Double summation in abelian group

How can we prove that in abelian group the following equality holds? $$\sum_{\nu=1}^n\sum_{\mu=1}^ma_{\mu\nu}=\sum_{\mu=1}^m\sum_{\nu=1}^na_{\mu\nu}$$

-
Could you explain your notation? – amWhy Apr 12 '13 at 15:01
@amWhy To each pair of indexes $(\nu,\mu)$ such that $1 \leq \nu \leq n, ~~ 1 \leq \mu \leq m$ an element $a_{\mu \nu}$ of a group is assigned. – Igor Apr 12 '13 at 15:06
@amWhy The following definition is given in the book: $\sum_{\nu = 1}^{n} \sum_{\mu = 1}^{m} a_{\mu \nu} = \sum_{\nu = 1}^{n} ( \sum_{\mu = 1}^{m} a_{\mu \nu} )$ – Igor Apr 12 '13 at 15:09
Got it...thanks. – amWhy Apr 12 '13 at 15:10

Note that $\prod$ is (or can be) defined recursively as $$\tag1\prod_{k=1}^0 a_k = 1$$ $$\tag2\prod_{k=1}^{N+1} a_k = \left(\prod_{k=1}^{N} a_k\right)\cdot a_{N+1}$$

Lemma 1. $$\tag3\prod_{k=1}^N 1=1.$$

Proof: The case $N=0$ follows from $(1)$, the induction step by $$\prod_{k=1}^{N+1} 1\stackrel{(2)}= \left(\prod_{k=1}^{N} 1\right)\cdot 1=1\cdot 1=1.$$

Lemma 2. $$\tag4\left(\prod_{k=1}^N a_k\right)\cdot \left(\prod_{k=1}^N b_k\right)=\prod_{k=1}^N (a_kb_k).$$

Proof: If $N=0$, this follows from $1\cdot 1=1$. For the induction step, note that \begin{align}\left(\prod_{k=1}^{N+1} a_k\right)\cdot \left(\prod_{k=1}^{N+1} b_k\right) &\stackrel{(2)}=\left(\prod_{k=1}^{N} a_k\right)\cdot a_{N+1}\cdot \left(\prod_{k=1}^{N} b_k\right)\cdot b_{N+1}\\&\stackrel{\text {ab.}}= \left(\prod_{k=1}^{N} a_k\right)\cdot \left(\prod_{k=1}^{N} b_k\right)\cdot a_{N+1} b_{N+1}\\&\stackrel{(4)}= \left(\prod_{k=1}^N (a_kb_k)\right)\cdot a_{N+1}b_{N+1}\\&\stackrel{(2)}=\prod_{k=1}^{N+1} (a_kb_k).\end{align}

Now for the desired claim:

First case: $n=0$ In this case, the left hand side is $$\prod_{\nu=1}^0\prod_{\mu=1}^m a_{\mu\nu}\stackrel{(1)}=1.$$ The right hand side is $$\prod_{\mu=1}^m \prod_{\nu=1}^0 a_{\mu\nu}\stackrel{(1)}=\prod_{\mu=1}^m 1 \stackrel{(3)}=1.$$

For the induction step, assume that $$\prod_{\nu=1}^n\prod_{\mu=1}^m a_{\mu,\nu}=\prod_{\mu=1}^m\prod_{\nu=1}^n a_{\mu,\nu}$$ for all $m$ and all choices of $a_{\mu,\nu}$. Then \begin{align} \prod_{\nu=1}^{n+1}\prod_{\mu=1}^m a_{\mu,\nu} &\stackrel{(2)}= \left(\prod_{\nu=1}^{n}\prod_{\mu=1}^m a_{\mu,\nu}\right)\cdot \prod_{\mu=1}^m a_{\mu,{N+1}} \\ &\stackrel{(4)}= \prod_{\mu=1}^m\left(\prod_{\nu=1}^{n} a_{\mu,\nu}\right)\cdot \prod_{\mu=1}^m a_{\mu,{N+1}}\\ &\stackrel{(4)}= \prod_{\mu=1}^m\left(\prod_{\nu=1}^{n} a_{\mu,\nu}\right)\cdot \prod_{\mu=1}^m a_{\mu,{N+1}}\\ &\stackrel{(3)}= \prod_{\mu=1}^m\left(\left(\prod_{\nu=1}^{n} a_{\mu,\nu}\right)\cdot a_{\mu,N+1}\right) \\ &\stackrel{(2)}= \prod_{\mu=1}^m\prod_{\nu=1}^{n+1} a_{\mu,\nu}. \end{align}

-
Thank you very much! It's just what i was looking for! – Igor Apr 12 '13 at 15:47

To expand on Daniel Robert-Nicoud's answer, one induction actually suffices:

For $m=1$ the result is clear. Suppose $\sum_{\nu=1}^n\sum_{\mu=1}^k a_{\mu\nu} = \sum_{\mu=1}^k\sum_{\nu=1}^n a_{\mu\nu}$ for all $n\in\mathbb{N}$. $$\sum_{\mu=1}^{k+1}\sum_{\nu=1}^n a_{\mu\nu} = \sum_{\mu=1}^k\sum_{\nu=1}^n a_{\mu\nu} + \sum_{\nu=1}^n a_{k+1,\nu} = \sum_{\nu=1}^n\sum_{\mu=1}^k a_{\mu\nu} + \sum_{\nu=1}^n a_{k+1,\nu} = \sum_{\nu=1}^n\left(\sum_{\mu=1}^k a_{\mu\nu}+a_{k+1,\nu}\right) = \sum_{\nu=1}^n\sum_{\mu=1}^{k+1} a_{\mu\nu}.$$

The general statement now follows by induction.

-
Indeed you do. Although induction on $n$ would work just as well of course. – Abel Apr 12 '13 at 15:26
Thank you. It's clear now. – Igor Apr 12 '13 at 15:40

You can easily prove it by induction on $n$ and $m$.

-