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[parent] proof of Cayley's theorem (Proof)

Let $ G$ be a group, and let $ S_G$ be the permutation group of the underlying set $ G$. For each $ g\in G$, define $ \rho_g : G\rightarrow G$ by $ \rho_g (h) = gh$. Then $ \rho_g$ is invertible with inverse $ \rho_{g^{-1}}$, and so is a permutation of the set $ G$.

Define $ \Phi :G\rightarrow S_G$ by $ \Phi (g) = \rho_g$. Then $ \Phi$ is a homomorphism, since

$\displaystyle (\Phi (gh))(x) = \rho_{gh}(x) = ghx = \rho_g(hx) = (\rho_g\circ\rho_h)(x) = ((\Phi (g))(\Phi (h)))(x)$

And $ \Phi$ is injective, since if $ \Phi (g) = \Phi (h)$ then $ \rho_g = \rho_h$, so $ gx = hx$ for all $ x\in X$, and so $ g=h$ as required.

So $ \Phi$ is an embedding of $ G$ into its own permutation group. If $ G$ is finite of order $ n$, then simply numbering the elements of $ G$ gives an embedding from $ G$ to $ S_n$.



"proof of Cayley's theorem" is owned by Evandar.
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Cross-references: order, finite, embedding, injective, homomorphism, permutation, inverse, invertible, permutation group, group

This is version 2 of proof of Cayley's theorem, born on 2002-03-03, modified 2003-05-29.
Object id is 2751, canonical name is ProofOfCayleysTheorem.
Accessed 4693 times total.

Classification:
AMS MSC: 20B99 (Group theory and generalizations :: Permutation groups :: Miscellaneous)
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