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 October 17th, 2007, 08:01 AM #1 Newbie   Joined: Sep 2007 Posts: 10 Thanks: 0 bijection I have the question for a review problem For b in a group G, the function G→G sending each x to bx is one-to-one and onto (bijective). What i am thinking is that this is just a permutation, and permutations are always bijective, but i can't think of a way to prove it formally.
 October 17th, 2007, 08:51 AM #2 Site Founder     Joined: Nov 2006 From: France Posts: 824 Thanks: 7 Watch out, a permutation on a set S with n elements is any bijection f: S->S by definition of a permutation. Therefore you are right to say that x->bx is a permutation, but this is trivially equivalent to saying that it is bijective. Therefore you are back to the same problem, which I let you solve. Note that your remark is useful to show that your group G is isomorphic to a subgroup of Permutations(n) by the mapping b->(x->bx). Whence Cayley's theorem: every finite group is isomorphic to a subgroup of a permutation group.
 October 17th, 2007, 07:05 PM #3 Newbie   Joined: Sep 2007 Posts: 10 Thanks: 0 suppose we have bx1=bx2, then (b^-1)bx1=(b^-1)bx2 ((b^-1)bx)1=((b^-1)bx)2 ex1=ex2 x1=x2 Therefore the function is one-to-one Suppose one has a g in G, and one wants an x with bx=g and the solution is clearly x=b^-1g then one can check that b(b^-1g)=(bb^-10g=eg=g so indeed every g is in the image and we have that the function is a bijection.
 October 17th, 2007, 09:51 PM #4 Site Founder     Joined: Nov 2006 From: France Posts: 824 Thanks: 7 Yes, but you don't even need to check the surjectivity, because if a function is injective from a set of cardinal n into itself, then it is necessarily bijective.

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