Let $G$ be a group and $\mathcal{A}=\{S_i\subseteq G \mid i\in I \}$ ($S_i$ is a no-empty subset of $G$.)
Conditions:
- $G=\bigcup_{i\in I}S_i$;
- For any $i,j\in I$, we can find $S_k,\,\,s.t.\;S_iS_j\subseteq S_k$;
- For any $i,j\in I$, $S_i\subsetneqq S_j$ is impossible.
Prove that $\mathcal{A}$ is a set of all cosets of some normal subgroup of $G$.
I have tried in this way:
Let $S_1$ contains $1_G$. Then I have proved $S_1$ is a subgroup of $G$. If we can prove that $S_i(i\neq 1)$ is exactly a right (resp. left) coset of $S_1$, then $S_1$ will be a normal subgroup. But I failed , so how can we prove that $S_i(i\neq 1)$ is exactly a right (resp. left) coset of $S_1$?
Consider $a\in G$; then $a\in S_i$ for some $i$. We know $S_iS_1\subseteq S_j$ for some $j$. Since $1\in S_1$, this means $S_i=S_i\cdot 1\subseteq S_iS_1\subseteq S_j$, so $S_i=S_j$ and we conclude $S_iS_1\subseteq S_i$. Since $a\in S_i$, this implies $aS_1\subseteq S_i$. From here $S_1\subseteq a^{-1}S_i$ as well.
Now, $a^{-1}\in S_k$ for some $k$, so $S_1\subseteq a^{-1}S_i\subseteq S_kS_i\subseteq S_l$ for some $l$, and again $S_1=S_l$. Therefore, $S_1\subseteq a^{-1}S_i\subseteq S_1$, so $aS_1\subseteq S_i\subseteq aS_1$, so $S_i=aS_1$. Similarly, $S_i=S_1a$. Thus $aS_1=S_1a$, so $S_1\lhd G$, and moreover every $S_i$ is a coset of $S_1$.