This question is a continuation from this question.
I.N.Herstein in Page 34 of his book "Topics in Algebra" defines a congruence relation like this:
DEFINITION: Let $G$ be a group , $H$ be a subgroup of $G$; for $a,b \in G$ we say $a$ is congruent to $b \mod H$ written as $ a \equiv b \mod H$ if $ab^{-1} \in H $
Further he goes on to prove this Lemma 2.6 in page 35
LEMMA : For all $a \in G $ $Ha = \left \{x \in G |a \equiv x \mod H \right \}$
PROOF: Let $[a] = \left \{x \in G |a \equiv x \mod H \right \}$. We first show that $Ha \subseteq [a]$. For if $h \in H$, then $a(ha)^{-1} = h^{-1}$ which is also in $H$. So by definition of congruence, this implies $ha \subseteq [a]$ for every $h \in H$ and so $Ha \subseteq [a]$.
My question:
To have $ab^{-1} \in H$ means $a,b \in G$ {since it is the necessity for congruence}. How does the proof go on to claim $Ha \subseteq [a]$ specifically? It should technically, claim $a,(ha) \in G$ which would imply $Ha \subseteq G$
Help much appreciated Soham