Trivial Quotient Group is Quotient Group
Jump to navigation
Jump to search
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 $p, q \in G$.
\(\ds \map \phi p\) | \(=\) | \(\ds \map \phi q\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds p \set {e_G}\) | \(=\) | \(\ds q \set {e_G}\) | Definition of $\phi$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \set p\) | \(=\) | \(\ds \set q\) | from above | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds p\) | \(=\) | \(\ds q\) | 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: $1.7$