Let $e$ be an idempotent in a semigroup $S$. The union of all subgroups of $S$ containing $e$ is called the maximal subgroup of $S$ containing $e$ and is denoted by $H(e)$. We know that $H(e)$ is actually a subgroup of $S$ with idedntity $e$. I want to prove: $$H(e)=\{t\in eSe: e\in St\cap tS\}$$
I think I should prove the right hand side is a group containing $H(e)$.
To show the RHS is a group, suppose that $t \in $RHS. Then $e \in St \cap tS$ implies that there exist $s_1,s_2 \in S$ with $s_1t =ts_2 = e$ and then $s_1e = s_1ts_2=es_2$.
Also $t \in eSe \Rightarrow te = et =t$, so $tes_2 = s_1et = e$ and hence $s_1e=es_2$ is a $2$-sided inverse to $t$.
For closure, clearly $t,u \in eSe \Rightarrow tu \in eSe$, and $tu$ has inverse $u^{-1}t^{-1}$.