- Fundamental theorem of cyclic groups
In
abstract algebra , the fundamental theorem of cyclic groups states that if G, is acyclic group of order n, then everysubgroup of G, is cyclic. Moreover, the order of any subgroup of G, is a divisor of n, and for each positive divisor k, of n, the group G, has at most one subgroup of order k,.Proof
Let G = langle a angle, be a cyclic group for some a in G and with identity e, and order n,, and let H, be a subgroup of G,.We will now show that H, is cyclic. If H = { e }, then H = langle e angle,. If H eq { e }, then since G, is cyclic every element in H, is of the form a^t,, where t, is a positive integer. Let m, be the least positive integer such that a^m in H.
We will now show that H = langle{a^m} angle,. It follows immediately from the closure property that langle {a^m} angle subseteq H.
To show that H subseteq langle {a^m} angle we let b in H. Since b in G we have that b = a^k, for some positive integer k,. By the
division algorithm , k = mq + r, with 0 le r < m,, and so a^k = a^{mq + r} = a^{mq}a^r,, which yields a^r = a^{-mq}a^k,. Now since a^k in H and a^{-mq} = (a^{m})^{-q} in H, it follows from closure that a^r in H. But m, is the least integer such that a^m in H and 0 le r < m,, which means that r = 0, and so b = a^k = a^{mq} = (a^m)^q in langle a^m angle. Thus H subseteq langle {a^m} angle.Since H subseteq langle a^m angle and langle {a^m} angle subseteq H it follows that H = langle a^m angle , and so H, is cyclic.
We will now show that the order of any subgroup of G, is a divisor of n,. Let H, be any subgroup of G,. We have already shown that H = langle a^m angle,, where m is the least positive integer such that a^m in H. Since e = a^n = a^m, it follows that n = mq, for some integer q,. Thus m | n,.
We will now prove the last part of the theorem. Let k, be any positive divisor of n,. We will show that langle a^{n/k} angle, is the one and only subgroup of langle a angle, of order k,. Note that langle a^{n/k} angle, has order nover{gcd(n, {n over {k) = {n over {n over k = k,. Let H, be any subgroup of langle a angle, with order k,. We know that H = langle a^m angle,, where m, is a divisor of n,. So m = operatorname{gcd}(n, m), and k = | langle a ^m angle | = | a ^m | = |a^{operatorname{gcd}(n, m)}| = {n over {operatorname{gcd}(n, m) = {n over m},. Consequently m = {n over k}, and so H = langle a^{n over k} angle,, and thus the theorem is proved.
Alternate proof
Let G = langle a angle be a cyclic group, and let H be a subgroup of G. Define a morphism varphi: mathbb{Z} ightarrow G by varphi(n) = a^n. Since G is cyclic generated by a, varphi is surjective. Let K = varphi^{-1}(H) subseteq mathbb{Z}. K is a subgroup of mathbb{Z}. Since varphi is surjective, the restriction of varphi to K defines a surjective morphism from K onto H, and therefore H is isomorphic to a quotient of K. Since K is a subgroup of mathbb{Z}, K is nmathbb{Z} for some integer n. If n = 0, then K = {0}, hence H = {0}, which is cyclic. Otherwise, K is isomorphic to mathbb{Z}. Therefore H is isomorphic to a quotient of mathbb{Z}, and is necessarily cyclic.
Converse
The following statements are equivalent.
* A group G of order n, is cyclic.
* For every divisor d, of n, a group G has exactly one subgroup of order d,.
* For every divisor d, of n, a group G has at most one subgroup of order d,.ee also
*
Cyclic group
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