Quotient of Group by Itself

Theorem
Let $G$ be a group.

Let $G / G$ be the quotient group of $G$ by itself.

Then $G / G \cong \{e\}$.

That is, the quotient of a group by itself is isomorphic to the trivial group.

Proof
Let the homomorphism $\phi:G \to \{e\}$ be defined as:


 * $\forall g \in G: \phi(g) = e$

Then $\ker(\phi) = G$ and $\operatorname{Im}(\phi) = \{e\}$.

By the First Isomorphism Theorem, $G / \ker(\phi) \cong \operatorname{Im}(\phi)$:


 * $G / G \cong \{e\}$