This question was asked in a recent exam and I am wondering if my response is correct:
Let $G$ be a group with identity $e$. Let $H$ be an abelian non-trivial subgroup with the property that $H \cap gHg^{-1}=\{ e \}$ for all $g \not \in H$. If $K=\{ g\in G\, \colon \, gh=hg \text{ for all } h\in H\}$, then which of the following is/are true?
- $K$ is a proper subgroup of $H$.
- H is a proper subgroup of $K$.
- $K=H$
- there exists no abelian subgroup $L\subseteq G$ such that $K$ is a proper subgroup of $L$.
My response to the question was option 3 and 4. Here's my attempt at proving this:
We claim that $H=K$. Let $h\in H$. Then $hh'=h'h$ for all $h' \in H$ because $H$ is abelian, hence, $h \in K$. Now, suppose $k \in K$. Assume the contrary that $k\not\in H$. By our assumption that $k\in K$, we obtain that $khk^{-1}=h$ for all $h\in H$, hence, $kHk^{-1}=H$. Furthermore, by hypothesis, $H\cap kHk^{-1}=H\cap H = H=\{e\}$. This is a contradiction to the fact that $H$ is non-trivial. This completes the proof of our claim.
Now, suppose that there exists an abelian subgroup $L \subseteq G$ such that $K$ is a proper subgroup of $L$. We have already shown that $H=K$. So , let $l \in L\setminus H$. Clearly, $lHl^{-1}=H$ because $L$ is an abelian subgroup containing $H$. This would force $H$ to be trivial, hence, a contradiction.
Is my proof and response to question correct? Thanks in advance.