# group of invertible elements

• Nov 8th 2012, 02:31 AM
MarcoRo
group of invertible elements
Hello!

Could anybody help me with the following problem?

Let K be a finite field and denote the (multipl.) group of invertible elements of K with K^*. We set G:={r^2:r is in K^*}.
How can one prove that G=K^* if K has characteristic 2?

Marco
• Nov 8th 2012, 06:59 AM
Deveno
Re: group of invertible elements
hint: it suffices to prove that the map r→r2 is injective (why?)

since K is of characteristic 2, show r2 - s2 = (r - s)2.
• Nov 8th 2012, 07:30 AM
MarcoRo
Re: group of invertible elements
Hi Deveno,

thank you for your reply, I think that I understood the idea.

In the same situation, if the characteristic is >2, then G seems to be a subgroup of K* with index 2. Could you also give a hint how to see this?
• Nov 8th 2012, 07:58 AM
Deveno
Re: group of invertible elements
fix r in K*, if there is ONE solution to x2 - r in K[x], call it a, then we have TWO (since K is a field there is at most two roots to a polynomial of degree 2), since -a is likewise a (distinct) solution (since char(K) ≠ 2, a ≠ -a) and:

(x - a)(x + a) = x2 + (a - a)x + -a2 = x2 - r.

so the mapping a→a2 sends both a and -a to the same image.

now, let's show that this map is a homomorphism:

(ab)2 = abab = a(ba)b = a(ab)b (fields have an abelian multiplicative group)

= a2b2.

and the kernel of this homomorphsim is {-1,1} (why?)

the question now is: is this homomorphism surjective onto the subgroup G of squares of K*? (the squares of K* IS a subgroup because K* is finite, and we have proven closure:

a2b2 = (ab)2).

clearly (?) it is, so the subgroup G of squares of K* is isomorphic to K*/{1,-1} which has order |K*|/2, so:

[K*:G] = |K*|/|G| = |K*|/(|K*|/2) = 2.
• Nov 8th 2012, 08:24 AM
MarcoRo
Re: group of invertible elements
What exactly is the grouphomomorphism a->a^2 you are applying the group isomorphy theorem to?
It is a map from K* to G, right?
• Nov 8th 2012, 08:41 AM
Deveno
Re: group of invertible elements
yes.