In fact, you can show that for all n 1, An has no subgroup of...

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In fact, you can show that for all n ≥ 1, An has no subgroup of index 2. Assume to the contrary that there exists H< An with [An: H] = 2. An. Let o An be any 3-cycle. Consider oH E A₁/H. Try to show H. Now H contains all 3-cycles ... Then H that o In fact, you can show that for all n ≥ 1, An has no subgroup of index 2. Assume to the contrary that there exists H< An with [An: H] = 2. An. Let o An be any 3-cycle. Consider oH E A₁/H. Try to show H. Now H contains all 3-cycles ... Then H that o

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Here is the proof Lets assume for contradiction that there exists a subgroup H An of index 2 where ... View the full answer