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nope. i think an average 7th year can solve a simple problem is group theory,.

 

I would definately like to know what you call simple in that case.

 

Even the basic examples of groups such as symmetric groups and quternian groups would be hard to explain.

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I agree that the proofs to such results may appear to most to be quite trivial, but they certainly don't constitute any substantial part of group theory. The fact is you won't be able to prove something that you can't fully understand, and to make a school student understand and appreciate fully the beauty of group theory is far from an easy task.

 

As I said earlier, even the simplest examples are beyond the scope of school curiculum.

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i dont know. something like showing that a cyclic group with finite order is Abelian. or something like that.

 

Think about it: if you'd been given that problem at A-level, you'd probably have gone "wtf?!?" We've had a couple of terms to get used to the idea of university mathematics, and it's a completely different style to what you get at A-level.

 

It took me long enough to get my head around delta-epsilon arguments, and I'm still not sure I completely understand them.

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It took me long enough to get my head around delta-epsilon arguments, and I'm still not sure I completely understand them.
When you get into more rigorous proof classes (Real Analysis, Topology, etc.), [math]\epsilon-\delta[/math] arguments will become the easiest type of proof you will encounter. :)
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The theorem that Pulkit is referring to is often known as the Odd Order Theorem and is Feit, W. and Thompson, J. G. "Solvability of Groups of Odd Order." Pacific J. Math. 13, 775-1029, 1963

 

As its title says, the paper proves that all groups of odd order are soluble (the British version of solvable) and as an immediate consequence you get the fact that a finite simple non-cyclic group has even order.

 

Humorous proofs in maths are hard to come by but this impressive theorem is the source of one that, as students, we thought of many years ago.

 

Theorem

n! is even for all positive integers n

Proof

Suppose, for a contradiction, n! is odd.

It follows that the symmetric group [math]S_n[/math], which has order n!, is soluble by the Odd Order Theorem. But for n>4 this is well-known to be false, so n! must be even.

This leaves the case where n<5 which can be checked individually

 

:)

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I heard its a 250 page proof. So much for simplicity !

 

Theres a similar goliath of a proof in group theory for the theorem "all finite simple non-abelian groups are of even order".

 

Yeah but it is whole theory more general than just FMT. The FMT is just a consequence of whole the theory set up by the guy. There are actually people understanding all of this theory since they pointed out an error in the original reasoning and helped to repair it.

 

In any case it is not like on the first page you see proof : and then 250 pages further you see QED or something like that. It is an entire theory with many results.

 

Mandrake

 

If i remember correctly it was Andrew Wiles who proved FMT in 1994 and corrected his proof in 1995.

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The theorem that Pulkit is referring to is often known as the Odd Order Theorem and is Feit' date=' W. and Thompson, J. G. "Solvability of Groups of Odd Order." Pacific J. Math. 13, 775-1029, 1963

 

As its title says, the paper proves that all groups of odd order are soluble ([i']the British version of solvable[/i]) and as an immediate consequence you get the fact that a finite simple non-cyclic group has even order.

 

Humorous proofs in maths are hard to come by but this impressive theorem is the source of one that, as students, we thought of many years ago.

 

Theorem

n! is even for all positive integers n

Proof

Suppose, for a contradiction, n! is odd.

It follows that the symmetric group [math]S_n[/math], which has order n!, is soluble by the Odd Order Theorem. But for n>4 this is well-known to be false, so n! must be even.

This leaves the case where n<5 which can be checked individually

 

:)

 

 

n=1 anybody?

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