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