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The thing I don't understand about the proof is that we are using the fact that $A_5$ contains all its three cycles. I can't see why this doesn't prove $A_4$ is simple. How can we use this argument when it should also work for $A_4$?

Also, why can we just break down $A_5$ into three types?

  1. $2^2$
  2. $1^2 3^1$
  3. $5^1$

Surely you have to include $4^1 1^1$ and $3^1 2^1$ too. However, in written notes it doesn't do this.

Also, does $A_4$ contain all three cycles? Maybe I'm incorrectly assuming this.

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I'm trying to understand this proof. It's written notes that I have scanned.

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    @GeoffRobinson Okay that cleared it up. Hmm. Yes I sort of understand your comment. I suppose it makes sense now.2012-01-23

1 Answers 1

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Every three-cycle is a product of two transpositions. So yes, $A_4$ also contains all three-cycles. But the crucial point is that every normal subgroup of $A_5$ that has more than one element also contains all three-cycles. Since the three-cycles generate $A_5$, every normal subgroup of $A_5$ with more than one element is all of $A_5$.

To show that every subgroup of $A_5$ with more than one element contains all three cycles, you go by cases: every non-identity element of $A_5$ is a 3-cycle, or a 5-cycle, or a product of two disjoint transpositions. By conjugation and multiplication you can cook up every three-cycle from such an element. But this needs space, i.e., if the set that your permutations act on has only 4 elements, you don't have the permutations that you want to conjugate with.

My answer is assuming you have a full proof of the simplicity of $A_5$ in front of you. If not, you will find one searching the internet.

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    Ok, look at the handwritten notes. What happens is that you get the required 3-cycles as $g^{-1}(edcba)g(abcde)$ and $g^{-1}(dc)(ba)g(ab)(cd)$. If $(abcde)$ is in the normal subgroup, then so is $g^{-1}(edcba)g(abcde)$ and similarly for $(ab)(cd)$.2016-04-25