There are exactly 116 different groups P where $7\mathbf{Z}^{3} \subset P \subset \mathbf{Z}^{3}$ There are exactly 116 different groups P  where $7\mathbf{Z}^{3} \subset P \subset \mathbf{Z}^{3}$
I don't know how to prove this. Is it provable at all? How? 
 A: By the Fourth (or Lattice, or whatever numbering you use) Isomorphism Theorem, the subgroups of $G$ that contain a normal subgroup $N$ are in one-to-one correspondence with the subgroups of $G/N$. 
Here, $G=\mathbf{Z}^3$ is abelian, so $7\mathbb{Z}^3$ is a normal subgroup. Thus, asking for subgroups $P$ that contain $N=7\mathbb{Z}^3$ is equivalent to asking for subgroups of $(\mathbb{Z}^3)/(7\mathbb{Z}^3) \cong (\mathbb{Z}/7\mathbb{Z})^3$.
The latter is a 3-dimensional vector space over $\mathbb{Z}/7\mathbb{Z}$, the field with $7$ elements; the subgroups are the subspaces. Count the subspaces.
A: The number of $k$-dimensional subspaces of a vector space of dimension $n$ over a finite field  of $q = p^{m}$ elements is the product
\begin{align}
\binom{n}{k}_{q} = \frac{(q^{n} - 1) \cdots (q^{n} - q^{k-1})}{(q^{k} - 1) \cdots (q^{k} - q^{k-1})}.
\end{align}
To prove this consider the following. A $k$-dimensional subspace is specified by $k$ linearly independent vectors, say, $\{ v_1, \dots, v_k \}$. There are $q^{n}-1$ ways to choose $v_1$, $q^{n} - q$ ways to choose $v_2$ (so as not to lie in a subspace spanned by $v_1$), and so on. Continuing in this manner, there are $q^{n} - q^{j}$ ways to choose $v_{j+1}$ (so as not to lie in the subspace spanned by any preceding vectors). The number of $k$ linearly independent vectors of an $n$-dimensional space is therefore the product 
\begin{align}
(q^{n} - 1) \cdots (q^{n} - q^{k-1}).
\end{align}
Setting $n = k$ gives the number of possible bases of each $k$-dimensional subspace. Therefore, we normalize the former by the latter and this gives the rational function which counts the number of $k$-dimensional subspaces. 
The total number of subspaces of an $n$-dimensional vector space over a finite field of $q$ elements is therefore the sum
\begin{align}
\sum _{k = 0}^{n} \binom{n}{k} _{q}. 
\end{align}
For your example, as my astute colleagues Mariano and Arturo suggest, $n = 3$ and $q = 7$, and the total number of said subspaces is the sum
\begin{align}
\binom{3}{0}_7 + \binom{3}{1}_7  + \binom{3}{2}_7  + \binom{3}{3}_7   = 1 + 57 + 57 + 1 = 116.
\end{align}
Thus, there are $116$ groups $P$ such that $7 \mathbb{Z}^{3} \subset P \subset \mathbb{Z}^{3}$. 
