The Center of a Group

In mathematics, the center of a group refers to a specific subset within $G$ that includes all elements capable of commuting with every other member of the group. This pivotal subset is usually represented by $Z(G)$ when referring to a group $G$.

Put simply, the center comprises all elements $z$ in $G$ for which $zg = gz$ holds true for any $g$ within $G$.

$$ Z(G) = \{z \in G | zg = gz \quad \forall g \in G\} $$

This conveys that for any element $z$ within the center and any element $g$ in the group, the result of multiplying $z$ and $g$ remains constant regardless of the sequence of multiplication.

The center of a group invariably forms a subgroup within that group and is also an abelian subgroup, as by definition, its elements are commutative.

Exploring the center offers insights into the group's internal structure and hints at how far the group deviates from being abelian. If a group's center encompasses the entire group itself, then the group is abelian.

Example

Let's delve into an intuitive example of a group's center: the group of rotations in a plane.

Imagine a set encompassing all possible rotations around the origin in a two-dimensional plane.

Each rotation can be characterized by an angle $ \theta $, indicating the degree to which an object is rotated around the origin.

For instance, $ \theta = 0° $, $ \theta = 10° $, $ \theta = 20° $, and so on.

In this context, the "center" of the group consists of all rotations that commute with any other rotation. Essentially, we're looking for rotations that yield the same outcome whether applied before or after any other rotation.

Upon reflection, the only rotation meeting this criterion is the $0$ degree rotation, which effectively leaves everything unchanged.

This is because a $0$ degree rotation followed by any other rotation $ \theta $ does not alter the angle $ \theta $, and similarly, applying a rotation of $ \theta $ followed by a $0$ degree rotation does not affect the outcome of the first rotation.

Example 2

Consider the group of $2 \times 2$ invertible matrices with real number components under the operation of matrix multiplication.

Our goal is to uncover the center of this group. The center, denoted as $Z(GL(2, \mathbb{R}))$, is the collection of matrices $A$ within $GL(2, \mathbb{R})$ such that $AB = BA$ holds true for any $B$ within $GL(2, \mathbb{R})$.

To be a part of the center, a matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ must commute with any other matrix $B = \begin{pmatrix} e & f \\ g & h \end{pmatrix}$ within $GL(2, \mathbb{R})$.

Let's calculate the products $AB$ and $BA$ and equate them to pinpoint the requirements for $a$, $b$, $c$, and $d$.

\[AB = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} e & f \\ g & h \end{pmatrix} = \begin{pmatrix} ae+bg & af+bh \\ ce+dg & cf+dh \end{pmatrix}\]

\[BA = \begin{pmatrix} e & f \\ g & h \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} ea+fc & eb+fd \\ ga+hc & gb+hd \end{pmatrix}\]

In the realm of $2\times2$ invertible matrices, for $AB = BA$ to be true, the center consists of matrices that seamlessly integrate with others, without altering their collective impact when multiplied.

$$ AB = BA $$

$$ \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} e & f \\ g & h \end{pmatrix} = \begin{pmatrix} e & f \\ g & h \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix} $$

This is true for non-zero scalar matrices, as multiplying by a scalar matrix preserves the overall impact of the original matrix, simply scaling its effect.

Hence, $A$ must be a diagonal matrix of the form $\begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix}$ where $a$ is not zero, ensuring the matrix is invertible.

$$ AB = BA $$

$$ \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix}\begin{pmatrix} e & f \\ g & h \end{pmatrix} = \begin{pmatrix} e & f \\ g & h \end{pmatrix}\begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix} $$

Here, $ a $ represents any nonzero real number.

In essence, the center of $GL(2, \mathbb{R})$ comprises all non-zero scalar matrices $aI$, where $I$ is the $2 \times 2$ identity matrix and $a$ is a nonzero real number.

Such matrices commute with every other matrix in $GL(2, \mathbb{R})$, as the act of scalar multiplication does not affect the order in which matrices are multiplied.

The Center of Any Group Invariably Forms a Subgroup

The center of a group $G$, denoted by $Z(G)$, invariably forms a subgroup of $G$ by naturally satisfying certain key properties.

A set must meet three criteria to be considered a subgroup: it must encompass the group's identity element, be closed under the group's operation, and include the inverse of each of its elements.

Let’s explore how the center of a group adheres to these criteria:

Incorporates the identity element
Intrinsically, the identity element $e$ of a group $G$ interacts with all elements of $G$ seamlessly, given $eg = ge = g$ for every $g \in G$. This establishes that $e$ is inherently part of the center $Z(G)$.
Exhibits closure under the group's operation
When we select any two elements $a$ and $b$ from $Z(G)$, it stands that for any $g \in G$, $ag = ga$ and $bg = gb$. The crux of the matter is to show that their combination $ab$ also commutes with every element $g$ in $G$, or in other words, $g(ab) = (ab)g$. Since $a$ and $b$ both commute with $g$, it follows $$ g(ab) = (ga)b = (ag)b = a(gb) = a(bg) = (ab)g $$ This elegantly demonstrates that $ab$ is a member of $Z(G)$, affirming that $Z(G)$ is closed under the group operation.
Encompasses the inverse of each element
Should $a$ reside in $Z(G)$, then $a$ naturally commutes with every $g$ in $G$, as $ag = ga$. The challenge lies in showing that the inverse of $a$, denoted as $a^{-1}$, also finds its place in $Z(G)$, implying $a^{-1}$ must commute with every $g \in G$. $$ ag = ga $$ Let's multiply both sides by $ a^{-1} $ on the left $$ a^{-1}ag = a^{-1}ga $$ $$ eg = a^{-1}ga $$ $$ g = a^{-1}ga $$ Now, let's multiply both sides by $ a^{-1} $ on the right $$ ga^{-1} = a^{-1}gaa^{-1} $$ $$ ga^{-1} = a^{-1}ge $$ $$ ga^{-1} = a^{-1}g $$ This logic flows smoothly to reveal that $ga^{-1} = a^{-1}g$, thereby including $a^{-1}$ within $Z(G)$.

These attributes firmly establish that the center of a group fulfills all necessary conditions to be acknowledged as a subgroup within that group.

Its definition intrinsically guarantees the inclusion of elements that exhibit commutative properties with every other element of the group, thus seamlessly qualifying it as a subgroup by the principles of group theory.