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In a group $G$, if there is a member $x$ of $G$ s.t. $G=\{x^n, n \in \mathbb{N} \text{ or } \mathbb{Z}\}$, is there a name for such $x$? Thanks!

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Such an element is a generator. –  Andrew Nov 19 '12 at 4:19
    
Thanks! (1) is the exponent in $\mathbb{N}$ or $\mathbb{Z}$? (2) When does a generator exist for a group? –  Tim Nov 19 '12 at 4:20
    
Saying that a generator exists is equivalent to saying that the group is cyclic. See en.wikipedia.org/wiki/Generating_set_of_a_group. You want the exponents to live in $\Bbb Z$ so that we get inverses. Otherwise we are dealing with monoids, or some such. –  Andrew Nov 19 '12 at 4:21
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When it is isomorphic to one of $\mathbb{Z}$ or some $\mathbb{Z}_n$. But that's just a mild restatement. –  André Nicolas Nov 19 '12 at 4:23

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up vote 1 down vote accepted

As you have seen above, such an element is called a generator.

Note, however, that if $n \in \mathbb{N} = \{0, 1, 2, 3, \ldots\}$ then $G$ would have to be a finite group. (Why?)

For this reason, it is preferable to consider groups $G$ such that $G = \{x^n | n \in \mathbb{Z}\}$ for some $x \in G$.

In such a case, $x$ is called the generator and the group is called cyclic; however, using $\mathbb{Z}$ allows you to talk about both finite and infinite cyclic groups with the same notation.

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Thanks! Why "if n∈N={0,1,2,3,…} then G would have to be a finite group"? –  Tim Nov 19 '12 at 7:03
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@Tim, if so, then the inverse element of any element, $x$ say, would have to be a positive power of $x,$ so we must have $x^n\cdot x=x^{n+1}=1$ for some $n\in\Bbb N,$ implying the group has only $n+1$ elements. –  Andrew Nov 19 '12 at 22:59
    
@Andrew Right, as long as we pick the inverse $x^n$ where $n \in \mathbb{N}$ is minimal (otherwise we can't conclude the group has $n+1$ elements). –  Benjamin Dickman Nov 20 '12 at 9:09
    
Thanks, @Andrew! –  Tim Nov 20 '12 at 11:50
    
@B.D, yes good point, I should have mentioned that! –  Andrew Nov 20 '12 at 16:10

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