Problem 23 - Algebra: Pure and Applied

Question

Let \(Z(G)\) be the center of a group \(G\). Show that (a) \(Z(G)

Step-by-Step Solution

Answer

(a) Generally, \(Z(G) < G\). (b) If \(G / Z(G)\) is cyclic, \(G\) is Abelian.

Step 1: Define the Center of a Group

The center of a group \(G\), denoted as \(Z(G)\), is the set of elements in \(G\) that commute with every other element of \(G\). Mathematically, \(Z(G) = \{ z \in G \mid zg = gz \text{ for all } g \in G \}\).

Step 2: Show That the Center is a Subgroup

To show that \(Z(G) < G\), we first note that \(Z(G)\) is non-empty because it always contains the identity element \(e\) of \(G\). To be a subgroup, \(Z(G)\) must be closed under the group operation and inverses, which it is, as the product and inverse of two commuting elements also commute with all of \(G\). Thus, \(Z(G)\) is a subgroup of \(G\).

Step 3: Prove the Proper Subgroup Case

In general, \(Z(G) eq G\) except when \(G\) is Abelian itself, meaning all elements commute (but this is outside the strict scope of (a), which implicitly assumes \(G\) is not entirely commutative). Therefore, typically \(Z(G) < G\).

Step 4: Assume \(G / Z(G)\) is Cyclic

Now assume that \(G / Z(G)\) is a cyclic group, meaning there exists an element \(gZ(G)\) such that every element of \(G/Z(G)\) is a power of \(gZ(G)\). This means any element \(aZ(G) \in G/Z(G)\) can be written as \((gZ(G))^n = g^nZ(G)\).

Step 5: Show G is Commutative Under the Given Assumption

For any elements \(x, y \in G\), since \(G/Z(G)\) is cyclic, \(xZ(G) = g^aZ(G)\) and \(yZ(G) = g^bZ(G)\) for some integers \(a\) and \(b\). Thus, \(xyZ(G) = (g^aZ(G))(g^bZ(G)) = g^{a+b}Z(G)\) and \(yxZ(G) = (g^bZ(G))(g^aZ(G)) = g^{b+a}Z(G)\). Since \(xyZ(G) = yxZ(G)\), and this holds for any cosets, elements of \(G\) commute, making \(G\) Abelian.

Key Concepts

Center of a Group

The center of a group, denoted as \(Z(G)\), is a fascinating concept in group theory. It includes all elements in a group \(G\) that commute with every other element of the group. This means for any element \(z\) in \(Z(G)\), and any element \(g\) in \(G\), the equation \(zg = gz\) holds true.

Think of \(Z(G)\) like a calm center that balances out the actions within the group. It ensures certain important symmetries.

• The center is always non-empty because it contains at least the identity element \(e\) of \(G\), as \(eg = ge\) for any element \(g\).
• \(Z(G)\) is closed under the group operation. If \(a\) and \(b\) are in \(Z(G)\), then so is their product \(ab\), as \((ab)g = a(bg) = a(gb) = (ag)b = (ga)b = g(ab)\).
• Inverses are also part of \(Z(G)\). If \(z\) is in \(Z(G)\), its inverse \(z^{-1}\) must also be in \(Z(G)\) because \(z^{-1}g = g z^{-1}\).

These properties ensure that \(Z(G)\) is a subgroup of \(G\) and typically, \(Z(G) eq G\) unless \(G\) is an Abelian group, where all elements naturally commute.

Cyclic Group

A cyclic group is a type of group that is generated by a single element. Imagine it as a circle where traveling around it a certain number of times gets you to every element in the group. If you have an element \(a\) in group \(G\), and you can create every element of \(G\) by simply using repeated operations on \(a\), then \(G\) is cyclic.

• Every finite cyclic group can be represented as \( \{e, a, a^2, \, ..., \, a^{n-1}\} \), where \(n\) is the order (number of elements) of the group, and \(a^n = e\).
• A significant property of cyclic groups is that they are always abelian. In other words, for any two elements \(a^m\) and \(a^n\), \(a^m a^n = a^{m+n} = a^n a^m\). This makes calculations very straightforward.

When \(G/Z(G)\) (the quotient group) is cyclic, it implies a high degree of symmetry in the set of cosets, leading us to conclude that the original group \(G\) is indeed abelian.

Abelian Group

An Abelian group is a group where the order in which you perform operations does not matter. Named after the mathematician Niels Henrik Abel, these groups are inherently commutative. If \(G\) is Abelian, for any elements \(a\) and \(b\) in \(G\), you have \(ab = ba\), meaning the product is the same whether you multiply \(a\) by \(b\) or \(b\) by \(a\).

• Abel’s insight helps tremendously in simplifying the analysis of groups, as many complex properties become much simpler when you do not have to worry about the order of operations.
• This property becomes particularly interesting in group theory because it serves as a hallmark of structure and predictability.
• An Abelian group is somewhat related to the real numbers and vectors in terms of addition – their sums do not depend on the order of the addends.

The real beauty of Abelian groups is seen when trying to solve problems like understanding the structure of \(G\) from \(G/Z(G)\). If \(G/Z(G)\) is cyclic, it simplifies the understanding not only of the quotient's behavior but also clarifies the commutative nature of the entire group \(G\).