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text: 'Just as we can [currying curry] functions, so we can "curry" [47t homomorphisms] and [3t9 actions].\n\nGiven an action $\\rho: G \\times X \\to X$ of group $G$ on set $X$, we can consider what happens if we fix the first argument to $\\rho$. Writing $\\rho(g)$ for the induced map $X \\to X$ given by $x \\mapsto \\rho(g, x)$, we can see that $\\rho(g)$ is a [499 bijection].\n\nIndeed, we claim that $\\rho(g^{-1})$ is an inverse map to $\\rho(g)$.\nConsider $\\rho(g^{-1})(\\rho(g)(x))$.\nThis is precisely $\\rho(g^{-1})(\\rho(g, x))$, which is precisely $\\rho(g^{-1}, \\rho(g, x))$.\nBy the definition of an action, this is just $\\rho(g^{-1} g, x) = \\rho(e, x) = x$, where $e$ is the group's identity.\n\nWe omit the proof that $\\rho(g)(\\rho(g^{-1})(x)) = x$, because it is nearly identical.\n\nThat is, we have proved that $\\rho(g)$ is in $\\mathrm{Sym}(X)$, where $\\mathrm{Sym}$ is the [-497]; equivalently, we can view $\\rho$ as mapping elements of $G$ into $\\mathrm{Sym}(X)$, as well as our original definition of mapping elements of $G \\times X$ into $X$.\n\n# $\\rho$ is a homomorphism in this new sense\n\nIt turns out that $\\rho: G \\to \\mathrm{Sym}(X)$ is a homomorphism.\nIt suffices to show that $\\rho(gh) = \\rho(g) \\rho(h)$, where recall that the operation in $\\mathrm{Sym}(X)$ is composition of permutations.\n\nBut this is true: $\\rho(gh)(x) = \\rho(gh, x)$ by definition of $\\rho(gh)$; that is $\\rho(g, \\rho(h, x))$ because $\\rho$ is a group action; that is $\\rho(g)(\\rho(h, x))$ by definition of $\\rho(g)$; and that is $\\rho(g)(\\rho(h)(x))$ by definition of $\\rho(h)$ as required.',
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