# Trivial Quotient Group is Quotient Group

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## Theorem

Let $G$ be a group.

Then the trivial quotient group:

- $G / \set {e_G} \cong G$

where:

- $\cong$ denotes group isomorphism
- $e_G$ denotes the identity element of $G$

is a quotient group.

## Proof

From Trivial Subgroup is Normal:

- $\set {e_G} \lhd G$

Let $x \in G$.

Then:

- $x \set {e_G} = \set {x e_G} = \set x$

So each (left) coset of $G$ modulo $\set {e_G}$ has one element.

Now we set up the quotient epimorphism $\psi: G \to G / \set {e_G}$:

- $\forall x \in G: \map \phi x = x \set {e_G}$

which is of course a surjection.

We now need to establish that it is an injection.

Let $p, q \in G$.

\(\displaystyle \map \phi p\) | \(=\) | \(\displaystyle \map \phi q\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle p \set {e_G}\) | \(=\) | \(\displaystyle q \set {e_G}\) | Definition of $\phi$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \set p\) | \(=\) | \(\displaystyle \set q\) | from above | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle p\) | \(=\) | \(\displaystyle q\) | Definition of Set Equality |

So $\psi$ is a group isomorphism and therefore:

- $G / \set {e_G} \cong G$

$\blacksquare$

## Sources

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): $\S 7.4$. Kernel and image: Example $140$ - 1978: John S. Rose:
*A Course on Group Theory*... (previous) ... (next): $1$: Introduction to Finite Group Theory: $1.7$