Maybe it will help you to first look at a different quotient group, and then look at $G/G$.
Suppose that you have a subgroup $H$ of $G$ which is exactly half the size of $G$. The factor group $G/H$ is the set of cosets of H in G. What are these cosets?
If you multiply everything in $H$ by an element that is in $H$, you get back $H$. If you do the same with some fixed $g_0\in G$ such that $g_0\notin H$, the resulting coset $g_0H=\{g_0h : h \in H\}$ has no elements in common with $H$. (If it did have some $g_0h\in H$, by closure $g_0hh^{-1}=g_0\in H$, a contradiction.) Since $g_0H$ is the same size as $H$, which is half the size of $G$, $H$ and $g_0H$ together make up all the elements of $G$. This means that for any $g'\in G$ you pick, either $g'H=H$ or $g'H=g_0H$; in other words, these are the only two cosets.
So we have that $G/H=\{H, g_0H\}$. The operation for a factor group is $(aH)(bH)=(ab)H$, for any two cosets $aH,bH\in G/H$. So, for example, $(g_0H)(g_0H)=(g_0^2)H$. (For the coset "$H$" it's easiest to think of it as having an invisible $\text{id}_G$, so $(g_0H)(H)=(g_0\text{id}_G)H=g_0H$. Thus we have that $H=\text{id}_{G/H}$.) In particular, $G/H$ is a group with $2$ elements, so we know that it is isomorphic to $\mathbb{Z}_2$ because that is the only group of order $2$.
Now let's try doing it with $G/G$.
Well, $gG=G$ for any $g\in G$. So that is the only coset. Thus $G/G=\{G\}$, and $\text{id}_{G/G}=G$. It has only one element, so it must be isomorphic to the only group with one element, the trivial group.