15. If F:G+ G Is A Homomorphism Of Groups, Then Prove That F = {a E Gf(a) = A} Is A Subgroup Of G

15. If F:G+ G Is A Homomorphism Of Groups, Then Prove That F = {a E Gf(a) = A} Is A Subgroup Of G

Understanding the structure of groups and group homomorphisms is fundamental in abstract algebra. The statement under consideration explores a crucial property: given a group homomorphism \(f: G \to G\), the set \(F = \{a \in G \mid f(a) = A\}\) forms a subgroup of \(G\). In this article, we will delve deep into the proof of this statement, exploring the necessary concepts, definitions, and logical steps involved.

Preliminaries and Definitions

Before diving into the proof, it's essential to understand some core concepts related to groups and homomorphisms.

What is a Group?

A group is a set \(G\) equipped with a binary operation \(\cdot\) satisfying the following properties:

• Closure: For all \(a, b \in G\), \(a \cdot b \in G\).


• Associativity: For all \(a, b, c \in G\), \((a \cdot b) \cdot c = a \cdot (b \cdot c)\).


• Identity element: There exists an element \(e \in G\) such that for all \(a \in G\), \(e \cdot a = a \cdot e = a\).


• Inverse element: For each \(a \in G\), there exists \(a^{-1} \in G\) such that \(a \cdot a^{-1} = a^{-1} \cdot a = e\).

What is a Group Homomorphism?

A homomorphism between two groups \(G\) and \(H\) is a function \(f: G \to H\) satisfying:

\[
f(a \cdot b) = f(a) \cdot' f(b) \quad \text{for all } a, b \in G,
\]

where \(\cdot\) and \(\cdot'\) are the group operations in \(G\) and \(H\), respectively. In the context of the problem, the homomorphism is from a group \(G\) to itself, i.e., \(f: G \to G\).

Key Properties of Group Homomorphisms

• Preservation of identity: \(f(eG) = eG\) or possibly other elements if \(f\) is not an isomorphism.
• Preservation of inverses: \(f(a^{-1}) = (f(a))^{-1}\).
• Kernel of \(f\): \(\ker f = \{a \in G \mid f(a) = e\}\), which is always a subgroup of \(G\).

In this context, the notation \(f(a) = A\) indicates that the homomorphism maps \(a\) to a specific element \(A\) in \(G\).

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Understanding the Set \(F = \{a \in G \mid f(a) = A\}\)

Given a homomorphism \(f: G \to G\), and a fixed element \(A \in G\), the set

\[
F = \{a \in G \mid f(a) = A\}
\]

is called a fiber or preimage of \(A\) under \(f\). The main goal is to demonstrate that \(F\) is a subgroup of \(G\).

Proving that \(F\) is a Subgroup of \(G\)

To establish that \(F\) is a subgroup, we need to verify the subgroup criteria:

1. Non-empty: \(F \neq \emptyset\)


2. Closure under the group operation: For any \(a, b \in F\), \(a \cdot b \in F\)


3. Closure under inverses: For any \(a \in F\), \(a^{-1} \in F\)

Let's analyze these in detail.

Step 1: Show that \(F\) is Non-Empty

Since \(f\) is a homomorphism, it maps the identity element \(e \in G\) to some element \(f(e)\).

• Observation: For any homomorphism \(f: G \to G\),

\[
f(e) = e,
\]

if \(f\) is a group homomorphism from \(G\) to itself (which is often the case).

• Implication: If \(A = f(e)\), then \(e \in F\), because

\[
f(e) = A,
\]

making the set \(F\) non-empty.

• General case: If \(A \neq f(e)\), then \(F\) could be empty unless \(A\) is in the image of \(f\). For the purposes of the proof, we consider that \(A \in \operatorname{Im}(f)\), ensuring \(F\) is non-empty.

Conclusion: There exists at least one element \(a \in G\) such that \(f(a) = A\), so \(F \neq \emptyset\).

Step 2: Closure Under the Group Operation

Suppose \(a, b \in F\). Then:

\[
f(a) = A, \quad f(b) = A.
\]

We need to verify whether \(a \cdot b \in F\), i.e., whether \(f(a \cdot b) = A\).

• Using the homomorphism property:

\[
f(a \cdot b) = f(a) \cdot f(b) = A \cdot A.
\]

• Analysis:

\[
f(a \cdot b) = A \cdot A,
\]

which equals \(A\) if and only if \(A \cdot A = A\).

Implication:

• For \(F\) to be closed under the group operation, the element \(A\) must satisfy:

\[
A \cdot A = A,
\]

which implies \(A\) is an idempotent element under the group operation.

• Note: In a group, the only idempotent element is the identity \(e\), because:

\[
A \cdot A = A \Rightarrow A \cdot A = A \Rightarrow A \cdot A = A \Rightarrow A \cdot A \cdot A^{-1} = A \cdot A^{-1} \Rightarrow A = e,
\]

since multiplying both sides on the right by \(A^{-1}\) yields:

\[
A \cdot A = A \Rightarrow A \cdot A \cdot A^{-1} = A \cdot A^{-1} \Rightarrow A = e.
\]

Therefore:

• If \(A = e\), then for any \(a, b \in F\),

\[
f(a) = e, \quad f(b) = e,
\]

and

\[
f(a \cdot b) = f(a) \cdot f(b) = e \cdot e = e,
\]

so \(a \cdot b \in F\).

• If \(A \neq e\), the set \(F\) contains elements mapped to \(A\), but the closure may not hold unless \(A\) is the identity.

Conclusion:

• The set \(F\) is closed under the group operation if and only if \(A = e\).

Step 3: Closure Under Inverses

Suppose \(a \in F\), so \(f(a) = A\). To show \(a^{-1} \in F\), we examine:

\[
f(a^{-1}) = (f(a))^{-1} = A^{-1}.
\]

• Implication:

\[
a^{-1} \in F \quad \text{if and only if} \quad f(a^{-1}) = A,
\]

which is equivalent to:

\[
A^{-1} = A.
\]

• Therefore, for \(a^{-1} \in F\), the element \(A\) must satisfy:

\[
A^{-1} = A,
\]

meaning \(A\) is its own inverse.

• In a group, elements satisfying \(A = A^{-1}\) are elements of order 1 or 2, i.e., elements of order dividing 2.

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Summary of Conditions for \(F\) to be a Subgroup

From the above analysis, the set

\[
F = \{a \in G \mid f(a) = A\}
\]

is a subgroup of \(G\) if and only if:

1. \(A = e\), the identity element, ensuring non-emptiness and closure under the group operation.


2. \(A = A^{-1}\), which is automatically true if \(A = e\).