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I will change this statement.

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 You need to be more clear. Trivially if $L$ is a lattice ordered group and $T$ a totally ordered group and $L \hookrightarrow T^n$ for some $n$, then $L\hookrightarrow T^{m}$ also for all $m \geq n$. So the naive interpretation of your question gives a resounding "no". I assume you have some additional conditions in mind which led you to suspect that the embedding can be unique. – Willie Wong Mar 1 2012 at 9:59
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 Do you mean whether there is only one embedding between $L$ and $T^n$? If this is your question, then the answer is also no. Let me build a trivial counterexample. Suppose $T$ is a totally ordered group, then $T$ is embeddable into $T^2$ using at least two different embeddings: one given by the mapping which assigns $x$ to $(x,e)$, and the other assigning $x$ to $(e,x)$. – boumol Mar 1 2012 at 10:28
In view of trivial counterexamples to the question as stated --- both of the preceding comments as well as "increase $T$" --- I'll vote to close as "not a real question", with the hope that Rajnish can edit it to become a real question. – Andreas Blass Mar 1 2012 at 14:54

closed as not a real question by Andreas Blass, Vladimir Dotsenko, Simon Thomas, Mark Sapir, Andy Putman Mar 2 2012 at 19:46

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