Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

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

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.

share|cite|improve this answer

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.