Question

Let \( G \) be an abelian group and \( \Phi: G \to (\mathbb{Z},+) \) be a surjective group homomorphism. Let \( 1 = \Phi(a) \) for some \( a \in G \).
Consider the following statements:
\( P \): For every \( g \in G \), there exists an \( n \in \mathbb{Z} \) such that \( g a^n \in \ker(\Phi) \).
\( Q \): Let \( e \) be the identity of G and \( \langle a \rangle \) be the subgroup generated by \( a \). Then \( G = \ker(\Phi) \langle a \rangle \) and \( \ker(\Phi) \cap \langle a \rangle = \{e\} \).
Which of the following statements is/are correct?

• A. \( P \) is TRUE
• B. \( P \) is FALSE
• C. \( Q \) is TRUE
• D. \( Q \) is FALSE

Hint:

When you see a surjective homomorphism \( \Phi: G \to \mathbb{Z} \), think about the First Isomorphism Theorem, which states \( G/\ker(\Phi) \cong \mathbb{Z} \). This tells you that \( \ker(\Phi) \) is a normal subgroup and the cosets are indexed by integers. The element \( a \) with \( \Phi(a)=1 \) essentially picks out a representative for the coset corresponding to 1, which can then generate all other coset representatives.

Answer 1

The Correct Option is A, C

Step 1: Understanding the Concept:
This problem explores the structure of an abelian group \( G \) that has a surjective homomorphism onto the integers \( \mathbb{Z} \). It relates the group to the kernel of the homomorphism and the subgroup generated by a specific element \( a \) that maps to 1. This is an application of the First Isomorphism Theorem and properties of group structure, particularly related to direct products.
Step 2: Key Formula or Approach:
- A homomorphism \( \Phi \) satisfies \( \Phi(g_1 g_2) = \Phi(g_1) + \Phi(g_2) \) (operation in G is written multiplicatively, in \( \mathbb{Z} \) additively).
- The kernel is \( \ker(\Phi) = \{g \in G \mid \Phi(g) = 0\} \).
- To check statement P, we must find an integer \( n \) for any given \( g \).
- To check statement Q, we must verify two conditions for an internal direct product (since G is abelian):
1. \( G = K H \), where \( K=\ker(\Phi) \) and \( H=\langle a \rangle \). This means every \( g \in G \) can be written as \( g=kh \) for some \( k \in K, h \in H \).
2. \( K \cap H = \{e\} \).
Step 3: Detailed Explanation or Calculation:
Analysis of Statement P:
Let \( g \) be an arbitrary element of \( G \). Since \( \Phi \) is surjective, \( \Phi(g) \) is an integer in \( \mathbb{Z} \). Let's call this integer \( m \). \[ \Phi(g) = m \in \mathbb{Z} \] We are looking for an integer \( n \) such that \( g a^n \in \ker(\Phi) \). This means we need \( \Phi(g a^n) = 0 \).
Using the homomorphism property: \[ \Phi(g a^n) = \Phi(g) + \Phi(a^n) = \Phi(g) + n \Phi(a) \] We are given \( \Phi(g) = m \) and \( \Phi(a) = 1 \). So we need: \[ m + n . 1 = 0 \] This equation gives \( n = -m \). Since \( m \) is an integer, \( n = -m \) is also a well-defined integer. Thus, for any \( g \in G \), if we choose \( n = -\Phi(g) \), then \( g a^n \in \ker(\Phi) \). Therefore, statement P is TRUE.
Analysis of Statement Q:
This statement claims that \( G \) is the internal direct product of \( \ker(\Phi) \) and \( \langle a \rangle \). We check the two conditions.
Condition 1: \( G = \ker(\Phi) \langle a \rangle \)
From the analysis of P, we showed that for any \( g \in G \), there exists an \( n \in \mathbb{Z} \) such that \( g a^n \in \ker(\Phi) \). Let \( k = g a^n \). By definition, \( k \in \ker(\Phi) \). We can write \( g \) as \( g = k a^{-n} \). Here, \( k \in \ker(\Phi) \) and \( a^{-n} \in \langle a \rangle \) (since \( \langle a \rangle \) is a subgroup, it contains inverses and powers). So, any element \( g \in G \) can be written as a product of an element from \( \ker(\Phi) \) and an element from \( \langle a \rangle \). Thus, \( G = \ker(\Phi) \langle a \rangle \). The first condition holds.
Condition 2: \( \ker(\Phi) \cap \langle a \rangle = \{e\ \)}
Let \( x \) be an element in the intersection, \( x \in \ker(\Phi) \cap \langle a \rangle \). Since \( x \in \langle a \rangle \), we can write \( x = a^k \) for some integer \( k \in \mathbb{Z} \). Since \( x \in \ker(\Phi) \), we have \( \Phi(x) = 0 \). Substituting \( x = a^k \): \[ \Phi(a^k) = 0 \] Using the homomorphism property: \[ k \Phi(a) = 0 \] We are given \( \Phi(a) = 1 \). \[ k . 1 = 0 \implies k = 0 \] Therefore, \( x = a^0 = e \), where \( e \) is the identity element of \( G \). The only element in the intersection is the identity element. The second condition holds.
Since both conditions are satisfied, statement Q is TRUE.
Step 4: Final Answer:
Both statements P and Q are TRUE. Therefore, options (A) and (C) are correct.
Step 5: Why This is Correct:
The argument relies on the fundamental properties of group homomorphisms. Statement P is a direct consequence of the surjectivity of \( \Phi \) and the existence of an element \( a \) mapping to the generator of \( \mathbb{Z} \). Statement Q shows that the group \( G \) "splits" into the direct product of its kernel and the infinite cyclic group generated by \( a \). This is a result known as the Splitting Lemma for short exact sequences of abelian groups.