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    Upon using that all knowing source wikipedia I have found that the normaliser is defined as

    "The normalizer is defined as N(S) = {x e G : x^(-1)Sx = S}." which goes on to say basically that S is a normal subgroup. Now if it is a normal subgroup the right coset must equal the left coset

    Now looking at the case of S_3 which has 6 elements e = (1)(2)(3), a_1 = (1)(2 3), a_2 = (1 3)(2), a_3 = (1 2)(3), a_4 = (1 2 3) and a_5 = (1 3 2)

    now finding the normalisers can't be done in this case right? cause say we do {e, a_1} a_1*a_2  \not = a_2*a_1 and it goes on! This is the case for all of the elements. I can only conclude that any of those elements is not a normaliser of the group, right?
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    So much wrong here...

    It does not say that S is a normal subgroup. So I don't know what you mean by "if it is a normal subgroup the right coset must equal the left coset".

    If you think e S e \neq S you're doing something *very* wrong.

    More generally, if we have x^{-1}Sx = S this does NOT mean that x^{-1}sx = s \forall s \in S which is what I *think* you think to be the case.
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    (Original post by DFranklin)
    So much wrong here...

    It does not say that S is a normal subgroup. So I don't know what you mean by "if it is a normal subgroup the right coset must equal the left coset".

    If you think e S e \neq S you're doing something *very* wrong.

    More generally, if we have x^{-1}Sx = S this does NOT mean that x^{-1}sx = s \forall s \in S which is what I *think* you think to be the case.

    hmmmm, you make a good point.......let me go over this again
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    (Original post by DFranklin)
    So much wrong here...

    It does not say that S is a normal subgroup. So I don't know what you mean by "if it is a normal subgroup the right coset must equal the left coset".

    If you think e S e \neq S you're doing something *very* wrong.

    More generally, if we have x^{-1}Sx = S this does NOT mean that x^{-1}sx = s \forall s \in S which is what I *think* you think to be the case.
    {e,a_1} is a subset of S_3 so call that S and let x e G so the aim here is to show that s e S, sx = xs so we do a_1 against the group of S3 which is (a_1*e, a_1*a_1, a_1*a_2, a_1*a_3, a_1*a_4, a_1*a_5) and show that this is equal to (e*a_1, a_1*a_1, a_2*a_1, a_3*a*1, a_4*a_1, a_5*a_1)

    Now my problem arises that it is not equal cause a_2*a_1 is not the same as a_1*a_2 same as a_1*a_3 is not the same as a_3*a_1 .....so xs=xs is not satisfied?
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    (Original post by boromir9111)
    {e,a_1} is a subset of S_3 so call that S and let x e G so the aim here is to show that s e S, sx = xs
    No it isn't. You want to show that xS = Sx. This is not the same thing.
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    Remember that xS and Sx are sets. Sx = xS is not shorthand for sx = xs for all s in S.
 
 
 
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