Genady Posted January 29 Posted January 29 I.e., show that A4 has a normal subgroup. I did it by brutal force and would like to know if there is a more elegant way. My solution: A4 consists of 4!/2=12 permutations: e = identity, 8 permutations of the kind (1 2 3)=(1 2)(2 3), (1 3 2)=(1 3)(3 2), etc., and 3 permutations with separated cycles: a = (1 2)(3 4) b = (1 3)(2 4) c = (1 4)(2 3) Because of the separation, the cycles in a, b, and c commute, and thus a2 = b2 = c2 = e. I've checked manually that ab = ba = c, ac = ca = b, and bc = cb = a. So, {e, a, b, c} is an abelian group. An abelian subgroup is normal. Thus, A4 is not simple. Let me know if any of the above need elaboration.
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