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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 \(2\).

Theorem

Let \(G\) be a group and \(a\) be any element in \(G\). Then the set \(\langle a \rangle\) cyclic group generated by \(a\) \[\begin{aligned}\end{aligned}\] is a subgroup of \(G\). Furthermore, \(\langle a \rangle\) is the smallest subgroup of \(G\) that contains \(a\).

Proof

The identity is in \(\langle a \rangle\) since \(a^0 = e\). If \(g\) and \(h\) are any two elements in \(\langle a \rangle\), then by the definition of \(\langle a \rangle\) we can write \(g = a^m\) and \(h = a^n\) for some integers \(m\) and \(n\). So \(gh = a^m a^n = a^{m+n}\) is again in \(\langle a \rangle\). Finally, if \(g = a^n\) in \(\langle a \rangle\), then the inverse \(g^{-1} = a^{-n}\) is also in \(\langle a \rangle\). Clearly, any subgroup \(H\) of \(G\) containing \(a\) must contain all the powers of \(a\) by closure; hence, \(H\) contains \(\langle a \rangle\). Therefore, \(\langle a \rangle\) is the smallest subgroup of \(G\) containing \(a\).

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?

Theorem

Every subgroup of a cyclic group is cyclic.

Proof

The main tools used in this proof are the division algorithm and the Principle of Well-Ordering. Let \(G\) be a cyclic group generated by \(a\) and suppose that \(H\) is a subgroup of \(G\). If \(H = \{ e \}\), then trivially \(H\) is cyclic. Suppose that \(H\) contains some other element \(g\) distinct from the identity. Then \(g\) can be written as \(a^n\) for some integer \(n\). Since \(H\) is a subgroup, \(g^{-1} = a^{-n}\) must also be in \(H\). Since either \(n\) or \(-n\) is positive, we can assume that \(H\) contains positive powers of \(a\) and \(n \gt 0\). Let \(m\) be the smallest natural number such that \(a^m \in H\). Such an \(m\) exists by the Principle of Well-Ordering.

We claim that \(h = a^m\) is a generator for \(H\). We must show that every \(h' \in H\) can be written as a power of \(h\). Since \(h' \in H\) and \(H\) is a subgroup of \(G\), \(h' = a^k\) for some integer \(k\). Using the division algorithm, we can find numbers \(q\) and \(r\) such that \(k = mq +r\) where \(0 \leq r \lt m\); hence, \[\begin{aligned}\end{aligned}\]. So \(a^r = a^k h^{-q}\). Since \(a^k\) and \(h^{-q}\) are in \(H\), \(a^r\) must also be in \(H\). However, \(m\) was the smallest positive number such that \(a^m\) was in \(H\); consequently, \(r=0\) and so \(k=mq\). Therefore, \[\begin{aligned}\end{aligned}\] and \(H\) is generated by \(h\).

Condensed: the full section is in Judson, Abstract Algebra: Theory and Applications.

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Câu hỏi người ta hỏi

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.

Một số phần của trang này được chuyển từ Judson, Abstract Algebra: Theory and Applications (GFDL 1.3). Được nén lại và giải thích lại ở đây; lỗi là của chúng tôi.

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