Quotient of Group by Itself
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Theorem
Let $G$ be a group.
Let $G / G$ be the quotient group of $G$ by itself.
Then:
- $G / G \cong \set e$
That is, the quotient of a group by itself is isomorphic to the trivial group.
Proof
Let the homomorphism $\phi: G \to \set e$ be defined as:
- $\forall g \in G: \map \phi g = e$
Then:
- $\map \ker \phi = G$
and:
- $\Img \phi = \set e$
By the First Isomorphism Theorem:
- $G / \map \ker \phi \cong \Img \phi$
Hence the result:
- $G / G \cong \set e$
$\blacksquare$
Sources
- 1978: John S. Rose: A Course on Group Theory ... (previous) ... (next): $1$: Introduction to Finite Group Theory: $1.7$