Formula on the sum of all the elements of $G=\Bbb Z_{p_1^{\alpha_1}}\times \dots \times \Bbb Z_{p_n^{\alpha_n}}$

Question

Let $G=\Bbb Z_{p_1^{\alpha_1}}\times \dots \times \Bbb Z_{p_n^{\alpha_n}}$ be listed as $G=\{a_1,\dots,a_k\}$, where:

• $p_1\dots,p_n$ are primes;
• $\alpha_1,\dots,\alpha_n$ are nonnegative integers;
• $k:=\prod_{i=1}^np_i^{\alpha_i}$.

As an example, let's take:

\begin{alignat}{1} \Bbb Z_{2^2} \times \Bbb Z_3\times \Bbb Z_{2}= &\{(0,0,0),(0,1,0),(0,2,0),(1,0,0),(1,1,0),(1,2,0),\\ &(2,0,0),(2,1,0),(2,2,0),(3,0,0),(3,1,0),(3,2,0),\\ &(0,0,1),(0,1,1),(0,2,1),(1,0,1),(1,1,1),(1,2,1),\\ &(2,0,1),(2,1,1),(2,2,1),(3,0,1),(3,1,1),(3,2,1)\} \\ \end{alignat}

Then we get:

\begin{alignat}{1} \sum_{i=1}^{24}a_i &= \bigl(3\cdot 2\cdot(0+1+2+3),\space\space 2^2\cdot2\cdot (0+1+2)),\space\space 2^2\cdot 3\cdot (0+1)\bigr) \\ &= \biggl(3\cdot 2\cdot\sum_{i=0}^{2^{2}-1}i,\space\space 2^2\cdot2\cdot\sum_{i=0}^{3-1}i,\space\space 2^2\cdot 3\cdot \sum_{i=0}^{2-1}i\biggr) \\ \end{alignat}

This suggests the following:

Claim. Let $G=\Bbb Z_{p_1^{\alpha_1}}\times \dots \times \Bbb Z_{p_n^{\alpha_n}}$ be listed as $G=\{a_1,\dots,a_k\}$, where $k:=\prod_{i=1}^np_i^{\alpha_i}$. Then:

\begin{alignat}{1} \sum_{i=1}^{k}a_i= \biggl(p_2^{\alpha_2}\dots p_n^{\alpha_n}\sum_{i=0}^{p_1^{\alpha_1}-1}i,\space\space\dots,\space\space p_1^{\alpha_1}\dots p_{n-1}^{\alpha_{n-1}}\sum_{i=0}^{p_n^{\alpha_n}-1}i\biggr) \\ \tag 1 \end{alignat}

I've tried to prove $(1)$ by induction on $(\alpha_1,\dots,\alpha_n)$. The claim holds for $(\alpha_1,\dots,\alpha_n)=(0,\dots,0)$; in fact, in this case $k=1$ and $G=\{a_1\}=\{(\underbrace{0,\dots,0}_{n\space slots})\}$, whence:

\begin{alignat}{1} \sum_{i=1}^{k}a_i= a_1=(\underbrace{0,\dots,0}_{n\space slots}) \end{alignat}

and

\begin{alignat}{1} &\biggl(p_2^{\alpha_2}\dots p_n^{\alpha_n}\sum_{i=0}^{p_1^{\alpha_1}-1}i,\space\space\dots,\space\space p_1^{\alpha_1}\dots p_{n-1}^{\alpha_{n-1}}\sum_{i=0}^{p_n^{\alpha_n}-1}i\biggr)= \\ &\biggl(1\dots 1\sum_{i=0}^{0}i,\space\space\dots,\space\space 1\dots 1\sum_{i=0}^{0}i\biggr)= \\ &(1\dots 1\cdot 0,\space\space\dots,\space\space 1\dots 1\cdot 0)= \\ &(\underbrace{0,\dots,0}_{n\space slots}) \\ \end{alignat}

As inductive step, I've assumed $(1)$ to hold for $\{\alpha_1,\dots,\alpha_{j-1},\alpha_j,\alpha_{j+1},\dots,\alpha_n\}$ and checked what happens for $\{\alpha_1,\dots,\alpha_{j-1},\alpha_j+1,\alpha_{j+1},\dots,\alpha_n\}$, for an arbitrary $j\in\{1,\dots,n\}$. This is where I get stuck, and I don't know whether I get lost in the algebra, or I'm applying induction incorrectly, or simply the claim is false.

Answer 1

I think you should use induction over the number of groups in the crossproduct. I will simply write $G = G_1 \times G_2 \times \cdots \times G_n$.

$q_k := |G_k|$ and $G_k = \{a^k_1 , \dots , a^k_{q_k}\}$.

For $n = 1$ the proposition is trivially true.

Now proof $n$ assuming the claim is true for $n-1$. The sum looks like this:

$$\sum_{a\in G} a = \sum_{j_1=1}^{q_1} \sum_{j_2=1}^{q_2} \cdots \sum_{j_n=1}^{q_n} (a_{j_1}^1, a_{j_2}^2, \dots, a_{j_{n-1}}^{n-1} , a_{j_n}^n)$$

Because of the way addition is defined on product groups it follows: \begin{equation} \sum_{j_n=1}^{q_n} (a_{j_1}^1, a_{j_2}^2, \dots, a_{j_{n-1}}^{n-1} , a_{j_n}^n) = (q_n a_{j_1}^1, q_n a_{j_2}^2, \dots, q_n a_{j_{n-1}}^{n-1} ,\sum_{j_n=1}^{q_n} a_{j_n}^n) \end{equation}

If we take this result and plug it back into the original equation we get:

$$\sum_{a\in G} a = \sum_{j_1=1}^{q_1} \sum_{j_2=1}^{q_2} \cdots \sum_{j_n=1}^{q_n} (a_{j_1}^1, a_{j_2}^2, \dots, a_{j_{n-1}}^{n-1} , a_{j_n}^n) = \sum_{j_1=1}^{q_1} \sum_{j_2=1}^{q_2} \cdots \sum_{j_{n-1}=1}^{q_{n-1}} (q_n a_{j_1}^1, q_n a_{j_2}^2, \dots, q_n a_{j_{n-1}}^{n-1} ,\sum_{j_n=1}^{q_n} a_{j_n}^n) $$

But as you may realize now we can use our original induction result, as it is now only a sum over $n_1$ summation signs. $\square$

Remark: We did neither use that we were using the integer groups nor that our subsets over which we add even are groups. In fact the claim in your questions holds for the product of subsets of groups. All we needed was the way addition is defined on products.

Hope it helps.

Answer 2

I think you are applying induction incorretly, by applying induction on $(\alpha_1,\dots,\alpha_n)$ you are assuming that $\alpha_1=\dots=\alpha_n$ but it can happen otherwise. I think it would be easier to prove the claim for all natural numbers by induction instead of only for powers of primes.

New Claim. Let $G=\Bbb Z_{l_1}\times \dots \times \Bbb Z_{l_n}$ be listed as $G=\{a_1,\dots,a_k\}$, where $k:=\prod_{i=1}^nl_i$. Then:

\begin{alignat}{1} \sum_{i=1}^{k}a_i= \biggl(l_2\dots l_n\sum_{i=0}^{l_1-1}i,\space\space\dots,\space\space l_1\dots l_{n-1}\sum_{i=0}^{l_n-1}i\biggr) \\ \tag 2 \end{alignat} Proof: \begin{alignat}{2} \sum_{i=1}^{k}a_i= \biggl(\sum_{i=1}^{k}i,...,\sum_{i=1}^{k}i\biggr) \end{alignat}, where the $i$ is an element of $ \Bbb Z_{l_j}$.

therefore it suffices to prove that for each coordinate j, $\sum_{i=1}^{k}i = l_1...l_{j-1}l_{j+1}... l_n\sum_{i=0}^{l_j-1}i$.

For all $1\le j\le n$,

there are $l_1\times...l_{j-1}\times l_{j+1}...\times l_n$ elements in $\Bbb Z_{l_1}\times ... \times \Bbb Z_{l_j-1} \times \Bbb Z_{l_j+1} \times ... \times \Bbb Z_{l_n}$ therefore for each $i$ in $\Bbb Z_{l_j}$ there are $l_1\times...l_{j-1}\times l_{j+1}...\times l_n$ elements with $j-th$ coordinate $i$ in $G=\Bbb Z_{l_1}\times \dots \times \Bbb Z_{l_n}$.

Therefore $\sum_{i=1}^{k}i = (1\times l_1\times...l_{j-1}\times l_{j+1}...\times l_n) + (2\times l_1\times...l_{j-1}\times l_{j+1}...\times l_n) + ... + (l_{j-1}\times l_1\times...l_{j-1}\times l_{j+1}...\times l_n) = (l_1\times...l_{j-1}\times l_{j+1}...\times l_n) \times (1+2+...+l_j-1) = l_1...l_{j-1}l_{j+1}... l_n\sum_{i=0}^{l_j-1}i$

Since (2) holds for all numbers it also holds for powers of primes.