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The multiplicative group of a finite field is cyclic

(This is an extremely useful fact with a straightforward proof. It follows from this, for instance, that any finite extension of a finite field is simple).

Let K be a finite field, so |K|=pn for some prime p and n0. Denote by K× the multiplicative group consisting of the non-zero elements of K. We show that K× is cyclic, i.e., generated by a single element.

Let d be the lowest common multiple (LCM) of the orders of all the elements aK×. Then d|K×|=pn1, and by Lagrange’s theorem, ad=a for all aK, i.e. (xa)xd+1xaK. Hence, pnd+1, since xd+1x can have at most d+1 roots. Thus pn1dpn1 i.e. d=pn1.

It remains to show that there exists an element whose order is the LCM of the orders of all group elements. Indeed, this is generally true for abelian groups, as shown by the following.

Proposition: Let G be an abelian group and let a,bG be elements with finite orders α,β (respectively). Then there exists cG of order LCM(α,β).

Proof: First, suppose α and β are co-prime and set c=ab. Then if ci=1, then ai=bi and so (b1)iα=aiα=(aα)i=1, so βiα which implies βi by co-primality. Similarly, αi. So αβi since α and β are co-prime, so c has order αβ.

Now consider the general case and write α=ipimi,β=ipini where the pi are distinct primes. For each i, define (Mi,Ni) by (Mi,Ni)={(mi,0)if mini(0,ni)otherwise and let α=ipiMi,β=ipiNi. Then αα and ββ, while α and β are co-prime and αβ=LCM(α,β). We have ord(aα/α)=αandord(bβ/β)=β. Thus by the coprime order case first considered, their product ab has order αβ=LCM(α,β).

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