9.2 Cyclic groups and cyclic subgroups
An important family of (sub)groups are those arising as collections of powers of a single element.
Given a (multiplicatively-written) group G and an element x\in G, we’ll define \langle x\rangle to be the subset \{x^i:i\in\mathbb{Z}\} of G consisting of all powers of x.
It is easy to check that:
The set \langle x\rangle is a subgroup of G.
We need to check the conditions of Theorem 9.3.
If x^i,x^j are powers of x, then x^ix^j=x^{i+j} is a power of x, so \langle x\rangle is closed.
We have e=x^0 is a power of x, so e\in\langle x\rangle.
If x^i\in\langle x\rangle, then (x^i)^{-1}=x^{-i}\in\langle x\rangle, so \langle x\rangle is closed under taking inverses.
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The following explicit description of \langle x\rangle follows immediately from Propositions 8.31 and 8.33.
If \operatorname{ord}(x)=\infty then \langle x\rangle is infinite with x^i\neq x^j unless i=j.
If \operatorname{ord}(x)=n then \langle x\rangle=\{e,x,\dots,x^{n-1}\} is finite, with n distinct elements.This example shows that there may be more than one possible choice of generator. However, not every element is a generator, since, for example, \langle 0\rangle=\{0\} and \langle 2\rangle = 2\mathbb{Z} \neq \mathbb{Z}.
Every cyclic group is abelian.
Suppose G=\langle x\rangle is cyclic with a generator x. Then if g,h\in G, g=x^i and h=x^j for some integers i and j. So gh=x^ix^j=x^{i+j}=x^jx^i=hg, and so G is abelian.
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Of course, this means that not every group is cyclic, since no non-abelian group is. But there are also abelian groups, even finite ones, that are not cyclic.
Let G be a finite group with |G|=n. Then G is cyclic if and only if it has an element of order n. An element x\in G is a generator if and only if \operatorname{ord}(x)=n.
Suppose x\in G. Then |\langle x\rangle|=\operatorname{ord}(x), and since \langle x\rangle\leq G, \langle x\rangle=G if and only if \operatorname{ord}(x)=|\langle x\rangle|=|G|=n.
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Every subgroup of a cyclic group is cyclic.
Let G=\langle x\rangle be a cyclic group with generator x, and let H\leq G be a subgroup.
If H=\{e\} is the trivial subgroup, then H=\langle e\rangle is cyclic. Otherwise, x^i\in H for some i\neq0, and since also x^{-i}\in H since H is closed under taking inverses, we can assume i>0.
Let m be the smallest positive integer such that x^m\in H. We shall show that H=\langle x^m\rangle, and so H is cyclic.
Certainly \langle x^m\rangle\subseteq H, since every power of x^m is in H. Suppose x^k\in H and write k=mq+r where q,r\in\mathbb{Z} and 0\leq r<m. Then x^r=x^{k-mq}=x^k(x^m)^q\in H, since x^k\in H and x^m\in H. But 0\leq r<m, and m is the smallest positive integer with x^m\in H, so r=0 or we have a contradiction. So x^k=(x^m)^q\in\langle x^m\rangle. Since x^k was an arbitrary element of H, H\subseteq\langle x^m\rangle.□