$G/H$ is called a quotient group as others have said. Here's some intuitive motivation; I'll leave the rigorous details to textbooks.
A good motivating example is $\mathbb{Z}/5\mathbb{Z}$, the integers mod $5$ under addition. Think of it as the integers with the extra relation $5=0$, so there are just the numbers $\{0, 1, 2, 3, 4\}$ with eg. $3+4=2$. Setting $5=0$ forced us to set $5n=0$ for all $n \in \mathbb{Z}$. Indeed, we're setting precisely the elements of $\langle 5 \rangle = 5 \mathbb{Z} \subset \mathbb{Z}$ equal to zero and then fixing up the addition table to be consistent with this requirement.
More generally, given a group $G$ and an element $h \in G$, setting $h=1$ (I'll use multiplicative notation now) forces us to set $h^n = 1$ for $n \in \mathbb{Z}$, so all of $\langle h \rangle$ is $1$. Indeed, for any $g \in G$, we also need $ghg^{-1} = gg^{-1} = 1$, so we're typically setting "more" than $\langle h \rangle$ to $1$. If $G$ is abelian, as in the previous example, $ghg^{-1} = gg^{-1}h = h$, so we can get away with just setting all of $\langle h \rangle$ to $1$.
In general, setting some elements of $G$ to $1$ has the effect of setting an entire "normal subgroup" $H \subset G$ to $1$; this is a subgroup such that $ghg^{-1} \in H$ for all $g \in G$. Any normal subgroup can be set to $1$ without setting anything not in that subgroup to $1$. So, when considering which groups arise as quotients of some initial group, we just look for normal subgroups. Note that $\{1\}$ and $G$ are trivially normal subgroups of $G$, corresponding to the fact that we can always set only $1$ to $1$ (i.e. do nothing) or we can set everything to $1$ (resulting in the one-element group nobody wants).
A very important class of groups are those for which the only way to set some elements to $1$ is trivially, i.e. whose normal subgroups are just the one-element group and the entire group. Such groups are called "simple". The finite simple groups have been classified (most likely; it's extremely long).
