7.10

1. What was the hardest part of the reading for you?
Honestly, and I am embarrassed, I struggled with the whole thing. I'm just not exactly sure what this chapter is saying. It's obvious that it's all aimed at proving theorem 7.52, but I'm not exactly sure what that theorem is saying (because I can read) but I really struggled to follow the proofs of the lemmas and the meaning behind the theorem.

2. Figure out what Theorem 7.52 is saying
Okay, so I'm going to break this down and looks stuff up. The theorem states "For each n =/ 4, the alternating group An is a simple group."
Alright, I know the chapter addresses this, but I'm still not sure about n not equaling 4... but let's move on from there since it's just the assumption. Well, let's review "Alternating Groups". Chapter 7.9 says that an alternating group is the set of all even permutations (even permutations can be written as the product of an even number of transpositions [2-cycles]) in Sn. The set of these permutations is written An... And a simple group is a group whose only normal subgroups (or subgroups that have left and right cosets equal to each other) are <e> and itself (G). So, what the theorem is saying, as far as I understand it, is if the set of all permutations (not with 4 elements... or unique mappings containing the numbers 1-4, since A4 really has 24 elements...) that has permutations that are even, or can be written as an even number of 2-cycles, is a subgroup that only has the normal subgroups of <e> and itself (the alternating group of elements of Sn that has all even permutations.)