H is an abelian proper subgroup of G with $H \cap gHg^{−1} = \{e\}$ for all $ g \notin H $ then $H$ is equal to centralizer of $H$ in $G$

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 $K = \{ g \in G : gh = hg \, \, \, \forall h \in H\}$. I want to prove the following. $$ \\ (1) H = K \\ (2) \text {there exists no abelian subgroup $L$ $\subset$ G such that $K$ is a proper subgroup of $L$ } $$. I tried the following . To prove part $(1)$ it is trivial to see that $ H \subset K $ as $H$ is an abelain group. Now we have to show that $ K \subset H $ for this let $ x \in K \implies xh = hx \, \, \forall h \in H \, \, so \, \, xHx^{-1} = H $. I got stuck here how to proceed further. To prove part $(2)$ i tried way of contradiction suppose there exist an abelian subgroup $L$ of $G$ such that K is a proper subgroup of $L$ . Then there exist $x \in K \text {such that} x \notin L $ so $x$ commute with all elements of $ H$ . Now how to proceed further. Thank you for help.

Answer 1

Clearly, $H\subseteq K$. Suppose $H\subsetneq K$. Then there is a $k\in K\setminus H$. Since $k\notin H$, we have $H\cap kHk^{-1}=\{e\}$. Now, choose $h(\neq e)\in H$(possible because $H$ is non-trivial). Then $khk^{-1}=h\in H\cap kHk^{-1}=\{e\},$ a contradiction.