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: Show $H \subseteq K$

Since $H$ is abelian, for all $h_1, h_2 \in H$: $$h_1 h_2 = h_2 h_1$$

This means every element of $H$ commutes with all elements of $H$, so $H \subseteq K$.

Step 2: Show $K \subseteq H$

Let $g \in K$. Then $gh = hg$ for all $h \in H$.

This means $ghg^{-1} = h$ for all $h \in H$, which implies $h \in gHg^{-1}$.

Therefore, $h \in H \cap gHg^{-1}$.

Case 1: If $g \in H$, then clearly $g \in K$ is consistent.

Case 2: If $g \notin H$, then by the given property: $$H \cap gHg^{-1} = {e}$$

But we just showed that all $h \in H$ satisfy $h \in H \cap gHg^{-1}$.

Since $H$ is non-trivial (contains elements other than $e$), this would contradict $H \cap gHg^{-1} = {e}$.

Therefore, $g \notin H$ is impossible for $g \in K$.

Thus, $K \subseteq H$.

Step 3: Conclude $K = H$

From Steps 1 and 2: $K = H$

(C) is TRUE

Step 4: Check if there exists abelian $L \subseteq G$ with $K \subsetneq L$

Suppose such an abelian subgroup $L$ exists with $K \subsetneq L$.

Then there exists $\ell \in L$ with $\ell \notin K = H$.

Since $\ell \in L$ and $H \subseteq K \subseteq L$, and $L$ is abelian: $$\ell h = h \ell \text{ for all } h \in H$$

This means $\ell \in K$, which contradicts $\ell \notin K$.

Therefore, no such abelian subgroup $L$ exists.

(D) is TRUE

Evaluating other options:

(A) $K$ is a proper subgroup of $H$:

Since $K = H$, this is FALSE

(B) $H$ is a proper subgroup of $K$:

Since $K = H$, this is FALSE

Answer: (C) and (D) are true