Which compact Euclidean (flat) 3 manifolds can be written as a coset space $ G/H $ for some three dimensional subgroup $ G $ of $ Iso(E^3) \cong \mathbb{R}^3 \rtimes O_3 $ and a discrete subgroup $ H $ of $ G $?
I can already construct 2 examples this way
$$
G= \{
\begin{bmatrix}
1 & 0 & 0 & x \\
0 & 1 & 0 & y \\
0 & 0 & 1 & z \\
0 & 0 & 0 & 1
\end{bmatrix}
: x,y,z \in \mathbb{R} \}
\, , \,
H= \{
\begin{bmatrix}
1 & 0 & 0 & k \\
0 & 1 & 0 & n \\
0 & 0 & 1 & m \\
0 & 0 & 0 & 1
\end{bmatrix}
: k,n,m \in \mathbb{Z} \}
$$
yielding the three torus $ T^3 \cong G/H $. And
$$
G= \{
\begin{bmatrix}
a & b & 0 & x \\
-b & a & 0 & y \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}
: a^2+b^2=1 \}
\, , \,
H= \{
\begin{bmatrix}
1 & 0 & 0 & k \\
0 & 1 & 0 & n \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}
: k,n \in \mathbb{Z} \}
$$
yielding an example with $ \pi_1(G/H) \cong \mathbb{Z} \rtimes \mathbb{Z}^2 $ and first Betti number 1 (thus distinct from the torus).
Some relevant facts:
Every compact flat 3 manifold $ M $ (there are 10 of these, 6 orientable and 4 non-orientable) is the quotient of the 3 torus $ T^3 $ by some finite group $ \Gamma $. Moreover $ \pi_1(M) $ is torsion free and virtually $ \mathbb{Z}^3 $ fitting into the short exact sequence
$$
1 \to \mathbb{Z}^3 \to \pi_1(M) \to \Gamma \to 1
$$
The ten distinct $ \pi_1(M) $ that show up here are exactly the 10 torsion free crystallographic groups of 3 space.
Best Answer
The answer to the title question is that there are 18 flat three manifolds, 6 closed orientable, 4 closed non orientable, 4 open orientable, 4 open non orientable, and of these exactly 5 of the 6 closed orientable, 2 of the 4 closed non-orientable, 4 of the 4 open orientable, and 3 of the 4 open non orientable can be expressed as coset manifolds $ G/H $ for some Lie group $ G $ and subgroup $ H $.
However the text of the question asks something much narrower than the title; it asks which of the ten closed Euclidean three manifolds can be realized as a coset manifold $ G/H $ for $ G $ a subgroup of $ SE_3 $ and $ \Gamma $ a discrete subgroup of $ H $ (it turns out the answer is the same if we let $ G $ be an Lie group and $ H $ a discrete subgroup). The answer is only the five orientable manifolds $$ T^3, MT(\begin{bmatrix} -1 & 0 \\0&-1 \end{bmatrix}),MT(\begin{bmatrix} 0 & -1 \\1& 0 \end{bmatrix}),MT(\begin{bmatrix} 0 & -1 \\1& 1 \end{bmatrix}),MT(\begin{bmatrix} 0 & -1 \\1&-1 \end{bmatrix}) $$
where $ MT $ denotes the mapping torus of $ T^2 $ corresponding to that mapping class in $ MCG(T^2)=SL_2(\mathbb{Z}) $. This and other details can be found in https://deepblue.lib.umich.edu/handle/2027.42/24381
Note that if we still ask that $ H $ is a subgroup of $ SE_3 $ but non longer require $ \Gamma $ to be discrete then I believe we can realize both the trivial $ S^! \times K^2 $ and nontrivial $ U_1 $ principal bundles over the Klein bottle $ K^2 $ as $ H/\Gamma $ for certain choices of $\Gamma \leq H \leq SE_3 $.