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Is there a $SL(2,\mathbb{Z})$-action on $\mathbb{Z}$?

I read this somewhere without proof and I am not sure if this is true.

Thank you for your help.

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Sure, the trivial action. You may want to add some qualifiers to your question. –  Najib Idrissi Nov 27 '12 at 21:06
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I am looking for a non-trivial action. –  berl13 Nov 27 '12 at 21:07
    
$\mathrm{SL}(2, \mathbb{Z}/2\mathbb{Z})$ is a quotient of the modular group, and its cardinal is 6. There aren't many groups of order 6, and in fact it's easily shown to be $\mathfrak{S}_3$, which can be mapped nontrivially to $\mathfrak{S}(\mathbb{Z})$. This gives you a nontrivial action of the modular group on $\mathbb{Z}$. –  Najib Idrissi Nov 27 '12 at 21:15
    
I was thinking that maybe you wanted an action with endomorphisms of $\mathbb{Z}$, but $\mathfrak{S}_3$ has $\mathbb{Z}/2\mathbb{Z}$ as a quotient, and this group acts nontrivially on $\mathbb{Z}$ by sending the generator on $-I$ (where $I$ is the identity), which is a group endomorphism. –  Najib Idrissi Nov 27 '12 at 21:24
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Im a looking for an action on $\mathbb{Z}$ as a group. –  berl13 Nov 28 '12 at 8:45

2 Answers 2

Let $SL(2,\mathbb Z)$ act on $\mathbb Z^2$ act in the obvious way, now chooose your favourite bijection between $\mathbb Z^2$ and $\mathbb Z$ and you are done.

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(This is completely different to my first 'answer', which was simply wrong.)

Denote by $\text{End}(\mathbb{Z})$ the semi-group of group endomorphisms of $\mathbb{Z}$. Since $\mathbb{Z}$ is cyclic, any endomorphism is determined by the image of the generator $1$, and since $1 \mapsto n$ is an endomorphism for any $n\in\mathbb{Z}$, this is all of them.

Since $SL(2,\mathbb{Z})$ is a group, all of its elements are invertible, so must map to invertible endomorphisms, i.e. automorphisms. Obviously these are given only by $n = \pm 1$ in the notation above. So $\text{Aut}(\mathbb{Z}) \cong \mathbb{Z}_2$. So the question becomes: is there a non-trivial homomorphism $\phi : SL(2,\mathbb{Z}) \to \mathbb{Z}_2$?

Since $\mathbb{Z}_2$ is Abelian, $\phi$ must factor through the Abelianisation of $SL(2,\mathbb{Z})$, which is$^*$ $\mathbb{Z}_{12}$. There is a unique surjective homomorphism $\mathbb{Z}_{12} \to \mathbb{Z}_2$, and therefore a unique surjective $\phi$, which gives a unique non-trivial action of $SL(2,\mathbb{Z})$ on $\mathbb{Z}$. Unfortunately, I can't see an easy way to decide whether a given $SL(2,\mathbb{Z})$ matrix maps to $1$ or $-1$, but maybe somebody else can.

$^*$A proof of this can be found at the link provided in the comments.

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See Theorem 3.8 on page 6 in Keith Conrad's blurb on $\mathrm{SL}_2(\mathbf{Z})$ for a proof of the footnote. –  commenter Nov 28 '12 at 15:27

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