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Suppose a no trivial totally ordered group .This group has maximum element?

A totally ordered group is a totally ordered structure (G,∘,≤) such that (G,∘) is a group.I couldnt find a more exact definition

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thanks for any help !! –  art Mar 22 '13 at 14:47
    
Does $(\mathbb Z,{+},{\le})$ satisfy your assumption? That certainly doesn't have a maximal element. (But it looks strange that your assumption doesn't relate the order to the group operation at all). –  Henning Makholm Mar 22 '13 at 14:51
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I assume you want the ordering to be compatible with the group operation, such that if $a \geq b$ and $c\geq d$ then $ac\geq bd$.

In this case, the group cannot have a maximal element, which we can see as follows: Assume $g$ is such a maximal element and let $h\in G$ with $h\geq 1$.

Now we have that $g\geq g$ and $h\geq 1$ so $gh\geq g$ which by maximality would mean $gh = g$ so $h = 1$.

But if $G$ is not trivial, it has an element $h$ with $h\geq 1$ and $h\neq 1$, which gives us our contradiction.

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Yes Tobias Kildetoft i agree with your observation . Thanks again to all the help! :) –  art Mar 22 '13 at 15:16
    
@art: If you found Tobias’s answer useful, you should accept it; for information on why and how to accept answers, see this. –  Brian M. Scott Mar 22 '13 at 17:35
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