What is a simple example of a non-abelian partially ordered group?
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$\begingroup$ Question in the title and question in the question's body to not match. $\endgroup$– Git GudApr 6, 2013 at 18:20
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2$\begingroup$ $S_3$ is nonabelian and if we give it the trivial ordering (every element is $\leq$ every other element) we're done. If you want a nontrivial ordering, let $x \leq y$ if $x$ and $y$ have the same sign. $\endgroup$– Julien ClancyApr 6, 2013 at 18:23
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$\begingroup$ @Julien Clancy: This is not a partial order, because it is not an antisymmetric relation. $\endgroup$– Gejza JenčaApr 8, 2013 at 16:30
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$\begingroup$ @GejzaJenča See en.wikipedia.org/wiki/Partially_ordered_group $\endgroup$– Julien ClancyApr 8, 2013 at 20:15
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$\begingroup$ @Julien Clancy I referenced to the second part of your comment. "To have the same sign" is clearly a nontrivial equivalence relation, so it is not antisymmetric. $\endgroup$– Gejza JenčaApr 8, 2013 at 20:43
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2 Answers
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All strictly increasing functions $\mathbb R\to \mathbb R$, equipped with composition as the group operation and with the usual partial order given by $f\leq g$ if and only if $f(x)\leq g(x)$ for every $x\in\mathbb R$. Note that this is a lattice ordered group.
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$\begingroup$ @Gejza, Very nice example. Thank you very much. $\endgroup$ Apr 8, 2013 at 23:20
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Free group over two generators $F(a,b)$ is non-abelian and is poset w.r.t dictionary order induced by $a<b$.
Yeah although, if you wanted a non total partial ordered group, than Gejza's example is best.
answered Jul 16, 2015 at 13:53