You cannot embed the Poincare group or the Lorentz group into a compact Lie group $G$. Indeed, denote the Lie algebra of $G$ as $\mathfrak{g}$ and the Lorentz algebra as $\mathfrak{l}=\mathfrak{o}(1,3)\cong\mathfrak{sl}_2(\mathbf{C})$.

The Killing form on $\mathfrak{g}$ is non-positive-definite, but then so is its restriction to $\mathfrak{l}$. Restriction of the Killing form on $\mathfrak{g}$ to $\mathfrak{l}$ is $\mathfrak{l}$-invariant and is therefore proportional to the Killing form on $\mathfrak{l}$, since the latter is a simple real Lie algebra. Finally, the Killing form on $\mathfrak{l}$ has signature $(3,3)$, contradiction.

By the same reasoning, you cannot mod out by a discrete subgroup of the Lorentz group and get a compact group: the Lie algebra does not change, so the Killing form cannot become non-positive-definite.

On the other hand, there are well-known 'compactifications' of the translation group. You can either mod out $\mathbf{R}/\mathbf{Z}=S^1$ or immerse $\mathbf{R}\rightarrow T^2$ as an irrational winding depending on what kind of compactifications you are interested in.

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