**Edited** (this question contains two versions of a similar question)


Is there some finitely presented group $G$ generated by $g\_1,...,g\_n$ such that 
there is an element $g\in G$ expressed as a finite word in $g\_i$'s so that it is impossible to prove neither $g=1$ nor $g\ne 1$?


Is such a group $G$ exists, what would be a relatively simple example? 


**Adjusted question.** Is there $G,g$ so that $g\ne 1$ in $G$ but is it impossible to prove this in *finite time*? 



An element $g$ in a group such that neither $g=1$ nor $g\ne 1$ can be proved.
Let me offer another way to explain a similar idea as in
Benjamin's answer.


The answer to the original question, as well as the updated
question, is Yes, there is such a finite group presentation.


Since you are asking whether or not a given assertion is
"provable", you should be explicit about the theory in which you
are making such proofs. Let's suppose that you intend us to use
one of the standard default background theories, such as ZFC or
ZFC + large cardinals or whatever. Such theories have a computably
presentable list of axioms.


Now, consider the standard proof that the word problem is not
decidable. The way this is proved, one fixes any Turing machine
$M$, and writes down a finitely presentation of a group $G\_M$,
whose words in effect simultate the comptuation of $M$. The result
is that there is a generator $g$ of the presentation such that
$g=1$ in the group if and only if $M$ halts (on empty input).
Thus, we reduce the halting problem, which is undecidable, to the
word problem, and so the word problem is undecidable.


But now for the given background theory $T$, there will be an
explicit Turing machine $M$, such that the question of whether $M$
halts is not provable in $T$ (for example, imagine the Turing
machine that searches for a proof of a contradiction in $T$).
Since $T$ does not prove whether $M$ halts or not, it follows that
$T$ will neither prove nor refute the corresponding assertion
about whether $g=1$ or not in the group presentation $G\_M$.


In this example, there will be no proof that $g=1$ and no proof that $g\neq 1$. 


