Is this a sufficient flow of logic?


Consider a fifth degree polynomial that has a formulaic solution. Then we have a radical extension with a string of subgroups of the Galois group which each have a quotient that is abelian. Within the string, the subgroups must have all 3-cycles in $S_5$

Suppose there exists a subgroup with all 3 cycles. Since its quotient is abeleian, two 3-cycles and multiply by there inverses we can argue all 3-cycles are in the subgroup. However there are no cycles in the identity, so this is a contradiction. Thus we have no formula to solve this quintic polynomial.

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    $\begingroup$ It's a good start. But it's not that we have no formula, it's that we have no formula in radicals. Also, you don't need Abelian quotients, you just need a normal series with each having prime indexes in the next (that's how Galois thought about it). $\endgroup$ – Gregory Grant May 17 '15 at 6:18
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    $\begingroup$ A normal subgroup with prime index would almost immediately imply cyclic (thus abelian) quotient group, since the quotient group would have prime order? Not really that important though. Two ways of saying the same thing. $\endgroup$ – Kaj Hansen May 17 '15 at 6:36
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    $\begingroup$ You start out with the wrong assumption. Some quintics do allow solution by radicals. You need to make an assumption about the Galois group (and prove that there is a quintic with that Galois group). $\endgroup$ – Tobias Kildetoft May 17 '15 at 8:13
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    $\begingroup$ @KajHansen Yes it's equivalent, it's just that it requires much less machinery to state it in terms of prime indexes, Galois did not think in terms of quotient groups and commutativity like we do now, he had a much more down-to-earth perspective. Not to take away from his achievement. But Edwards' book on Galois Theory shows how he thought about it, he goes through Galois actual proof, and it takes a lot of the mystery out of how a 19 year old could have done that. $\endgroup$ – Gregory Grant May 17 '15 at 12:24

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