Welcome to NRICH.

 
Field Algebra and Simple Groups
By Edwin Koh on Sunday, July 01, 2001 - 03:21 pm:

Let x be an element of a field F of characteristic not equal to 2. Can we always find elements a and b in F such that a + b = 1 and a - b = x?

Suppose a (non-abelian) group G has only one proper normal subgroup N. Is it true that G/N is simple?

By James Lingard on Wednesday, October 10, 2001 - 12:09 am:

This is a very old question, but if you're still interested...

The answer to the first question is "yes". Let a = (1 + x)/2 and let b = (1 - x)/2. (You are allowed to do this because you know that the field is not of characteristic 2 and so 2 ¹ 0 in F.) Then it is easy to see that these values of a and b satisfy the equations you want.

In general, doing this kind of problem in an arbitrary field is very similar to doing it with the real numbers, except you have to be careful whenever you're dividing to make sure you're not dividing by something which is actually zero. Pretty much everything else works the same as you'd expect.

Now for the second question. The answer is also "yes". There may be an easier way of doing this, but here's how I went about it:

Suppose that H is a normal subgroup of G/N. So the elements of H look like Nx for some x in G. Now if H is normal in G/N then that means that

(Ny)-1(Nx)(Ny)

is an element of H for every element Ny of G/N and every element Nx of H. But

(Ny)-1(Nx)(Ny) = (Ny-1)(Nx)(Ny) = N(y-1xy)

Now define K to be the set of elements x of G such that Nx is an element of H. Then K is a subgroup of G which contains N (can you prove this?) and we have just shown that for every x in K and for every y in G, y-1xy is an element of K, which is precisely saying that K is a normal subgroup of G.

So either K = G, in which case H = G/N, or K = N, in which case H = 1. Therefore G/N is simple.

I hope that's clear - as always, please tell me if it isn't and I'll try to explain better.

James.