Quotient Group G/G = {identity}? I know this is a basic question, but I'm trying to convince myself of Wikipedia's statement. "The quotient group $G / G$ is isomorphic to the trivial group."
I write the definition for left multiplication because left cosets = right cosets. $ G/G = \{g \in G : gG\} $ But how is this isomorphic to the trivial group, $ \{id_G\} $? $gG$ can't be simplified to $id_G$ ?
Thank you.

The notation of quotient group such $G/N$ is beautiful, but often misleads us to forget that $G/N$ is no more than a group where its elements are cosets $gN$ defined as:

For $N \unlhd G$ and $g \in G$ then $gN = \{gn \: | \: n \in N\}$.

Identity of the group is $N$. For elements $aN, bN \in G/N$, where $a, b \in G$, are considered the same element $aN = bN$ if $ab^{-1} \in N$.
In the case of $N = G$, we have that the identity of the group is element $G$. For $a, b \in G$, we ALWAYS have that $ab^{-1} \in G$. That means no matter which elements we choose they are always the same element $G$. So this new group $G/G$ has only one element $G$ which is also its identity. Written explicitly $G/G = \{G\} \cong \{1\}$.
