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Question: An infinite cyclic group has exactly two generators.

Answer: Suppose $G=\langle a\rangle$ is an infinite cyclic group. If $b=a^{n}\in G$ is a generator of $G$ then as $a\in G,\ a=b^{m}={(a^{n})}^{m}=a^{nm}$ for some $m\in Z$.

$\therefore$ We have $a^{nm-1}=e.$

(We know that the cyclic group $G=\langle a\rangle$ is infinite if and only if $0$ is the only integer for which $a^{0}=e$.) So, we have, $nm-1=0\Rightarrow nm=1.$ As $n$ and $m$ are integers, we have $n=1,n=-1.$

Now, $n=1$ gives $b=a$ which is already a generator and $n=-1$ gives $$H=\langle a^{-1}\rangle =\{(a^{-1})^{j}\mid j\in Z\} =\{a^{k}\mid k\in Z\}=G$$ That is $a^{-1}$ is also generator of $g$

My question is that am I approaching this question correctly?

  • 0
    In your conclusion, I assume you mean to say $a^{nm-1} = e$? Other than that, you are absolutely right, not much to say here.2012-04-20
  • 1
    Just a tiny formatting thing: in the fourth line you mean $a^{nm - 1}$ rather than $a^{nm} - 1$. I couldn't edit it because it was too small an edit.2012-04-20
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    snap =] $\hspace{1cm}$2012-04-20
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    You are writing your conclusions wrong. You don't want to conclude that $a^{-1}$ is also a generator (this is *easy*). You want to conclude that the **only** generators are $a$ and $a^{-1}$. So you want to show that **if** $a^n$ is a generator, **then** $n=1$ or $n=-1$. You are done once you get there. The rest is confused given what you are trying to show. Otherwise, the approach is fine.2012-04-20
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    @Arturo: Your comment answers the question, so shouldn't it be an answer rather than a comment?2012-04-21
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    @TaraB: Fair point.2012-04-21

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