Fundamental theorem of cyclic groups.html
 
| ca | de | en | es | fr | it | nl | no | pl | pt | ru | ro | fi | sv | tr | vo |


 

In abstract algebra, the fundamental theorem of cyclic groups states that if G\, is a cyclic group of order n\, then every subgroup 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\,.

Contents

Proof

Let G = \langle a\rangle\, 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 \rangle\,. If H \neq \{ 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}\rangle\,. It follows immediately from the closure property that \langle {a^m}\rangle \subseteq H.

To show that H \subseteq \langle {a^m}\rangle 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 \rangle. Thus H \subseteq \langle {a^m}\rangle.

Since H \subseteq \langle a^m\rangle and \langle {a^m}\rangle \subseteq H it follows that H = \langle a^m\rangle \, 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 \rangle\,, 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} \rangle\, is the one and only subgroup of \langle a\rangle\, of order k\,. Note that \langle a^{n/k} \rangle\, has order {n\over{gcd(n, {n \over {k}})}} = {n \over {n \over k}} = k\,. Let H\, be any subgroup of \langle a \rangle\, with order k\,. We know that H = \langle a^m \rangle\,, where m\, is a divisor of n\,. So m = \operatorname{gcd}(n, m)\, and k = | \langle a ^m \rangle | = | 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} \rangle\,, and thus the theorem is proved.

Alternate proof

Let G = \langle a \rangle be a cyclic group, and let H be a subgroup of G. Define a morphism \varphi: \mathbb{Z} \rightarrow 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 n\mathbb{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\,.

Generalization

Suppose that R is a right Ore domain in which every left ideal is principal, and let M be a left R-module which is generated by n elements. Then each submodule of M can also be generated by n elements (and possibly fewer). This result implies the fundamental theorem of cyclic groups by observing that the ring R=\mathbb{Z} of integers satisfies these conditions, and a cyclic group is precisely a left \mathbb{Z}-module which is generated by one element. (Its submodules are its subgroups.)

See also
Gry komputerowe • Warzywa • Zakupy online • Wikipedia • www.kurtka.pikawka.pl • disto d3 • długi szalik • znak.powiadaj.pl • Artykuly • Artykuly • Tenis • Wojny • Obory • Plock • Wzmacniacze All Right Reserved © 2007, Designed by Stylish Blog.