Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

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ADVANCED ABSTRACT ALGEBRA
a mu = a
a mu−1 = e
O(a) | (mu − 1)
n | (mu−1)
Hence, there exists an integer v such that
nv = mu−1
mu − nv = 1
gcd(m, n) = 1 .
This completes the proof of the theorem.
Theorem 11. Every subgroup H of a cyclic group G is cyclic.
Proof. If H = {e}, then H is obviously cyclic. So, let us suppose that H ≠ {e}. If a λ ∈ H , then a −λ ∈ H.
So, we can find a smallest positive integer m such that a m ∈ H. Therefore
< a m > ⊆ H (i)
Moreover,
Therefore
a λ ∈ H λ = qm, q ∈ Z
a λ = a qm
= (q m ) q ∈ < a m >
< a λ > ⊆ < a m >
It follows from (i) and (ii) that
H = < a m >
And hence H is cyclic.
H ⊆ < a m > (ii)
Theorem 12. Let G = < a > be a cyclic group of order n and H be a subgroup of G generated by a m , m
≤ n. Then
n
O(H) =
gcd(m,n)
Proof. We are given that
H = < a m >
Let gcd(m,n) = d, then we can find an integer q such that
m = qd
a m = a qd
But a qd ∈ < a d >, where < a d > is a subgroup generated by a d . Therefore
a m ∈ < a d >
H = < a m > ⊆ < a d > ….
Since gcd (m, n) = d, we can find u, v ∈ Z such that
d = un + vm
a d = a un+vm
(i)
= a un . a vm
= a vm (Θ a un = e)
But a vm ∈ < a m > = H . Therefore
a d ∈ H
< a d > ⊆ H …..
(ii)