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Given a finite abelian p-group: $G = \displaystyle\prod_{i=1}^n p^{k_i}\mathbb{Z}_{p^k}$ for some integers $k,k_1,...,k_n$.

Regarding elements of G as tuples $(x_1,...,x_n) \in G$, I can get subgroups by imposing one or more conditions of the forms:

(a) $x_i \equiv 0\pmod{p^m}$ for some integer $m$
(b) $x_i \equiv x_j\pmod{p^m}$ for some integer $m $

Are all subgroups of this form?

Edit: The answer is no as Alex Bartel commented. So let me extend condition (b) to:
(b) $x_i \equiv b*x_j\pmod{p^m}$ for some integers $m,b$ (or even, $\alpha_1x_1 + ... + \alpha_nx_n \equiv 0\pmod{p^m}$)
Also, let me extend the question: If the answer is no, can the conditions be extended a bit for the answer to be yes? More generally, is there an easy way to describe all those subgroups?

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No. For example the cyclic subgroup of $\mathbb{Z}/p\mathbb{Z}\times \mathbb{Z}/p\mathbb{Z}$ generated by $(1,2)$ is not covered by any of your conditions. – Alex B. Apr 13 '11 at 10:59
@Alex: Thanks. Changed the question accordingly. – user3533 Apr 13 '11 at 11:09
@Lee: $(\mathbb{Z}/p\mathbb{Z})^n$ is a vector space over the field $\mathbb{Z}/p\mathbb{Z}$. You're looking for its subspaces. They can be described by equations "some coordinates are these linear combinations of the other coordinates". So you will need this type of conditions. – user8268 Apr 13 '11 at 11:35
To expand on the comment by user8268: sometimes, you will need more than one congruence. E.g., the set of $(x,y,z)$ in ${\bf Z}_p^3$ such that $x\equiv y\equiv z\pmod p$ is a subgroup that cannot be described by a single congruence. – Gerry Myerson Apr 13 '11 at 13:04
I don't understand the $p^{k_i}$ in front of the $\mathbb{Z}_{p^k}$. That just has the effect of making the cylic group a cyclic group of order $p^{k-k_i}$ instead of order $p^k$. Wouldn't it be much simpler to simply write that $G$ is a product of cyclic $p$-groups? – Arturo Magidin Apr 13 '11 at 15:40
up vote 1 down vote accepted

Rewritten and (I hope) clarified

Yes, every subgroup can be characterized as the set of tuples satisfying certain linear congruences in its entries. To be more precise, we have the following:

Let $p$ be a prime. Then every subgroup of $\displaystyle \prod_{i=1}^n \mathbb{Z}_{p^k}$ (including the subgroups of a $G$ as given in the statement of the question) can be described as the set of elements $(h_1,\ldots,h_n)$ which satisfy a set of (at most) $n$ congruences of the following form: $$\begin{align*} h_{\sigma(1)} &\equiv 0 \pmod{p^{j_1}}\\ h_{\sigma(2)} &\equiv \alpha_{21}h_{\sigma(1)} \pmod{p^{j_2}}\\ h_{\sigma(3)} &\equiv \alpha_{31}h_{\sigma(1)} + \alpha_{32}h_{\sigma(2)} \pmod{p^{j_3}}\\ &\vdots\\ h_{\sigma(n)} &\equiv \alpha_{n1}h_{\sigma(1)} + \alpha_{n2}h_{\sigma(2)} + \cdots + \alpha_{n(n-1)}h_{\sigma(n-1)} \pmod{p^{j_n}} \end{align*}$$ where $\sigma\in S_n$ is a permutation of $\{1,\ldots,n\}$, $0\leq j_1\leq\cdots\leq j_n\leq k$ are nonnegative integers, and $\alpha_{ij}$ are integers.

What follows is essentially the proof that submodules of free modules over PIDs are free, as described in this previous answer, particularized for this situation.

Let $H$ be a subgroup of $\displaystyle \prod_{i=1}^n \mathbb{Z}_{p^k}$.

We proceed by induction on $n$. If $n=1$, then $H$ is a subgroup of $\mathbb{Z}_{p^k}$, and therefore is of the form $p^{j_1}\mathbb{Z}_{p^k}$ for some $j_1$, $0\leq j_1 \leq k$; thus, $(h_1)\in H$ if and only if $h_1\equiv 0\pmod{p^{j_1}}$, which gives the result we want.

Asume the result holds for subgroups of a product of $n$ copies of $\mathbb{Z}_{p^k}$, and let $H$ be a subgroup of $\displaystyle \prod_{i=1}^{n+1}\mathbb{Z}_{p^k}$.

If $H=\{e\}$, then we are done: $H$ is described by the congruences $h_i\equiv 0\pmod{p^{k}}$, $i=1,\ldots,n$. So we may assume $H$ is nontrivial.

Consider the projections onto the $n+1$ coordinates; if any projection, say the projection on the $j$th component is trivial, then $H$ is a subgroup of a product of $n$ copies of $\mathbb{Z}_{p^k}$, and can be described as in the proposition, together with the condition $h_j\equiv 0\pmod{p^k}$.

Assume then that all projections are nontrivial. Each projection gives a subgroup of $\mathbb{Z}_{p^k}$, hence of the form $p^{r_i}\mathbb{Z}_{p^k}$, with $0\leq r_i\lt k$. Let $\sigma(1)$ be the index with the largest subgroup (i.e., smallest $r_i$, possibly $0$); set $j_1 = r_{\sigma(1)}$. Let $\mathbf{a}_1$ be an element of $H$ whose $\sigma(1)$ coordinate is $p^{j_1}$. The minimality of $j_1$ means that $\mathbf{a}_{1}$ can be written as $$\mathbf{a}_1 = (a_1,\ldots,a_n) = p^{j_1}(b_1,\ldots,b_n) = a_{\sigma(1)}(b_1,\ldots,b_n)$$ with $b_{\sigma(1)}=1$, all $b_i$ integers.

Now let $K$ be the subgroup of $H$ corresponding to the kernel of the projection onto the $\sigma(1)$-coordinate; that is, all elements that have $\sigma(1)$ coordinate congruent to $0$ modulo $p^{k}$. By induction, there is a bijection $\sigma\colon \{2,\ldots,n+1\}\to\{1,\ldots,n+1\}-\{\sigma(1)\}$ and integers $\gamma_{rs}$, $2\leq r\leq n+1$, $1\leq s\lt r$ and nonnegative integers $j_2\leq\cdots\leq j_{n+1}\leq k$ such that the elements of $K$ can be described as $$\begin{align*} k_{\sigma(1)} &\equiv 0 \pmod{p^k}\\ k_{\sigma(2)} &\equiv 0 \pmod{p^{j_2}}\\ k_{\sigma(3)} &\equiv \gamma_{32}k_{\sigma(2)} \pmod{p^{j_3}}\\ &\vdots\\ k_{\sigma(n+1)} &\equiv \gamma_{(n+1)2}k_{\sigma(2)} + \cdots + \gamma_{(n+1)n}k_{\sigma(n)}\pmod{p^{j_{n+1}}}. \end{align*}$$ Moreover, the minimality of $j_1$ ensures that $j_1\leq j_2$.

Now, $K\cap\langle \mathbf{a}_1\rangle = \{e\}$, and $K+\langle \mathbf{a}_1\rangle\subseteq H$. In fact, $H = K\oplus\langle\mathbf{a}_1\rangle$: given $\mathbf{h}\in H$, we know that the $\sigma(1)$-component of $\mathbf{h}$ is a multiple of $p^{j_1}$ (by choice of $\sigma(1)$), and so there exists $m$ such that $\mathbf{h}- m\mathbf{a}_1 \in K$. Hence, $h\in K\oplus\langle\mathbf{a}_1\rangle$.

Therefore, given $\mathbf{h}=(h_1,\ldots,h_{n+1})\in \displaystyle\prod_{i=1}^{n+1}\mathbb{Z}_{p^k}$, we have that $\mathbf{h}\in H$ if and only if:

  • (i) $h_{\sigma(1)}\equiv 0\pmod{p^{j_1}}$; and
  • (ii) if $h_{\sigma(1)} = mp^{j_1}$, then $\mathbf{h}-m\mathbf{a}_1\in K$.

Thus, we only need to show that this can be expressed by congruences of the form described. Note that $h_{\sigma(1)} = ma_{\sigma(1)} = mp^{j_1}$.

Condition (i) on $\mathbf{h}$ can be written as $$h_{\sigma(1)} \equiv 0 \pmod{p^{j_1}}.$$ By our prior description of $K$, condition (ii) can be described as: $$\begin{align*} h_{\sigma(2)} -ma_{\sigma(2)}&\equiv 0 \pmod{p^{j_2}}\\ h_{\sigma(3)} -ma_{\sigma(3)}&\equiv \gamma_{32}\bigl(h_{\sigma(2)}-ma_{\sigma(2)}\bigr) \pmod{p^{j_3}}\\ &\vdots\\ h_{\sigma(n+1)}-ma_{\sigma(n+1)} &\equiv \gamma_{(n+1)2}\bigl(h_{\sigma(2)}-ma_{\sigma(2)}\bigr) + \cdots + \gamma_{(n+1)n}\bigl(h_{\sigma(n)}-ma_{\sigma(n)}\bigr)\pmod{p^{j_{n+1}}} \end{align*}$$ with the problem that $m$ depends on $h_{\sigma(1)}$.

However, $a_{\sigma(i)} = p^{j_1}b_{\sigma(i)} = a_{\sigma(1)}b_{\sigma(i)}$, and since $ma_{\sigma(1)} \equiv h_{\sigma(1)} \pmod{p^{k}}$ then for each $i$, $2\leq i\leq n+1$ we have $$ma_{\sigma(i)} = ma_{\sigma(1)}b_{\sigma(i)} \equiv h_{\sigma(1)}b_{\sigma(i)}\pmod{p^{k}}.$$ Since $j_i\leq k$, we can then rewrite the congruences above as: $$\begin{align*} h_{\sigma(1)} &\equiv 0 \pmod{p^{j_1}}\\ h_{\sigma(2)} -h_{\sigma(1)}b_{\sigma(2)}&\equiv 0 \pmod{p^{j_2}}\\ h_{\sigma(3)} -h_{\sigma(1)}b_{\sigma(3)} &\equiv \gamma_{32}\bigl(h_{\sigma(2)}-h_{\sigma(1)}b_{\sigma(2)}\bigr) \pmod{p^{j_3}}\\ &\vdots\\ h_{\sigma(n+1)}-h_{\sigma(1)}b_{\sigma(n+1)} &\equiv \gamma_{(n+1)2}\bigl(h_{\sigma(2)}-h_{\sigma(1)}b_{\sigma(2)}\bigr) \\ &\qquad\qquad\mathop{+} \cdots \mathop{+} \gamma_{(n+1)n}\bigl(h_{\sigma(n)}-h_{\sigma(1)}b_{\sigma(n)}\bigr)\pmod{p^{j_{n+1}}} \end{align*}$$ which can be rewritten in the desired form by moving the $h_{\sigma(1)}b_{\sigma(k)}$ summand to the right hand side, and regrouping; this satisfies the conditions of the statement above, since $b_1,\ldots,b_n$ and $\gamma_{ij}$ are fixed, giving rise to fixed $\alpha_{ij}$ that do not depend on $\mathbf{h}$.

This establishes the induction step, and so the description.

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@Arturo, isn't that subgroup of ${\bf Z}^3$ given by the congruences $b\equiv6a\pmod9$, $c\equiv15a+2b\pmod{81}$? – Gerry Myerson Apr 14 '11 at 5:10
@Gerry: Hmmm... Yes. $2b = 9m+6a$, and plugging that for your $2b$ gives my $18m+27a$. So there may be a way, I'm just not seeing how to phrase it in general. Inductively, if I can figure out how to do it for the second coordinate, it should straightforward to go from there. I'll sleep on it. Thanks! – Arturo Magidin Apr 14 '11 at 5:20
If it's any help, here's how I got there: general element is $(1,6,27)r+(0,9,18)s+(0,0,81)t=(a,b,c)$ with $a=r$, $b=6r+9s=6a+9s$, $c=27r+18s+81t=15r+2(6r+9s)+81t=15a+2b+81t$, and the congruences pop out. – Gerry Myerson Apr 14 '11 at 6:55
@Gerry: Thanks. There's a few more simplifications that can be made (in the basis elements, you can always assume that any entry to the right of the "pivot" is a multiple of the pivot, for instance) which might make this doable in general. – Arturo Magidin Apr 14 '11 at 13:04
@Gerry: Thanks again; that was very useful, actually. I think I got it, if my "you may assume"s are not invalit somewhere around the line. – Arturo Magidin Apr 14 '11 at 21:40

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