Isomorphism of Finite Group with Permutations of Quotient with Subgroup
Jump to navigation
Jump to search
Theorem
Let $G$ be a finite group.
Let $H$ be a subgroup of $G$.
Let $H$ contain no non-trivial normal subgroup of $G$.
Let $G / H$ denote the left coset space of $G$ by $H$.
Then $G$ is isomorphic to a subgroup of the group of permutations $\map \Gamma {G / H}$ of $G / H$.
Proof
Let a homomorphism $\phi: G \to \map \Gamma {G / H}$ be defined as:
- $\forall x \in G: \map {\map \phi g} {x H} = \paren {g x} H$
By Kernel is Normal Subgroup of Domain, $\ker \phi$ is a normal subgroup of $G$.
So:
\(\ds g\) | \(\in\) | \(\ds \ker \phi\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \forall x \in G: \, \) | \(\ds \paren {g x} H\) | \(=\) | \(\ds x H\) | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \forall x \in G: \, \) | \(\ds x^{-1} g x\) | \(\in\) | \(\ds H\) |
That is, $\ker \phi$ is the intersection of $H$ and all its conjugates.
Thus:
- $\ker \phi \subseteq H$
But $H$ contains no non-trivial normal subgroup of $G$.
Thus $\ker \phi$ is the trivial subgroup of $G$
Thus by Kernel is Trivial iff Group Monomorphism:
- $G$ is isomorphic to $\Img \phi$.
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Symmetric Groups: $\S 85$