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Show that the set of all permutations on G={a,b,c,d,e} is a group. I know that there are 5!=120 permutations here. I also know that a particular permutation "I" is the identity element of the set of these permutations. I'm not sure if I should go through each permutation and show it's inverse gives the identity element. I know that since function composition is associative and closed, then this set of permutations is as well.

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Hint: Permutations are set-theoretic bijections...(by definition.) – user21436 Apr 8 '12 at 15:20
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Since you admit the the identity permutation (the bijection $Id: x \mapsto x$) is a neutral element, the rest is easy. Given and element $f$ of the permutations set the inverse of $f$ is just the inverse permutation , that is if $f(x)=y$ then we define $f^{-1}(y)=x$ for every $x,y$ in the set. It is easy to check that $f\circ f^{-1}=f^{-1}\circ f=Id$ because if $f(x)=y$ then $f^{-1} (f(x))=f^{-1}(y)=x$ and $f(f^{-1}(y))=f(x)=y$ for all $x,y$ such that $f(x)=y$ and that the composition law is associative (i.e. $f\circ (g\circ h) =(f\circ g) \circ h$ for all permutations $f,g,h$ of the universe. Note that every two permutations are composable (and their composition is another permutation) since they have the same range and domain (the universe).

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