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Let $r(m)$ denote the residue class $r+m\mathbb{Z}$, where $0 \leq r < m$. Given disjoint residue classes $r_1(m_1)$ and $r_2(m_2)$, let the class transposition $\tau_{r_1(m_1),r_2(m_2)}$ be the permutation of $\mathbb{Z}$ which interchanges $r_1+km_1$ and $r_2+km_2$ for every $k \in \mathbb{Z}$ and which fixes everything else.

The group ${\rm CT}(\mathbb{Z})$ generated by all class transpositions of $\mathbb{Z}$ is simple (cf. Math. Z. 264 (2010), no. 4, 927-938, http://dx.doi.org/10.1007/s00209-009-0497-8).

Find the outer automorphism group of ${\rm CT}(\mathbb{Z})$.

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The description should be split into two parts: first, determine whether every automorphism is spatial, i.e., is the automorphism group equal to the normalizer in the group of permutations of $\mathbf{Z}$. Then it remains to determine this normalizer. Note that since you restrict to $m>0$, you at least have $n\mapsto -n$ in this normalizer, which is a non-inner automorphism. But it might be the only one, I don't know. –  YCor Nov 14 '12 at 18:33
    
No, $n \mapsto -n$ is not an automorphism. The group ${\rm CT}(\mathbb{Z})$ stabilizes $\mathbb{N}_0$ setwise, but for example the conjugate of its element $\tau_{0(2),1(2)}$ under the mapping $n \mapsto -n$ doesn't (it moves 0 to -1). –  Stefan Kohl Nov 14 '12 at 20:00
    
But a little modification turns it into one: $n \mapsto -n-1$ is indeed an automorphism. –  Stefan Kohl Nov 14 '12 at 20:26
    
OK thanks. I hadn't noticed that the requirement $r\ge 0$ indeed forces $\mathbf{N}_0$ (why does the English convention exclude 0 from $\mathbf{N}$?) to be stable. So in my previous comment, the definition of <i>spatial</i> should rather be the normalizer in the group of permutations of $\mathbf{N}_0$. –  YCor Nov 15 '12 at 9:48
    
Thanks. - If the answer to Question <mathoverflow.net/questions/112469/>; is negative, this would yield further spatial outer automorphisms –  Stefan Kohl Nov 15 '12 at 10:34

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