Question

Let \( G \) be a group with identity \( e \). Let \( H \) be an abelian non-trivial proper subgroup of \( G \) with the property that \( H \cap gHg^{-1} = \{ e \} \) for all \( g \notin H \). If \( K = \{ g \in G : gh = hg \text{ for all } h \in H \}, \) then

• A. \( K \) is a proper subgroup of \( H \)
• B. \( H \) is a proper subgroup of \( K \)
• C. \( K = H \)
• D. There exists no abelian subgroup \( L \subseteq G \) such that \( K \) is a proper subgroup of \( L \)

Hint:

For an abelian subgroup \( H \), if \( H \cap gHg^{-1} = \{e\} \) for \( g \notin H \), then \( H \) must be equal to its own centralizer.

Answer 1

The Correct Option is C, D

Step 1: Interpret the given property.
Given \( H \cap gHg^{-1} = \{e\} \) for all \( g \notin H \), it means conjugates of \( H \) outside \( H \) intersect trivially with \( H \). This suggests \( H \) is abelian and self-centralizing.

Step 2: Define \( K \).
\( K = \{ g \in G : gh = hg \text{ for all } h \in H \} \) is the centralizer of \( H \) in \( G \). Since \( H \) is abelian, all its elements commute with each other, so \( H \subseteq K. \)

Step 3: Show equality.
If \( g \in K \setminus H \), then \( gHg^{-1} = H \) (since they commute elementwise), contradicting \( H \cap gHg^{-1} = \{ e \}. \) Thus, no such \( g \) exists, implying \( K = H. \)

Final Answer: \[ \boxed{K = H.} \]