This is probably a super simple question...I just wanted to make sure that I am thinking about it correctly.
I am in the process of learning about quotient groups in Pinter's "A Book of Abstract Algebra". After introducing coset multiplication (which I will denote as $*_{coset}$) for the set $G/H$, where $H \triangleleft G$, Pinter wants to demonstrate that the set $G/H$, defined as $\{Ha, Hb, Hc, ...\}$, along with $*_{coset}$, form a group.
He goes through the usual routine to demonstrate that a set and an operation exhibit group structure (associativity, identity, inverse) but does not mention closure. I looked around and found a few other examples of people defining $\langle G/H, *_{coset} \rangle$ as a group...but they do not mention closure either.
Because $Ha *_{coset} Hb$, by definition, equals $H(a\circ_G b)$, I just wanted to make sure that the reason Pinter ignores this step is because closure is implied by the defintion of the operation...that is to say, $a \circ_G b \in G$ because $G$ is a group and $a,b \in G$. Therefore, $H(a\circ_G b) \in G/H$...and therefore closure is satisfied.
Is that the correct understanding? Just wanted to make sure that I was not missing something.
Yes, exactly. We know that $(ab)H \in G/H$ because $ab \in G$. The important part is showing that $aHbH = (ab)H$. This is not usually considered a definition, but rather something you need to prove.