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 group by itself is isomorphic to the trivial group.

Proof
Let $G$ be a group.

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


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

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

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


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