Let $H$ be a normal subgroup of a group $G=\{0,a,b,c,...\}$, where $0$ is the identity element in $G$. Then if $H=\{0\}$ then the quotient group $\frac{G}{H}$ is equal to $G$.
This has occurred in many contexts in my textbook.
Here's what I've done so far - In this case wouldn't it be more correct to say that the quotient group will be like $\frac{G}{H}=\{ \{0\},\{a\},\{b\},\{c\},....\}$ where $\{0\},\{a\},\{b\},\{c\},....$ are the singleton cosets of the subgroup $H$ in the group $G$ ?
Even if whatever they're writing is right, are they saying that $\frac{G}{H}=G$ because they are isomorphic and in that sense the equality is occurring? Am I getting it right?
This is true. The groups are, in fact quite trivially, isomorphic. Map a singleton coset to its representative.