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This probably is a stupid question, but is there an infinite cyclic group generated by every single one of its nonidentity elements?

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If $a$ is a generator of this group, $a^2$ is not a generator since $a$ is not in the subgroup generated by $a^2$. –  Davide Giraudo May 28 '12 at 20:14
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Perhaps you mean every non-identity element? –  Alex Becker May 28 '12 at 20:14
    
@DavideGiraudo: that is not generally true, for example in $\mathbb{Z}/3\mathbb{Z}$. –  akkkk May 28 '12 at 20:16
    
@auke But (I think) it is when the group is infinite. –  Davide Giraudo May 28 '12 at 20:17
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If $a=a^{2k}$ for some integer $k$, then $a^{2k-1}=e=a^0$ (identity element). But we have $a^j\neq a^n$ if $j\neq n$ thanks to the infiniteness of the group. –  Davide Giraudo May 28 '12 at 20:20
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4 Answers 4

The infinite cyclic group is unique up to isomorphism. If $C$ is infinite cyclic and is generated by $x$ and by $y$, then $x=ky$ and $y=hx$ for some $k,h\in\mathbb Z$. But then $x=khx$, so $hk=1$ and either $h=k=1$ or $h=k=-1$. So either $x=y$ or $x=-y$.

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Up to isomorphism there is only one infinite cyclic group, the integers under addition; it has only two generators, $1$ and $-1$.

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No. The only infinite cyclic group up to isomorphism is $\mathbb{Z}$, which has only two generators.

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The only infinite cyclic group is $\mathbb{Z}$. So, no.

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