Isomorphism of Finite Group with Permutations of Quotient with Subgroup

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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 $\Img \phi$.

$\blacksquare$


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