Right Coset and the associated lemma from Herstein: Appreciating whats going on? I.N. Herstein in Page 34 (last line) and Page 35 of "Topics in Algebra" book goes on to explain a definition of right coset and a lemma like this:
Def: If $H$ is a subgroup of G, and $a \in G$, then $Ha = \left \{ha|h\in H  \right \}$;then $Ha$ is the right coset of $H$ in $G$
Lemma: FOr all $a \in G $  $Ha = \left \{x \in G |a \equiv x mod H  \right \}$
He goes on to define a set $[a]$ exactly like $Ha$ and trying to show $Ha \subseteq [a] $ 
My confusion:


*

*Whats going on here?

*More specifically, what the lemma trying to convey and why did the author go on to define $[a]$ exactly like $Ha$ and trying to show $Ha \subseteq [a] $ Isnt it trivial that every set is a subset of itself?

*If you have the proof of the lemma with you, can you help me understand it. I am not able to understand why exactly are we dealing with $a(ha)^{-1}$ which I understand as motivated from $a = ha mod H$
Thanks for your time and patience
Soham

The bracket notation is usually used in reference to equivalence relations. If $\sim$ is an equivalence relation on a set $G$, then $[a] = \{x \in G : x \sim a\}$. 
So if I had to guess what is going on here, I would say that he defines $Ha = \{ha : h \in H\}$. Then he would have defined an equivalence relation $\sim$ on $G$ by $a \sim b$ if and only if $ab^{-1} \in H$. 
Then it is easy (but not trivial) to show that $Ha \subset [a]$. 
