Homomorphism

In abstract algebra, a homomorphism is a structure-preserving map between two algebraic structures of the same type. Think of it as a function that takes elements from one group (or ring, or vector space) and maps them to elements in another group, in such a way that the operations in the original structure are maintained in the target structure.

Definition of Group Homomorphism

Let $(G, *)$ and $(G', \cdot)$ be two groups. A function $f: G \to G'$ is called a group homomorphism if for all $a, b \in G$, the following condition holds: $f(a * b) = f(a) \cdot f(b)$

This means that applying the operation $*$ in group $G$ to two elements $a$ and $b$, and then mapping the result to $G'$ using $f$, is the same as mapping $a$ and $b$ to $G'$ first using $f$, and then applying the operation $\cdot$ in $G'$.

Properties of Group Homomorphisms

Several important properties follow directly from the definition:

• Identity Element Mapping: If $f: G \to G'$ is a homomorphism, then $f(e_G) = e_{G'}$, where $e_G$ and $e_{G'}$ are the identity elements of $G$ and $G'$ respectively.
  Proof: Let $a \in G$. Then $f(a * e_G) = f(a)$ by the property of identity. By homomorphism property, $f(a * e_G) = f(a) \cdot f(e_G)$. So, $f(a) = f(a) \cdot f(e_G)$. Multiplying by the inverse of $f(a)$ (which exists in $G'$), we get $e_{G'} = f(e_G)$.
• Inverse Element Mapping: For any $a \in G$, $f(a^{-1}) = (f(a))^{-1}$.
  Proof: We know that $a * a^{-1} = e_G$. Applying $f$ to both sides, $f(a * a^{-1}) = f(e_G)$. Using the homomorphism property and the identity mapping property, we get $f(a) \cdot f(a^{-1}) = e_{G'}$. Multiplying by $(f(a))^{-1}$ on the left, we get $f(a^{-1}) = (f(a))^{-1}$.
• Image of a Subgroup: The image of a subgroup under a homomorphism is a subgroup. If $H$ is a subgroup of $G$, then $f(H) = \{f(h) \mid h \in H\}$ is a subgroup of $G'$.
• Kernel of a Homomorphism: The kernel of a homomorphism $f: G \to G'$ is defined as $\text{ker}(f) = \{a \in G \mid f(a) = e_{G'}\}$. The kernel of a group homomorphism is always a normal subgroup of $G$.
  Proof of Normality: We need to show that for any $g \in G$ and any $k \in \text{ker}(f)$, $gkg^{-1} \in \text{ker}(f)$. $f(gkg^{-1}) = f(g)f(k)f(g^{-1})$ (by homomorphism property) $= f(g)e_{G'}f(g^{-1})$ (since $k \in \text{ker}(f)$, $f(k) = e_{G'}$) $= f(g)f(g^{-1})$ $= f(gg^{-1})$ $= f(e_G)$ $= e_{G'}$ Thus, $gkg^{-1} \in \text{ker}(f)$.

Types of Homomorphisms

• Monomorphism: An injective (one-to-one) homomorphism.
• Epimorphism: A surjective (onto) homomorphism.
• Isomorphism: A homomorphism that is both injective and surjective. If an isomorphism exists between two groups, they are considered structurally identical.
• Endomorphism: A homomorphism from a group to itself (i.e., $f: G \to G$).
• Automorphism: An isomorphism from a group to itself.

Examples of Group Homomorphisms

1. Let $G = (\mathbb{Z}, +)$ be the group of integers under addition and $G' = (\mathbb{Z}_n, +_n)$ be the group of integers modulo $n$ under addition modulo $n$. The function $f: \mathbb{Z} \to \mathbb{Z}_n$ defined by $f(x) = x \pmod n$ is a homomorphism.
   Verification: $f(a+b) = (a+b) \pmod n$. And $f(a) +_n f(b) = (a \pmod n) +_n (b \pmod n) = (a+b) \pmod n$. Thus, $f(a+b) = f(a) +_n f(b)$.
2. Consider the group of non-zero real numbers under multiplication $({\mathbb{R}}^\times, \times)$ and the group of positive real numbers under multiplication $({\mathbb{R}}^+ , \times)$. The function $f: {\mathbb{R}}^\times \to {\mathbb{R}}^+$ defined by $f(x) = |x|$ is a homomorphism.
   Verification: $f(a \times b) = |a \times b| = |a| \times |b|$. And $f(a) \times f(b) = |a| \times |b|$. Thus, $f(a \times b) = f(a) \times f(b)$.
3. The determinant function $\det: GL_n(\mathbb{R}) \to \mathbb{R}^\times$, where $GL_n(\mathbb{R})$ is the group of invertible $n \times n$ real matrices under matrix multiplication, is a homomorphism. This is because $\det(AB) = \det(A) \det(B)$.

Homomorphisms for Other Algebraic Structures

The concept of homomorphism extends to other algebraic structures:

• Ring Homomorphism: A function $f: (R, +, \times) \to (R', +, \times)$ between two rings is a ring homomorphism if it preserves both addition and multiplication: $f(a+b) = f(a)+f(b)$ and $f(a \times b) = f(a) \times f(b)$ for all $a, b \in R$.
• Vector Space Homomorphism (Linear Transformation): A function $T: V \to W$ between two vector spaces $V$ and $W$ is a linear transformation if it preserves vector addition and scalar multiplication: $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$ and $T(c\mathbf{u}) = cT(\mathbf{u})$ for all $\mathbf{u}, \mathbf{v} \in V$ and scalar $c$.

Automorphism

An automorphism is a special type of homomorphism: an isomorphism from an algebraic structure to itself. It's essentially a symmetry of the structure that preserves its operations and relationships. For groups, an automorphism is a bijective map $f: G \to G$ such that $f(a * b) = f(a) * f(b)$ for all $a, b \in G$.

Inner Automorphisms

A very important class of automorphisms in group theory are the inner automorphisms. For a group $G$ and a fixed element $a \in G$, the inner automorphism induced by $a$ is the function $\phi_a: G \to G$ defined by: $\phi_a(g) = a g a^{-1}$ for all $g \in G$.

Let's verify that $\phi_a$ is indeed an automorphism:

1. Homomorphism Property: $\phi_a(g_1 * g_2) = a (g_1 * g_2) a^{-1}$ $= a g_1 * a g_2 * a^{-1}$ (using associativity of the group operation) $= (a g_1 a^{-1}) * (a g_2 a^{-1})$ $= \phi_a(g_1) * \phi_a(g_2)$ So, it is a homomorphism.
2. Injectivity: Suppose $\phi_a(g_1) = \phi_a(g_2)$. Then $a g_1 a^{-1} = a g_2 a^{-1}$. Multiplying by $a^{-1}$ on the left and $a$ on the right, we get $g_1 = g_2$. Thus, $\phi_a$ is injective.
3. Surjectivity: For any $h \in G$, we want to find a $g \in G$ such that $\phi_a(g) = h$. Let $g = a^{-1} h a$. Then $\phi_a(g) = \phi_a(a^{-1} h a) = a (a^{-1} h a) a^{-1} = (a a^{-1}) h (a a^{-1}) = e h e = h$. Thus, $\phi_a$ is surjective.

Since $\phi_a$ is a bijective homomorphism from $G$ to itself, it is an automorphism.

Properties of Inner Automorphisms

• The set of all inner automorphisms of $G$, denoted by $Inn(G)$, forms a group under the operation of function composition.
• The group $Inn(G)$ is isomorphic to the quotient group $G/Z(G)$, where $Z(G)$ is the center of the group $G$. The center $Z(G)$ is the set of elements that commute with all elements in $G$, i.e., $Z(G) = \{z \in G \mid zg = gz \text{ for all } g \in G\}$.
  Recall: The center $Z(G)$ is always a normal subgroup of $G$.
• An element $g \in G$ commutes with $a$ if and only if $\phi_a(g) = g$. This means that $\phi_a$ is the identity automorphism if and only if $a$ is in the center $Z(G)$.

Outer Automorphisms

Automorphisms that are not inner automorphisms are called outer automorphisms. The group of all automorphisms of $G$ is denoted by $Aut(G)$. The relationship between these groups is $Inn(G) \subseteq Aut(G)$. The quotient group $Aut(G)/Inn(G)$ is called the group of outer automorphisms.

Example of Automorphisms

Consider the group $G = (\mathbb{Z}_4, +)$, the integers modulo 4 under addition. The elements are $\{0, 1, 2, 3\}$. The identity is $0$.

Let's find the inner automorphisms:

• $\phi_0(g) = 0 + g - 0 = g$. This is the identity automorphism.
• $\phi_1(g) = 1 + g - 1$. $\phi_1(0) = 1+0-1 = 0$ $\phi_1(1) = 1+1-1 = 1$ $\phi_1(2) = 1+2-1 = 2$ $\phi_1(3) = 1+3-1 = 3$ (modulo 4). This is the identity automorphism.
• $\phi_2(g) = 2 + g - 2$. $\phi_2(0) = 2+0-2 = 0$ $\phi_2(1) = 2+1-2 = 1$ $\phi_2(2) = 2+2-2 = 2$ $\phi_2(3) = 2+3-2 = 3$ (modulo 4). This is the identity automorphism.
• $\phi_3(g) = 3 + g - 3$. $\phi_3(0) = 3+0-3 = 0$ $\phi_3(1) = 3+1-3 = 1$ $\phi_3(2) = 3+2-3 = 2$ $\phi_3(3) = 3+3-3 = 3$ (modulo 4). This is the identity automorphism.

In $\mathbb{Z}_4$, every element commutes with every other element, so $Z(\mathbb{Z}_4) = \mathbb{Z}_4$. Therefore, all inner automorphisms are the identity automorphism.

Are there any other automorphisms? Consider the function $f(x) = -x \pmod 4$. $f(0)=0, f(1)=3, f(2)=2, f(3)=1$. Let's check if it's a homomorphism: $f(a+b) = -(a+b) \pmod 4$. $f(a)+f(b) = (-a) + (-b) \pmod 4 = -(a+b) \pmod 4$. It is a homomorphism. It is also clearly bijective. So, $f(x) = -x$ is an automorphism. Is $f(x) = -x$ an inner automorphism? $\phi_a(g) = a + g - a \pmod 4$. If $a=1$, $\phi_1(g) = 1+g-1 = g$. If $a=2$, $\phi_2(g) = 2+g-2 = g$. If $a=3$, $\phi_3(g) = 3+g-3 = g$. None of the inner automorphisms (which are all the identity) match $f(x) = -x$ except for when $f(x)=x$. The automorphism $f(x) = -x$ is an outer automorphism for $\mathbb{Z}_4$.

Automorphisms of the Symmetric Group $S_n$

For $n \ge 3$, the outer automorphism group of $S_n$ is trivial, meaning all automorphisms of $S_n$ are inner. However, for $n=1, 2$, the situation is different. $S_1$ and $S_2$ are abelian, so $Inn(S_n)$ is trivial. For $S_3$, $Inn(S_3) \cong S_3$, and $Aut(S_3) \cong S_3$. For $n \ge 3$, $Aut(S_n) = Inn(S_n)$.

Cayley's Theorem

Cayley's Theorem is a fundamental result in group theory that states every finite group $G$ is isomorphic to a subgroup of the symmetric group acting on the set of elements of $G$. In simpler terms, it says that any group can be represented as a group of permutations. This is significant because it allows us to study abstract groups using the concrete and well-understood structure of permutation groups.

Statement of Cayley's Theorem

For any group $G$, there exists an injective homomorphism $\phi: G \to Aut(G)$. Furthermore, if $G$ is a finite group of order $n$, then $G$ is isomorphic to a subgroup of $S_n$.

Proof of Cayley's Theorem (for Finite Groups)

Let $G$ be a finite group of order $n$. We want to show that $G$ is isomorphic to a subgroup of $S_n$.

We will construct a homomorphism $\phi: G \to S_n$. Let $G = \{g_1, g_2, \dots, g_n\}$. For each element $a \in G$, define a function $f_a: G \to G$ by $f_a(x) = ax$ for all $x \in G$.

Step 1: Show $f_a$ is a permutation of $G$.

1. Well-defined: Since $a$ and $x$ are elements of $G$, and $G$ is a group, the product $ax$ is also an element of $G$. So $f_a$ maps elements of $G$ to elements of $G$.
2. Injective: Suppose $f_a(x) = f_a(y)$. Then $ax = ay$. Multiplying by $a^{-1}$ on the left, we get $a^{-1}(ax) = a^{-1}(ay)$, which simplifies to $(a^{-1}a)x = (a^{-1}a)y$, so $ex = ey$, which means $x=y$. Thus, $f_a$ is injective.
3. Surjective: For any element $b \in G$, we want to find an $x \in G$ such that $f_a(x) = b$. Let $x = a^{-1}b$. Then $f_a(x) = f_a(a^{-1}b) = a(a^{-1}b) = (aa^{-1})b = eb = b$. Thus, $f_a$ is surjective.

Step 2: Show that the map $\phi: G \to S_n$ defined by $\phi(a) = f_a$ is a homomorphism. We need to show that $\phi(ab) = \phi(a) \circ \phi(b)$ for all $a, b \in G$, where $\circ$ denotes the composition of permutations. Recall that $\phi(a)$ is the function $f_a(x) = ax$, and $\phi(b)$ is the function $f_b(x) = bx$. The composition $(\phi(a) \circ \phi(b))(x)$ means applying $\phi(b)$ first, then $\phi(a)$. $(\phi(a) \circ \phi(b))(x) = \phi(a)(\phi(b)(x))$ $= \phi(a)(bx)$ (since $\phi(b)(x) = bx$) $= a(bx)$ (since $\phi(a)(y) = ay$) $= (ab)x$ (by associativity in $G$) Now, let's consider $\phi(ab)$. This corresponds to the function $f_{ab}(x) = (ab)x$. So, we have $(\phi(a) \circ \phi(b))(x) = (ab)x = \phi(ab)(x)$ for all $x \in G$. Therefore, $\phi(ab) = \phi(a) \circ \phi(b)$. This shows that $\phi$ is a group homomorphism.

Step 3: Show that $\phi$ is injective. We need to show that if $\phi(a) = \phi(b)$, then $a=b$. If $\phi(a) = \phi(b)$, then $f_a = f_b$. This means that $f_a(x) = f_b(x)$ for all $x \in G$. So, $ax = bx$ for all $x \in G$. Let's choose $x=e$ (the identity element of $G$). Then $ae = be$, which implies $a=b$. Thus, $\phi$ is injective.

Since $\phi: G \to S_n$ is an injective homomorphism, its image $\phi(G)$ is a subgroup of $S_n$ that is isomorphic to $G$. The order of $\phi(G)$ is the same as the order of $G$, which is $n$. However, the order of $S_n$ is $n!$. Since $n \le n!$ for $n \ge 1$, $G$ is isomorphic to a subgroup of $S_n$.

The Regular Representation

The homomorphism $\phi: G \to S_n$ constructed in the proof of Cayley's Theorem is called the regular representation of $G$. It provides a way to view the group $G$ as a group of permutations.

Example of Cayley's Theorem

Let's consider the Klein four-group $V = \{e, a, b, c\}$ with $a^2=b^2=c^2=e$, $ab=c$, $bc=a$, $ca=b$, and $V$ is abelian. The order of $V$ is $n=4$. According to Cayley's Theorem, $V$ is isomorphic to a subgroup of $S_4$.

Let's construct the permutations:

• $f_e(x) = ex = x$. This is the identity permutation: $(e)(a)(b)(c)$.
• $f_a(x) = ax$. $f_a(e) = ae = a$ $f_a(a) = aa = e$ $f_a(b) = ab = c$ $f_a(c) = ac = b$ So, $f_a$ corresponds to the permutation $(a e)(c b)$ in cycle notation (mapping $e \to a, a \to e, b \to c, c \to b$). Let's represent elements as $1, 2, 3, 4$. Let $e=1, a=2, b=3, c=4$. $f_2(1)=2, f_2(2)=1, f_2(3)=4, f_2(4)=3$. This is the permutation $(1 2)(3 4)$.
• $f_b(x) = bx$. $f_b(e) = be = b$ $f_b(a) = ba = c$ $f_b(b) = bb = e$ $f_b(c) = bc = a$ So, $f_b$ corresponds to the permutation $(b e a c)$. In numerical notation: $(1 3 2 4)$.
• $f_c(x) = cx$. $f_c(e) = ce = c$ $f_c(a) = ca = b$ $f_c(b) = cb = a$ $f_c(c) = cc = e$ So, $f_c$ corresponds to the permutation $(c e b a)$. In numerical notation: $(1 4 3 2)$.

The image of $V$ under $\phi$ is $\{ (1)(2)(3)(4), (1 2)(3 4), (1 3 2 4), (1 4 3 2) \}$. This set is a subgroup of $S_4$ and is isomorphic to the Klein four-group $V$.

Note that the permutations obtained depend on the ordering of elements chosen for $G$. However, the structure of the subgroup generated will always be isomorphic to $G$.

Significance of Cayley's Theorem

• Representation Theory: It provides a concrete representation for abstract groups.
• Understanding Group Structure: It shows that the properties of any group can be studied by looking at permutations.
• Foundation for Further Study: It is a stepping stone to more advanced topics like permutation group theory and representation theory.