Problem 26
Question
Show that for any element \(g \in G\) not equal to the identity e we have \(|C(g)| \geq 2\).
Step-by-Step Solution
Verified Answer
For any non-identity element \(g\), the centralizer \(C(g)\) contains at least \(e\) and \(g\), so \(|C(g)| \geq 2\).
1Step 1: Understanding Centralizers
The centralizer of an element \(g\) in a group \(G\), denoted as \(C(g)\), is the set of all elements in \(G\) that commute with \(g\). Formally, it is defined as \(C(g) = \{x \in G \mid xg = gx\}\). Our task is to show that if \(g\) is not the identity element \(e\), then \(|C(g)| \geq 2\).
2Step 2: Properties of Group Elements
Since \(g\) is not the identity element \(e\), this implies there are elements in \(G\) which do not commute with every element, especially for \(g\) which is not equal to \(e\). However, \(e\) commutes with every element in \(G\), including \(g\), which means \(e \in C(g)\).
3Step 3: Checking More Elements in Centralizer
Let us consider \(g\) itself: \(gg = gg\) trivially holds as it commutes with itself. Hence, \(g \in C(g)\). This gives us two distinct elements in \(C(g)\), namely \(e\) and \(g\), when \(g eq e\).
4Step 4: Conclusion
Since we have identified that both \(e\) and \(g\) are elements of \(C(g)\) and they are distinct because \(g eq e\), we can conclude the centralizer \(C(g)\) contains at least these two elements. Thus, \(|C(g)| \geq 2\) when \(g eq e\).
Key Concepts
CentralizerGroup ElementsIdentity ElementCommutativity
Centralizer
The centralizer of an element in a group is a fundamental concept in group theory. It relates to how some elements "interact" or "fit in" with others in that group. Specifically, the centralizer of an element \( g \) in a group \( G \), denoted \( C(g) \), is the set of all elements in \( G \) that commute with \( g \). This means that for any element \( x \) in \( C(g) \), the operation of \( x \) with \( g \) results in the same as \( g \) with \( x \). Formally, we define:
- \( C(g) = \{ x \in G \mid xg = gx \} \)
Group Elements
Group elements are the building blocks of a group structure. Each element follows certain rules defined by the group’s operation, such as multiplication or addition. In the context of centralizers, we look at elements that exhibit specific interactions within the group. For example, the identity element \( e \) and the element \( g \) itself when considered together show us behavior regarding commutativity. Such observations are vital in proving properties like \(|C(g)| \geq 2\) when \( g eq e \), because:
- \( e \) commutes with every element in \( G \) including \( g \).
- The element \( g \) trivially commutes with itself, always making it part of \( C(g) \).
Identity Element
The identity element \( e \) holds a unique position in any group. It acts as a neutral player in group operations, meaning for any element \( g \) in the group \( G \), the equation \( eg = ge = g \) holds true. This property ensures \( e \) commutes with every other element in the group. Because of this characteristic, the identity element is always part of any centralizer \( C(g) \) for any \( g \) in \( G \). Let's consider:
- \( e \in C(g) \) since \( eg = ge \) is always true.
Commutativity
Commutativity is a key feature in understanding centralizers and generally in group theory. When we say two elements \( x \) and \( g \) commute, we express this mathematically as \( xg = gx \). In a group structure, not all elements will commute with each other, which is why centralizers are interesting. Groups where every element commutes with every other element is known as abelian groups.However, in a non-abelian group, the centralizer \( C(g) \) tells us which elements pair smoothly with a given \( g \). In the context of the exercise, when \( g eq e \), commutativity helps identify:
- \( e \), as it always commutes with every element, thus \( e \in C(g) \).
- \( g \), because any element commutes with itself, ensuring \( g \in C(g) \).
Other exercises in this chapter
Problem 24
By definition, \(a, b \in G\) are conjugate if there exists a \(g \in G\) such that \(b=\mathrm{gag}^{-1}\). Give an example to show that this \(g\) need not be
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Let \(G\) be a finite group, \(P\) a Sylow \(p\) -subgroup of \(G,\) and \(H\) a subgroup of \(G\) with \(P \leq H \leq G\). Show that if \(P\) is normal in \(H
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Show that a group of order 8 is isomorphic to one of the following groups: \(\mathbb{Z}_{8}, \mathbb{Z}_{4} \times \mathbb{Z}_{2}, \mathbb{Z}_{2} \times \mathbb
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Let \(G\) be a group of order 60 . Suppose that \(G\) contains a normal subgroup of order 2. Show that \(G\) has normal subgroups of orders \(6,10,\) and 30 .
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