Question

Let \(G\) be a finite abelian group of odd order. Consider the following two statements: I. The map \( f : G \to G \) defined by \( f(g) = g^2 \) is a group isomorphism.
II. The product \( \displaystyle \prod_{g \in G} g = e. \)
Which one of the following statements is true?

• A. Both I and II are TRUE.
• B. I is TRUE but II is FALSE.
• C. II is TRUE but I is FALSE.
• D. Neither I nor II is TRUE.

Hint:

In finite abelian groups of odd order, the squaring map is bijective, and all non-identity elements cancel in pairs under multiplication.

Answer 1

The Correct Option is A

Step 1: Prove Statement I.
Since \(|G|\) is odd, \(\gcd(2, |G|) = 1\). In an abelian group, the map \(g \mapsto g^2\) is a homomorphism. Because \(2\) has a multiplicative inverse modulo \(|G|\), the map is bijective — hence an automorphism.
Step 2: Prove Statement II.
In a finite abelian group of odd order, every element \(g\) pairs with its inverse \(g^{-1}\). Their product is \(e\), and no element is self-inverse except \(e\). Hence, the total product over all elements equals \(e\).
Step 3: Conclusion.
Both statements are TRUE.