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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?

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Count the number of subgroups of the quotient $\mathbb Z^3/7\mathbb Z^3$. –  Mariano Suárez-Alvarez Dec 8 '11 at 2:23
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2 Answers

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.

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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 $7 \mathbb{Z}^{3} \subset P \subset \mathbb{Z}^{3}$.

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It might be instructive to explain how one arrives at the formula, e.g., with 1 or 2 dimensional subspaces. –  Arturo Magidin Dec 8 '11 at 4:14
I too would welcome a little more elaboration. Nonetheless: thank you. –  user20850 Dec 8 '11 at 15:25
Done. Thanks, friends. –  user02138 Dec 8 '11 at 16:53
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