Determine the normalizer in the symmetric group S4 of the subgroup H of all perm

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--well H is naturally isomorphic to S3, for example if h is: 1-->a 2-->b 3-->c 4-->4 (where {a,b,c} is {1,2,3} in perhaps a different order) then we can consider h' to be the element of S3: 1-->a 2-->b 3-->c you should verify that the map h-->h' is an isomorphism of H with S3. now certainly H is contained in N(H) (because H is normal in itself), so |N(H)| can only be 6,12 or 24. |N(H)| = 24, implies H is normal in S4, but: (1 4)(1 2 3)(1 4) = (2 3 4), which does not leave 4 fixed, so H is not normal. |N(H)| = 12, implies that N(H) = A4, since that is the ONLY subgroup of S4 of order 12. but this means that H must consist only of even permutations, which isn't true, because the transposition (1 2) is in H. therefore: N(H) = H is the only possibility left. ******** it might not be obvious that A4 is, in fact, the only subgroup of order 12 of S4. but note that any normal subgroup of S4 must contain either ALL of a cycle type, or NONE of a cycle type (since all permutations of the same cycle type are conjugate). let's look at the cycle types of A4: (1) - the identity (we will regard this as "the only" 1-cycle). (a b) - there are 6 transpositions (a b c) - there are 8 3-cycles (a b)(c d) - there are 3 double 2-cycles (a b c d) - there are 6 4-cycles so we need to solve: 12 = 1 + 6k + 8m + 3n + 6p where k,m,n,p are either 0 or 1. since 12 is even, and 1 is odd, n must be 1, or else we end up with an odd number. so we have: 12 = 1 + 3 + 6(k+p) + 8m = 4 + 6(k+p) + 8m that is: 8 = 6(k+p) + 8m. if m = 0, we have no solution, since 8 is not a multiple of 6. so we have: 8 = 6(k+p) + 8, or: 0 = 6(k+p), so that k = p = 0. thus 12 = 1 + 3 + 8 is the only combination that works. but this is: the identity, the 3 double 2-cycles, and the 8 3-cycles, that is: A4.

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