GATE 2023 | Maths | Sample Ques for CS-IT

Question

Let \(G\) be an abelian group and \(\Phi: G \rightarrow (\mathbb{Z}, +)\) 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 \(g^n a \in \text{ker}(\Phi)\).

\(Q\): Let \(e\) be the identity of \(G\) and \(<a>\) be the subgroup generated by \(a\). Then \(G = \text{ker}(\Phi) <a>\) and \(\text{ker}(\Phi) \cap <a> = \{e\}\).

Which of the following statements is/are correct?
(A) \(P\) is TRUE
(B) \(P\) is FALSE
(C) \(Q\) is TRUE
(D) \(Q\) is FALSE

Answer 1

(A) and (C).

Statement P:

Since \(\Phi\) is surjective, for every \(g \in G\), there exists an \(n \in \mathbb{Z}\) such that \(\Phi(g) = n\).
\[\Phi(g^n a) = \Phi(g)^n \cdot \Phi(a) = n \cdot 1 = 0\]
(due to \(\Phi\) being a homomorphism and \(\Phi(a) = 1\)).
This means \(g^n a \in \text{ker}(\Phi)\). Hence, statement P is TRUE.

Statement Q:

By the First Isomorphism Theorem, \(G/\text{ker}(\Phi) \cong \Phi(G) = \mathbb{Z}\).
Since \(a\) is mapped to 1, \(a\) generates \(G/\text{ker}(\Phi)\).
This implies \(G = \text{ker}(\Phi) <a>\) (every element of \(G\) can be written as a product of an element in \(\text{ker}(\Phi)\) and a power of \(a\)).
If an element is in both \(\text{ker}(\Phi)\) and <a>, it must be mapped to both 0 and 1 under \(\Phi\), which is impossible.
Therefore, \(\text{ker}(\Phi) \cap <a> = \{e\}\). Hence, statement Q is TRUE.

Therefore, both statements P and Q are correct.