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October 17th, 2007, 09:01 AM   #1
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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.
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October 17th, 2007, 09:51 AM   #2
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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.
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October 17th, 2007, 08:05 PM   #3
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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.
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October 17th, 2007, 10:51 PM   #4
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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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