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Cyclic Subgroups
Often a subgroup will depend entirely on a single element of the group; that is, knowing that particular element will allow us to compute any other element in the subgroup.
Cyclic Subgroups
Often a subgroup will depend entirely on a single element of the group; that is, knowing that particular element will allow us to compute any other element in the subgroup.
Example
Suppose that we consider \(3 \in {\mathbb Z}\) and look at all multiples (both positive and negative) of \(3\). As a set, this is \[\begin{aligned}\end{aligned}\]. It is easy to see that \(3 {\mathbb Z}\) is a subgroup of the integers. This subgroup is completely determined by the element \(3\) since we can obtain all of the other elements of the group by taking multiples of \(3\). Every element in the subgroup is generated by \(3\).
Example
If \(H = \{ 2^n : n \in {\mathbb Z} \}\), then \(H\) is a subgroup of the multiplicative group of nonzero rational numbers, \({\mathbb Q}^*\). If \(a = 2^m\) and \(b = 2^n\) are in \(H\), then \(ab^{-1} = 2^m 2^{-n} = 2^{m-n}\) is also in \(H\). By , \(H\) is a subgroup of \({\mathbb Q}^*\) determined by the element \(2\).
For \(a \in G\), we call \(\langle a \rangle\) the cyclic subgroup generated by \(a\). If \(G\) contains some element \(a\) such that \(G = \langle a \rangle\), then \(G\) is a cyclic group. In this case \(a\) is a generator of \(G\). If \(a\) is an element of a group \(G\), we define the order of \(a\) to be the smallest positive integer \(n\) such that \(a^n= e\), and we write \(|a| = n\). \(|a|\) the order of an element \(a\) If there is no such integer \(n\), we say that the order of \(a\) is infinite and write \(|a| = \infty\) to denote the order of \(a\).
Example
Notice that a cyclic group can have more than a single generator. Both \(1\) and \(5\) generate \({\mathbb Z}_6\); hence, \({\mathbb Z}_6\) is a cyclic group. Not every element in a cyclic group is necessarily a generator of the group. The order of \(2 \in {\mathbb Z}_6\) is \(3\). The cyclic subgroup generated by \(2\) is \(\langle 2 \rangle = \{ 0, 2, 4 \}\).
The groups \({\mathbb Z}\) and \({\mathbb Z}_n\) are cyclic groups. The elements \(1\) and \(-1\) are generators for \({\mathbb Z}\). We can certainly generate \({\mathbb Z}_n\) with \(1\) although there may be other generators of \({\mathbb Z}_n\), as in the case of \({\mathbb Z}_6\).
Example
The group of units, \(U(9)\), in \({\mathbb Z}_9\) is a cyclic group. As a set, \(U(9)\) is \(\{ 1, 2, 4, 5, 7, 8 \}\). The element \(2\) is a generator for \(U(9)\) since \[\begin{aligned}2^1 & = 2 \qquad 2^2 = 4 \\ 2^3 & = 8 \qquad 2^4 = 7 \\ 2^5 & = 5 \qquad 2^6 = 1\end{aligned}\].
Example
Not every group is a cyclic group. Consider the symmetry group of an equilateral triangle \(S_3\). The multiplication table for this group is . The subgroups of \(S_3\) are Notice that every subgroup is cyclic; however, no single element generates the entire group.
Subgroups of Cyclic Groups
We can ask some interesting questions about cyclic subgroups of a group and subgroups of a cyclic group. If \(G\) is a group, which subgroups of \(G\) are cyclic? If \(G\) is a cyclic group, what type of subgroups does \(G\) possess?
Condensed: the full section is in Judson, Abstract Algebra: Theory and Applications.
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What is a group, in plain words?
A set with one operation that is associative, has an identity, and lets every element be undone. Symmetries of any object form a group. That is where the idea came from.
What is the difference between a ring and a field?
A ring has addition and multiplication that behave like the integers (you cannot always divide); a field is a ring where every non-zero element has a reciprocal, like the rationals or the reals.
이 페이지의 일부는 다음에서 변경되었습니다. Judson, Abstract Algebra: Theory and Applications (GFDL 1.3). 여기서 압축하고 다시 설명; 오류는 우리의.
에 더 Abstract Algebra
GroupsSubgroups, cosets and Lagrange's theoremCyclic groups and permutation groupsHomomorphisms, normal subgroups and quotient groupsRings and fieldsGalois theory: why the quintic has no formula