Question

Consider the group $G = \{+1, -1, +i, -i\}$ under multiplication operation. Which of the following is not correct?

• A. $G$ is an abelian group
• B. $G$ is a cyclic group
• C. $\{1, -1\}$ is a subgroup of $G$
• D. $\{1, -i, i\}$ is a subgroup of $G$

Hint:

Lagrange's Theorem is your best friend for group theory MCQs. If the size of the "subgroup" doesn't divide the size of the main group, it's an impostor!

Answer 1

The Correct Option is D

Concept:
• A subset \(H\) of a group \(G\) is a subgroup if it is a group under the same operation.
• Fundamental properties of subgroups:
• Closure: If \(a, b \in H\), then \(a \cdot b \in H\).
• Identity: The identity of \(G\) (which is 1 here) must be in \(H\).
• Inverses: If \(a \in H\), then \(a^{-1} \in H\).
• Lagrange's Theorem: The order of a subgroup must divide the order of the group.

Step 1: Analyze the group G
The group \(G = \{1, -1, i, -i\}\) has 4 elements (\(|G| = 4\)).
It is abelian (multiplication of complex numbers is commutative).
It is cyclic because \(i^1=i, i^2=-1, i^3=-i, i^4=1\). Thus, \(i\) is a generator.
Statements (A) and (B) are correct.

Step 2: Analyze Option (C)
\(H_1 = \{1, -1\}\).
Closure: \(1\cdot 1=1, 1\cdot (-1)=-1, (-1)\cdot (-1)=1\). All results are in \(H_1\).
Identity \(1 \in H_1\). Inverses exist (\(1^{-1}=1, (-1)^{-1}=-1\)).
Statement (C) is correct.

Step 3: Analyze Option (D)
\(H_2 = \{1, -i, i\}\).
This subset has 3 elements. By Lagrange's Theorem, the order of a subgroup must divide the order of the group.
3 does not divide 4. Therefore, it cannot be a subgroup.
Alternatively, check closure: \(i \cdot i = -1\). Since \(-1 \notin \{1, -i, i\}\), the closure property fails.