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TheAbstractionist · #2 July 4th, 2009, 06:12 PM

$(1)\ \implies\ (2)$

Consider the map $\phi:G\times X\to X,\ \phi((g,x))=\psi(g)(x).$

$(2)\ \implies\ (1)$

For each $g\in G,$ define $\psi_g:X\to X$ by $\psi_g(x)=g\cdot x$ for all $x\in X.$ Then $\psi_g\in S(X)$ for all $g\in G$ $(\psi_g$ is a bijection from $X$ to itself) and the required homomorphism $\psi:G\to S(X)$ is defined by $\psi(g)=\psi_g$ for all $g\in G$ $(\psi_{hg}=\psi_h\psi_g$ for all $h,g\in G).$