# Subgroups of $\mathbb{Z}_p^n$

Is there a nice characterization or construction to list the subgroups of $\mathbb{Z}_p^n$, that is, $\mathbb{Z}_p \times \cdots \times \mathbb{Z}_p$ where $\mathbb{Z}_p$ is the cyclic group of prime order $p$?

Treat it as a vector space over $\Bbb Z/p\Bbb Z$. In this way you can see that it's a simple matter of counting numbers of choices of linearly independent vectors. All subspaces are sub-groups and vice versa since they will clearly be $\Bbb Z/p\Bbb Z$-modules given by descending the usual $\Bbb Z$ action.

But then the solution is clear, you need to count bases for $k$ dimensional subspaces for each $0\le k\le n$ and add up the numbers.

For $\{0\}$, clearly the empty set is your basis. Now any non-zero vector makes a one-dimensional subspace, so there are $p^n-1$, and of course we need to mod out by the scalar multiples, giving ${p^n-1\over p-1}$ one-dimensional subspaces. And continuing in this fashion, we know each time we need to count elements in the so-called Grassman manifold where here $q=p^n$ since they have the same structure as vector spaces.

$$|G(n,k)(\Bbb Z/p\Bbb Z)|=\prod_{i=1}^{k} {(q^{n-i+1}-1)\over (q^i-1)}$$

So the total number of subgroups will be

$$1+\sum_{k=1}^n \prod_{i=1}^{k} {(q^{n-i+1}-1)\over (q^i-1)}.$$

If you're wondering how we come up with these numbers, note that that the product without the denominator counts the number of linearly independent sets of vectors of size $k$ and the denominator gets rid of scalar multiple issues and permutations, so that each subspace is related to the set of all its possible bases.

All the proper subgroups are isomorphic to a k - fold direct product of $Z_p$where k < n. Obviously the cyclic decomposition of any proper subgroup only contains factors with order a power of p. Since every nonidentity element of any proper subgroup has order p, each factor must have order p. Thus the subgroups are nicely classified by their sizes. As for determining the size of a subgroup generated by a given set of m elements: the group is an $F_p$ vector space, so you can construct the n x m matrix with the elements as columns and perform row reduction to find the rank r. The size of the subgroup generated by these elements is $p^r$.

The number of one dimensional subspaces is $(p^n - 1)/(p - 1)$: there are $p^n - 1$ nonidentity elements and each generates a set of $p - 1$ nonidentity elements. I'm not sure how to count the higher dimensional subspaces but it seems like it might get more combinatorially complex with the increase in dimension, since the intersection of the distinct subspaces may have a wider range of dimensions.

• Yes, if you’re only interested in the isomorphism type. But if you want to count, say, how many $k$-dimensional subspaces of $(\Bbb Z/(p))^n$ there are, that’s another issue. – Lubin Apr 16 '16 at 17:28