}, +)$ with inverse $f^{-1}(a) = 3a$.