The identity has to remain fixed. So, either you didn't mention that, or the automorphism does not exist.
Hi guys,
I am looking an automorphism (isomorphism onto itself) of S7 (symmetric group on 7 letters). The extra condition I have been given is that no element is allowed to remain fixed after applying said isomorphism (eg 1 should not be sent back to 1). To make this clearer, I am looking for a function, f, on 7 letters that:
1. Is a bijection
2. Satisfies f(xy)=f(x)f(y)
3. Leaves no element of S7 fixed
I have tried fiddling around with a lot of modular arithmetic and conjugacy classes etc, but it is hard to find such a function. Perhaps such a function does not exist?
Thanks in advance
i think you are confusing the concepts of set-automorphism (bijection from a set to itself) and group-automorphism (isomorphism from a group to itself).
an element of Aut(S_{7}) maps permutations to permutations.
only elements of S_{7} map numbers to numbers (or letters to letters..whatever we decide the call the elements of our 7-element set).
however, it is true that every automorphism of S_{7} is inner, so Aut(S_{7}) is isomorphic to S_{7}
(that is every element of Aut(S_{7}) is a map: τ→στσ^{-1}, for some σ in S_{7}.
if we call this map f_{σ}, then σ→f_{σ} is a group isomorphism).
and there DO exist elements of S_{7} that do not leave any letter fixed, for example: 7-cycles.
however, conjugation by a 7-cycle will still be the identity map on all powers of that 7-cycle:
σ(σ^{k})σ^{-1} = σ^{1+k-1} = σ^{k}.
perhaps this will convince you that every automorphism of S_{7} fixes some element of S_{7} (which is the same as saying every σ in S_{7} is conjugate to itself, which should not be surprising).