Download Finite multiplicative subgroups of fields

Finite multiplicative subgroups of a field
1 Elements of finite order in an Abelian group
We need two results about the order of elements in an Abelian group. The first is easy.
(1.1) Lemma If x is an element of finite order m in a group G and d is a divisor of m, then xd has
order m/d.
Proof: Its order is the smallest positive k such that (xd )k = 1. (xd )k = xkd , so its order is the smallest
positive k such that m|kd. For any positive integer k, k < m/d then kd < m, whereas if k = m/d
then kd = m, since d|m, so the order is m/d. Q.E.D.
Apology. In the Lemma below, because the group A is Abelian, we use additive notation. When
the Lemma is applied, it is to a finite multiplicative subgroup of a field, and multiplicative notation is
used. This could be confusing.
(1.2) Lemma Given an Abelian group A with elements a and b of finite orders m and n respectively,
!a, b" contains an element of order lcm(m, n).
Proof. Part I: the case where gcd(m, n) = 1.
Choose integers r and s so that rm + sn = 1. Remember that lcm(m, n) = mn.
Let c = sa + rb. RTP: c has order mn. Now, mc = m(sa + rb) = rmb = (1 − sn)b = b, since
ma = nb = 0. If kc = 0, then mkc = kb = 0, so n divides k.
By the same argument, if kc = 0 the m divides k. Therefore mn divides the order of c. Conversely, mnc = 0, so c has order mn. This proves part I, where gcd(m, n) = 1.
In the general case, gcd(m, n) = g ≥ 2. Let x = ga, m! = m/g. Then x has order m! , and
gcd(m! , n) = 1. By part I, !x, b" contains an element of order m! n. But lcm(m, n) = m! n (and
!x, b" ⊆ !a, b"). Q.E.D.
2 Finite multiplicative subgroups of a field are cyclic
Let G be a finite multiplicative subgroup of a field K. Let N = |G|, By Lagrange’s Theorem, the
order of every element of G divides N .
Note: Multiplication is commutative in a field, but we do not therefore use additive notation. If
G is a finite multiplicative subgroup of a field, and a ∈ G, then the order of a is the smallest positive
integer n such that an = 1, not na = 0.
We use Lemma 1.2, but with altered notation.
(2.1) Theorem G is cyclic.
Proof: Let
n = max{|y| : y ∈ G}
Choose a ∈ G so |a| = n. If we can show that n = N , then |a| = |G|, which implies !a" = G, so
G is cyclic. As noted, n|N so n ≤ N . So RTP: n ≥ N .
Claim thyt for every g in G, |g||n, because by Lemma 1.2, G contains an element of order
lcm(|a|, |g|), so lcm(n, |g|) ≤ n, so |g||n, as claimed.
Thus g n = 1 for every g in G. By the Factor Theorem, x − g is a factor of the polynomial xn − 1
for every g in G. Therefore,
!
(x − g) divides xn − 1
g∈G
The degree of the left-hand side is N , so N ≤ n. Q.E.D.