It is easy to see that the totally ordered group $\mathbb{Z}$ (the integers) with the natural order has no non-trivial automorphisms. Is this also true for $\mathbb{Z}^n$ with the lexicographical order?
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2 Answers
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Here's a counterexample: on $\mathbb{Z}^2$, $f(x,y)=(x,y+x)$.
More generally, the order-preserving automorphisms of $\mathbb{Z}^n$ are exactly the upper triangular matrices with 1s on the diagonal (this should be easy to see by combining Charles's argument with my example in the case $n=2$, and then the generalization to arbitrary $n$ isn't too hard).
answered Nov 2, 2009 at 14:41
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$\begingroup$ Thanks, awesome! After Charles' answer I first thought that I was too dumb for the induction step. $\endgroup$– user717Commented Nov 2, 2009 at 14:45
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$\begingroup$ I also want to point out that the lexicographic order on $\mathbb{Z}^n$ has no minimal positive element, that is precisely the reason that the argument given by Eric holds. $\endgroup$– StinesCommented Oct 21, 2010 at 21:02
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Yes. Any group homomorphism fixes zero. Now, look at the set of things greater than zero. If the homomorphism is order preserving, then it must take the least thing there to the least thing in the image. However, as the image is surjective, it must then be fixed. Then, inductively, everything larger than zero is fixed. A similar argument works for things less than zero. In fact, this proof appears to merely require a totally ordered group, and not the lexicographic ordering, so this should work for any monomial ordering (that is, total order on Z^n.
answered Nov 2, 2009 at 14:32
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1$\begingroup$ This argument breaks down because the positive part of Z^n is not well-ordered. The argument shows that (0,0), (0,1), (0,2), etc must be fixed. But there's no next smallest element that has to be fixed after that. $\endgroup$ Commented Nov 2, 2009 at 14:40
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$\begingroup$ Ack, apologies. This is what I get for trying to do induction on no sleep, early in the morning. $\endgroup$ Commented Nov 2, 2009 at 18:29