Trivial Group is Cyclic Group

Theorem
The trivial group is a cyclic group and therefore abelian.

Proof
For a group $G = \left\{{e}\right\}$ to be a group, it follows that $e \circ e = e$.

Showing that $\left({G, \circ}\right)$ is in fact a group is straightforward:

G0: Closure
$G$ is closed:


 * $\forall e \in G: e \circ e = e$

G1: Associativity
$\circ $ is associative:


 * $e \circ \left({e \circ e}\right) = e = \left({e \circ e}\right) \circ e$

G2: Identity Element
$e$ is the identity:


 * $\forall e \in G: e \circ e = e$

G3: Inverse Elements
Every element of $G$ (all one of them) has an inverse:

This follows from the fact that the Identity is Self-Inverse, and the only element in $G$ is indeed the identity:


 * $e \circ e = e \implies e^{-1} = e$

Cyclic Group
In order for $G$ to be a cyclic group, every element $x$ of $G$ has to be expressible in the form $x = g^n$ for some $g \in G$ and some $n \in \Z$.

In this case, for every integer $n$, every element of $G$ can be expressed in the form $e^n$.

Thus $G$ is trivially a cyclic group.

Thus $G$ is abelian from Cyclic Group is Abelian.