Intro to Abstract Algebra

Abstract Algebra Intro to Groups Abelian The Group of Units Modulo \(n\) Multiplicative Group vs Additive Group Subgroups Cyclic Subgroups Center of a Group Centralizer of an element in a Group

Introduction

At this point, we are already familiar with the rules of linear algebra, and we have encountered many mathematical objects and operations in our school life. What we have done is essentially studying operations constrained by a particular environment, or structure. Now, we would like to discuss the "structure" of our mathematical world more formally. From now on, the term "algebra" expands beyond algebraic computation to mean the study of these underlying structures. In abstract algebra, we learn the fundamental laws that govern operations across mathematics. Our goal is to categorize these mathematical structures (Groups, Rings, and Fields), giving us a clear, unified view of the mathematical world. This rigorous approach is not just theoretical, which will contribute directly to advancements in computer science and engineering.

We start with a special kind of set, which is called a group.

Intro to Groups

Groups:

A binary operation on a set \(G\) assigns to each ordered pair \((a, b)\) of elements of \(G\) an element of \(G\), denoted by \(ab\). The requirement that \(ab \in G\) for all \(a, b \in G\) is called closure.

A set \(G\) equipped with a binary operation is said to be a group if the following three properties are satisfied:

  1. Associativity
    2. \[ \forall a, b, c \in G, \, (ab)c = a(bc). \]
  2. Identity
    3. \[ \exists \, e \in G \text{ s.t. } \forall a \in G, \, ae = ea = a. \]
  3. Inverses
    4. \[ \forall a \in G, \exists \, b \in G \text{ s.t. } ab = ba = e. \]

Remark. The group axioms guarantee the existence of an identity and of inverses, but in fact both are unique: a group has exactly one identity element, and every element \(a \in G\) has a unique inverse, which we henceforth denote by \(a^{-1}\). These uniqueness facts follow from the axioms by short standard arguments, and we will use them freely from now on.

For example, consider a square matrix: \[ A = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} \] assuming \(\det(A) = ad -bc \neq 0\) where \(a, b, c, d \in \mathbb{R}\). Then we know that we have the identity matrix: \[ I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \] such that \[ IA = AI = A \] and the inverse of \(A\): \[ A^{-1} = \begin{bmatrix} \frac{d}{\det (A)} & \frac{-b}{\det (A)} \\ \frac{-c}{\det (A)} & \frac{a}{\det (A)} \\ \end{bmatrix} \] such that \[ AA^{-1} = A^{-1}A = I \] We would like to say that this set of all \(2 \times 2\) invertible matrices forms the group under matrix multiplication: \[ GL(2, \mathbb{R}) = \left\{ \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} \mid a, b, c, d \in \mathbb{R}, \, ad-bc \neq 0 \right\}. \] Closure follows from the multiplicativity of the determinant, \(\det(AB) = \det(A)\det(B) \neq 0\), and the inverse \(A^{-1}\) also lies in \(GL(2, \mathbb{R})\) since \(\det(A^{-1}) = 1/\det(A) \neq 0\). Associativity is inherited from matrix multiplication, so all group axioms are satisfied. This group, the general linear group of degree 2 over the real numbers, is a fundamental object in abstract algebra.

Abelian

The general linear group is a typical example of a non-Abelian group.

Abelian Group:

If a group \(G\) satisfies the commutative property for every pair of its elements \(a\) and \(b\), the group \(G\) is called an Abelian group. \[ \forall a, b \in G, \, ab = ba \in G. \]

In general, matrix multiplication is non-commutative. For instance, take \[ A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \] in \(GL(2, \mathbb{R})\). A direct computation gives \[ AB = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} \neq \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} = BA. \] So there exist elements of \(GL(2, \mathbb{R})\) that do not commute, which shows that \(GL(2, \mathbb{R})\) is a non-Abelian group.

For example, a simple Abelian group can be the set of nonzero real numbers under ordinary multiplication, denoted by \(\mathbb{R}^*\). Let \(x \in \mathbb{R}^*\), then the identity is \(1\), the inverse is \(\frac{1}{x}\). Moreover, for every \(x, y \in \mathbb{R}^*\), \(xy = yx \in \mathbb{R}^*\).

The Group of Units Modulo \(n\)

In Computer Science, we frequently use modular arithmetic, which is fundamental to cryptography (e.g., RSA), hashing algorithms, and how computers handle integer representations. Here we introduce the Group of Units Modulo \(n\), denoted by \(U(n)\).

First, let's briefly review the definition of the modulo operation, which is based on the Division Algorithm:

Modular Arithmetic:

Given integers \(a\) (dividend) and \(n\) (divisor), we write \(a \bmod n = r\) if: \[ a = qn + r \] where \(q\) is the quotient and \(r\) is the remainder, and the remainder must satisfy the condition \(0 \leq r < n\).

For example, \[ \begin{align*} 11 \bmod 2 &= 1 \text{ since } 11 = 2 \cdot 5 + 1, \\\\ 24 \bmod 8 &= 0 \text{ since } 24 = 8 \cdot 3 + 0, \\\\ 9 \bmod 12 &= 9 \text{ since } 9 = 0 \cdot 12 + 9, \\\\ -5 \bmod 2 &= 1 \text{ since } -5 = (-3) \cdot 2 + 1, \\\\ \end{align*} \]

An important property in modular arithmetic is that an integer \(a\) has a multiplicative inverse modulo \(n\) if and only if \(a\) and \(n\) are relatively prime (i.e., their greatest common divisor is 1, \(\gcd(a, n) = 1\)).

For any positive integer \(n\), we define \(U(n)\) to be the set of all positive integers less than \(n\) that are relatively prime to \(n\). This set \(U(n)\) forms a group under the operation of multiplication modulo \(n\).

To see this, observe: associativity is inherited from ordinary integer multiplication; the element \(1\) is the identity since \(\gcd(1, n) = 1\); and the multiplicative inverse exists and lies in \(U(n)\) by the fact stated above. The only nontrivial point is closure: if \(\gcd(a, n) = \gcd(b, n) = 1\), then \(\gcd(ab, n) = 1\), and reducing modulo \(n\) preserves coprimality. So \(ab \bmod n \in U(n)\), and \(U(n)\) is indeed a group.

For example, when \(n = 12 = 2^2 \cdot 3\), we eliminate multiples of 2 and 3 from \(\{1, 2, \cdots, 10, 11\}\), we obtain \(U(12) = \{1, 5, 7, 11\}\). Moreover, Cayley Table for \(U(12) = \{1, 5, 7, 11\}\) is as follows:

\(\bmod \mathbf{12}\) \(\mathbf{1}\) \(\mathbf{5}\) \(\mathbf{7}\) \(\mathbf{11}\)
\(\mathbf{1}\) 1 5 7 11
\(\mathbf{5}\) 5 1 11 7
\(\mathbf{7}\) 7 11 1 5
\(\mathbf{11}\) 11 7 5 1

In this table, for example, \((5 \cdot 7) \bmod 12 = 11 \in U(12)\), and as you can see, this is an example of finite Abelian group.

The number of elements of a group \(G\) is called its order, denoted by \(|G|\). If \(G\) has finitely many elements, \(|G|\) is a positive integer; otherwise \(G\) is said to have infinite order. For example, \(|U(12)| = 4\), while \(GL(2, \mathbb{R})\) has infinite order. So \(GL(2, \mathbb{R})\) is an infinite non-Abelian group.

Multiplicative Group vs Additive Group

So far we have seen multiplicative groups whose binary operation is some sort of multiplication (like matrix multiplication or modular multiplication), but sometimes binary operation can be addition. A simple additive group is the set of integers: \[ \mathbb{Z} = \{\cdots, -2, -1, 0, 1, 2, \cdots\} \] This set forms a group under standard addition, which is a valid binary operation since the sum of two integers is an integer. For any \(a, b, c \in \mathbb{Z}\): \[ \begin{align*} &(a + b) + c = a + (b + c), \\\\ &a + 0 = 0 + a = a, \\\\ &a + (-a) = (-a) + a = 0. \\\\ \end{align*} \] In addition, this group is Abelian since standard addition is commutative: \(a + b = b + a\).

In abstract algebra, the same group axioms can be written using either multiplicative or additive notation, and switching between them can be confusing at first. The underlying group structure is the same; only the symbols differ.

In multiplicative notation, the operation is written \(ab\), the identity is \(e\) (or \(1\)), the inverse of \(a\) is \(a^{-1}\), and repeated application is written as an exponent: \[ a^n = \underbrace{a \cdot a \cdots a}_{n \text{ times}} \quad (n > 0), \qquad a^0 = e, \qquad a^{-n} = (a^{-1})^n. \] This gives the familiar exponent rules \(a^m a^n = a^{m+n}\) and \((a^m)^n = a^{mn}\) for all integers \(m, n\).

In additive notation, the operation is written \(a + b\), the identity is \(0\), the inverse of \(a\) is \(-a\), and repeated application is written as a multiple: \[ na = \underbrace{a + a + \cdots + a}_{n \text{ times}} \quad (n > 0), \qquad 0 \cdot a = 0, \qquad (-n)a = n(-a). \] Here the integer \(n\) in \(na\) is a multiplier, not a group element being combined with \(a\); the expression \(na\) is the additive counterpart of the multiplicative \(a^n\).

Care with non-Abelian groups. In multiplicative notation, the identity \((ab)^n = a^n b^n\) holds only when \(a\) and \(b\) commute. For a general group element, \[ (ab)^2 = abab, \quad \text{which equals } a^2 b^2 \text{ only if } ab = ba. \] Indeed, commutativity allows the rearrangement \[ (ab)^2 = a \cdot b \cdot a \cdot b = a \cdot a \cdot b \cdot b = a^2 b^2. \]

For negative exponents, we use the standard identity \((ab)^{-1} = b^{-1} a^{-1}\) (note the reversed order), so \[ (ab)^{-2} = \bigl((ab)^{-1}\bigr)^2 = (b^{-1} a^{-1})(b^{-1} a^{-1}). \]

Remark (Inverse of a product): For any elements \(a, b\) in a group, \((ab)^{-1} = b^{-1} a^{-1}\). This follows from \[ (ab)(b^{-1} a^{-1}) = a(bb^{-1})a^{-1} = a e a^{-1} = a a^{-1} = e, \] so \(b^{-1} a^{-1}\) satisfies the defining property of the inverse of \(ab\), and by the uniqueness of inverses (see Remark above) must equal \((ab)^{-1}\).

The same uniqueness-of-inverses argument gives another standard identity we will use freely: \[ (a^{-1})^{-1} = a \qquad \text{for every } a \in G, \] since \(a\) satisfies the defining property of the inverse of \(a^{-1}\).

We now define the notion of order at the element level.

Definition: Order of an Element The order of an element \(a\) in a group \(G\), denoted \(|a|\), is the smallest positive integer \(n\) such that \(a^n = e\) (or \(na = 0\) in additive notation). If no such \(n\) exists, we say \(a\) has infinite order, written \(|a| = \infty\).

For example, in \(U(12) = \{1, 5, 7, 11\}\), the identity \(1\) has order \(1\), and a direct computation shows \[ \begin{align*} 5^2 &= 25 \equiv 1 \pmod{12}, \\\\ 7^2 &= 49 \equiv 1 \pmod{12}, \\\\ 11^2 &= 121 \equiv 1 \pmod{12}, \end{align*} \] so every non-identity element of \(U(12)\) has order \(2\).

The order of an element controls every power that returns to the identity, in the following sense.

Theorem: Order Divides Any Annihilating Power Let \(g\) be an element of finite order in a group \(G\). If \(g^m = e\) for some integer \(m\), then \(|g|\) divides \(m\).
Proof:

Let \(n = |g|\). By the division algorithm, write \(m = qn + r\) with \(0 \leq r < n\). Then \[ g^r = g^{m - qn} = g^m (g^n)^{-q} = e \cdot e = e. \] Since \(0 \leq r < n\) and \(n\) is the smallest positive integer with \(g^n = e\), the only possibility is \(r = 0\). Therefore \(m = qn\), i.e., \(n \mid m\).

Subgroups

Subgroups:

If a subset \(H\) of a group \(G\) is itself a group under the same operation of \(G\), \(H\) is said to be a subgroup of \(G\) denoted as \(H \leq G\). (Also, we write \(H < G\) to denote a proper subgroup.)

For example, consider the group of integers under addition, \(\mathbb{Z}\). Let \(H\) be the set of all even integers: \[ H = \{\dots, -4, -2, 0, 2, 4, 6, \dots\}. \] This subset \(H\) forms a group under the same addition: the sum of two even integers is even (closure), \(0 \in H\) (identity), and the inverse \(-a\) of any even integer \(a\) is again even (inverses). Associativity is inherited from \(\mathbb{Z}\). Therefore \(H \leq \mathbb{Z}\).

Checking all group axioms every time can be tedious. A way to show that a subset \(H\) of a group \(G\) is a subgroup of \(G\) is One-Step Subgroup Test:

One-Step Subgroup Test:

A nonempty subset \(H\) of a group \(G\) is a subgroup of \(G\) if for all \(a, b \in H\), the element \(a b^{-1} \in H\). (In additive notation, the condition becomes \((a - b) \in H\).)

Proof:

Suppose \(H\) is a nonempty subset of \(G\) such that \(ab^{-1} \in H\) whenever \(a, b \in H\) (where \(b^{-1}\) denotes the inverse of \(b\) in \(G\)). We must verify that \(H\) possesses the identity element, inverses for every element, and closure. Associativity is inherited automatically from \(G\).

Since \(H\) is nonempty, there exists at least one element \(x \in H\). Applying the hypothesis with \(a = x\) and \(b = x\), we get: \[ a b^{-1} = x x^{-1} = e. \] Since the result must be in \(H\), the identity element \(\mathbf{e \in H}\).

Now we know \(e \in H\), let \(x\) be any element in \(H\). Apply the hypothesis with \(a = e\) and \(b = x\): \[ a b^{-1} = e x^{-1} = x^{-1}. \] Since the result must be in \(H\), for every \(x \in H\), its inverse \(\mathbf{x^{-1} \in H}\).

Finally, we must show that \(H\) is closed. Let \(x\) and \(y\) be any two elements in \(H\). We must show that \(xy \in H\). We have already shown that if \(y \in H\), then \(y^{-1} \in H\). Now, apply the hypothesis with \(a = x\) and \(b = y^{-1}\): \[ a b^{-1} = x (y^{-1})^{-1} = xy \] (using the standard identity \((y^{-1})^{-1} = y\); see the Remark below). Since the result must be in \(H\), \(\mathbf{xy \in H}\).

Therefore, \(H\) is a subgroup of \(G\).

Cyclic Subgroups

To construct a subgroup for a group \(G\), we start with a single element \(a \in G\) and keep applying the group operation to it. This generates a special set called a cyclic subgroup.

Cyclic Subgroup: Let \(G\) be a group and let \(a\) be any element of \(G\). The set of all integer powers of \(a\) is called the cyclic subgroup of \(G\) generated by \(a\), denoted by \(\langle a \rangle\): \[ \langle a \rangle = \{ a^n \mid n \in \mathbb{Z} \} \] (For additive operation, \(\langle a \rangle = \{ na \mid n \in \mathbb{Z} \}\)).

Is this really a subgroup of \(G\)? We can prove it easily using the One-Step Subgroup Test.

Proof:

Since \(a \in \langle a \rangle\), \(\langle a \rangle\) is not empty. For any integers \(m , n\), let \(a^m\) and \(a^n\) be any two elements in \(\langle a \rangle\).

We apply the One-Step Test condition: \[ a^m (a^n)^{-1} = a^m a^{-n} = a^{m-n}. \] Since \(m\) and \(n\) are integers, their difference \(m-n\) is also an integer. Therefore, \(a^{m-n}\) is another power of \(a\), so it must be in \(\langle a \rangle\).

Thus, \(\langle a \rangle\) is a subgroup of \(G\).

Let's revisit the group \(U(12) = \{1, 5, 7, 11\}\) under multiplication modulo 12. We can generate subgroups by picking elements and computing their powers modulo 12.

Probably, a more interesting group can be \(U(10) = \{1, 3, 7, 9\}\).

Definition: Cyclic Group and Generator A group \(G\) is called cyclic if there exists an element \(a \in G\) such that \(G = \langle a \rangle\). Any such element \(a\) is called a generator of \(G\).

By the definition above, \(U(10)\) is cyclic: the elements \(3\) and \(7\) each generate the whole group. On the other hand, since no single element of \(U(12)\) generates all four of its elements, \(U(12)\) is not cyclic.

Center of a Group

Here, we introduce another important subgroup.

Center of a Group: The center of a group \(G\) is the subset of elements in \(G\) that commute with every element of \(G\): \[ Z(G) = \{a \in G \mid ax = xa, \, \forall x \in G\}. \] (Note: The notation "\(Z(G)\)" comes from the German word for center, "Zentrum.")

Indeed, the center of a group \(G\) is a subgroup of \(G\):

Proof:

The identity element \(e \in G\) commutes with every element \(x \in G\): \[ ex = xe = x. \] So \(e \in Z(G)\), and thus \(Z(G)\) is nonempty. Now let \(a, b \in Z(G)\). We will apply the One-Step Subgroup Test and show that \(ab^{-1} \in Z(G)\), i.e. that \(ab^{-1}\) commutes with every \(x \in G\).

Step 1: \(b^{-1}\) commutes with every \(x \in G\).
Since \(b \in Z(G)\), for any \(x \in G\) we have \(bx = xb\). Multiplying both sides on the left and right by \(b^{-1}\): \[ b^{-1}(bx)b^{-1} = b^{-1}(xb)b^{-1}. \] The left side simplifies to \((b^{-1}b)xb^{-1} = xb^{-1}\), and the right side to \(b^{-1}x(bb^{-1}) = b^{-1}x\). Therefore \(xb^{-1} = b^{-1}x\) for every \(x \in G\).

Step 2: \(ab^{-1}\) commutes with every \(x \in G\).
Using Step 1 together with \(a \in Z(G)\) (so \(ax = xa\)): \[ \begin{align*} (ab^{-1})x &= a(b^{-1}x) \\\\ &= a(xb^{-1}) && \text{(by Step 1)} \\\\ &= (ax)b^{-1} \\\\ &= (xa)b^{-1} && \text{(since } a \in Z(G)\text{)} \\\\ &= x(ab^{-1}). \end{align*} \] Hence \(ab^{-1} \in Z(G)\), and by the One-Step Subgroup Test, \(Z(G) \leq G\).

Remark (Inverses commute with whatever the element commutes with): Step 1 of the proof above is an instance of a general fact: if \(a, x\) are elements of any group and \(ax = xa\), then \(ax^{-1} = x^{-1}a\). The derivation is identical — multiply \(ax = xa\) on the left and right by \(x^{-1}\). We will use this fact freely throughout the rest of this page.

The Center \(Z(G)\) is always an Abelian subgroup of \(G\): by definition every element of \(Z(G)\) commutes with every element of \(G\), and in particular with every other element of \(Z(G)\), so \(ab = ba\) for all \(a, b \in Z(G)\).

In fact, the center characterizes Abelian groups:

Theorem: Abelian \(\Leftrightarrow\) Z(G) = G \[ G \text{ is Abelian } \iff Z(G) = G. \]
Proof:

(\(\Rightarrow\)) Suppose \(G\) is Abelian. Then every element of \(G\) commutes with every other element, so every element of \(G\) lies in \(Z(G)\). Hence \(Z(G) = G\).

(\(\Leftarrow\)) Conversely, suppose \(Z(G) = G\). Then every element of \(G\) is in \(Z(G)\), so every element commutes with every other element. Hence \(G\) is Abelian.

For example, \(U(12) = \{1, 5, 7, 11\}\) is Abelian, so \(Z(U(12)) = U(12)\). On the other hand, \(GL(2, \mathbb{R})\) is non-Abelian, and its center consists only of nonzero scalar matrices: \[ Z(GL(2, \mathbb{R})) = \left\{ \begin{pmatrix} c & 0 \\ 0 & c \end{pmatrix} \,\Big|\, c \in \mathbb{R},\ c \neq 0 \right\}. \]

Proof sketch for \(Z(GL(2, \mathbb{R})) = \{cI \mid c \neq 0\}\): The inclusion \(\supseteq\) is immediate: any scalar matrix \(cI\) commutes with every matrix since \((cI)M = cM = M(cI)\). For the reverse inclusion, suppose \(A \in Z(GL(2, \mathbb{R}))\), so \(AM = MA\) for every invertible \(M\). Testing this against the elementary matrix \(E_{12} = \begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}\) forces the off-diagonal entries of \(A\) to vanish, and testing against a diagonal matrix with distinct nonzero entries forces the two diagonal entries of \(A\) to be equal. (A detailed computation is carried out for the shear matrix in the next section.) Hence \(A = cI\) for some \(c \neq 0\).

Centralizer of an element in a Group

It is possible that some elements in a non-Abelian group commute with only a proper subset of the group. The centralizer is a generalization of the Center \(Z(G)\) that allows us to precisely identify and study the set of elements that commute with a single, fixed element \(a \in G\). It is in non-Abelian groups that the centralizer becomes a powerful tool for exploring internal group structure.

Centralizer of \(a\) in \(G\): Let \(a \in G\) be fixed. The centralizer of \(a\) in \(G\) is the set of all elements in \(G\) that commute with \(a\): \[ C(a) = \{g \in G \mid ga = ag\}. \]

An important result is as follows:

\[ \forall a \in G, \, C(a) \leq G. \]
Proof:

The identity element \(e\) is in \(C(a)\) since \(ea = ae = a\), so \(C(a)\) is nonempty.

Let \(x, y \in C(a)\). We must show \(xy^{-1} \in C(a)\), meaning \((xy^{-1})a = a(xy^{-1})\).

We know \(xa = ax\) and \(ya = ay\). From the latter, we derive that \(y^{-1}\) also commutes with \(a\): From \(ya = ay\), applying left and right multiplications by \(y^{-1}\) on both sides of the equation: \[ y^{-1}(ya)y^{-1} = y^{-1}(ay)y^{-1} \implies ay^{-1} = y^{-1}a. \]

Now we verify that \(xy^{-1}\) commutes with \(a\): \[ \begin{align*} (xy^{-1})a &= x(y^{-1}a) \\\\ &= x(ay^{-1}) && \text{(Since } y \in C(a) \text{)} \\\\ &= (xa)y^{-1} \\\\ &= (ax)y^{-1} && \text{(Since } x \in C(a) \text{)} \\\\ &= a(xy^{-1}) \end{align*} \] Thus, \(xy^{-1}\) commutes with \(a\), and by the one-step subgroup test, \(C(a) \leq G\).

Note that:

\[ \forall a \in G, \, Z(G) \subseteq C(a). \]

This inclusion follows directly from the definitions: an element that commutes with every element of \(G\) certainly commutes with \(a\) in particular.

Taking the intersection over all \(a \in G\) yields a cleaner characterization of the center:

Theorem: Center as Intersection of Centralizers \[ Z(G) = \bigcap_{a \in G} C(a). \]
Proof:

(\(\subseteq\)) If \(g \in Z(G)\), then \(g\) commutes with every element of \(G\), so in particular \(g \in C(a)\) for every \(a \in G\). Hence \(g \in \bigcap_{a \in G} C(a)\).

(\(\supseteq\)) Conversely, if \(g \in \bigcap_{a \in G} C(a)\), then \(g \in C(a)\) for every \(a \in G\), meaning \(g\) commutes with every \(a \in G\). By the definition of the center, \(g \in Z(G)\).

Another important fact is as follows:

\[ G \text{ is Abelian } \iff \forall a \in G, \, C(a) = G. \]

This is also straightforward: if \(G\) is Abelian, every element commutes with every other, so \(C(a) = G\) for all \(a\). Conversely, if \(C(a) = G\) for all \(a \in G\), then every element of \(G\) commutes with \(a\), and since \(a\) is arbitrary, every pair of elements commutes, so \(G\) is Abelian.


Remember, the General Linear Group \(G = GL(2, \mathbb{R})\) is non-Abelian. We have shown that its center \(Z(G)\) consists only of scalar matrices. To verify the intersection formula, consider a non-scalar matrix \(A \in G\), for instance, the shear matrix: \[ A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}. \] Its centralizer \(C(A)\) is the set of all matrices \(M \in G\) that commute with \(A\) (\(MA = AM\)). Solving for \[ M = \begin{pmatrix} x & y \\ z & w \end{pmatrix} \] shows that \(C(A)\) requires \(z=0\) and \(w=x\): \[ C(A) = \left\{ \begin{pmatrix} x & y \\ 0 & x \end{pmatrix} \mid x, y \in \mathbb{R}, x \neq 0 \right\}. \] Notice that \(C(A)\) is a subgroup of \(G\), which fits the hierarchy \(Z(G) \subsetneq C(A) \subsetneq G\). If we calculate the centralizer for a different non-scalar matrix, say \[ B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}, \] we get a different subgroup \(C(B)\). The intersection of all such centralizers removes elements that fail to commute with some matrix, leaving only the center \(Z(G)\).

In contrast, for the Abelian group \(U(12)\), every element commutes with every other element. This means the centralizer of every element is the entire group: \[ C(1) = C(5) = C(7) = C(11) = U(12). \] Therefore, the intersection of all centralizers is simply the intersection of the set \(U(12)\) with itself, confirming that \(Z(U(12)) = U(12)\).