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$G$ is a subgroup of the Euclidean group which translational subgroup $N\triangleleft G$ (the isometries that represent pure translations) is a lattice in Euclidean space. If $H$ is a subgroup of $G$ so that $N\leq H\leq G$, does the following hold? $$\exists U\leq G\colon UN=H \text{ and } U \text{ finite}$$

The following does hold for sure: $$\exists K\subset G\colon KN=H \text{ and } K \text{ finite}$$ More precisely it holds iff $K$ is a set of coset representatives of $N$ in $H$ from which it follows that $KN=H$. The set $K$ is finite because $G/N$ is finite and $H/N\leq G/N$. Hence the question can be rephrased as "can we find a set of coset representatives which is a group". In other words, can we find $K$'s which are also $U$'s and if yes, how many can be find?

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This happens if and only if the the group is a semidirect product $G=H\rtimes U$. Any group that is not a semidirect product fails for any normal subgroup choice for $H$. – user641 Jan 21 '13 at 15:25
Do you mean $G=H\rtimes N$ or really $G=H\rtimes U$? Can you give me a hint as for why? – Wox Jan 21 '13 at 16:38

Edit: The original version of the question did not specidy that $G$ should be a group of isometries of a Euclidean space. Therefore the counterexample below is now obsolete.

I interpret he question as follows (largely based on the title). Given $N\le H\le G$, $N\lhd G$, does there exist a subgroup $U\le G$ such that $UN=H$ and with the property that all the elements of $U$ belong to distinct cosets of $N$? The latter requirement is equivalent to the condition $U\cap N=\{1_G\}$.

If that is what you are asking, then the answer is negative. A counterexample is formed by the choices $G=H=C_4=\langle t\rangle$, the cyclic group of four elements generated by $t$, and $N=\langle t^2\rangle$. A group $U$ with the prescribed properties would need to have order two, but $N$ is the only such subgroup of $G$, so we're out of luck.

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I rephrased it to make it more clear, but you interpreted it correctly. Thanks! – Wox Jan 21 '13 at 15:32
The group G is a space group. I'm not sure whether that makes a difference but I added it to my question. – Wox Jan 21 '13 at 15:42
In the case of a group of isometries with $N$ consisting of translation, the first idea that comes to my mind is that if you can find a point $x$ in the space such that all the cosets of $N$ contain an element that fixes the point $x$, then those representatives should for a group. I cannot comment, whether this is always possible though. – Jyrki Lahtonen Jan 21 '13 at 17:01

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