Double summation in abelian group How can we prove that in abelian group the following equality holds?
$$ \sum_{\nu=1}^n\sum_{\mu=1}^ma_{\mu\nu}=\sum_{\mu=1}^m\sum_{\nu=1}^na_{\mu\nu}$$
 A: 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.
A: You can easily prove it by induction on $n$ and $m$.
A: 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 
$$\begin{align}
\prod_{\nu=1}^{n+1}\prod_{\mu=1}^m a_{\mu,\nu}
&\stackrel{(2)}= \left(\prod_{\nu=1}^{n}\prod_{\mu=1}^m a_{\mu,\nu}\right)\cdot \prod_{\mu=1}^m a_{\mu,{N+1}} \\
&\stackrel{(4)}= \prod_{\mu=1}^m\left(\prod_{\nu=1}^{n} a_{\mu,\nu}\right)\cdot \prod_{\mu=1}^m a_{\mu,{N+1}}\\
&\stackrel{(4)}= \prod_{\mu=1}^m\left(\prod_{\nu=1}^{n} a_{\mu,\nu}\right)\cdot \prod_{\mu=1}^m a_{\mu,{N+1}}\\
&\stackrel{(3)}= \prod_{\mu=1}^m\left(\left(\prod_{\nu=1}^{n} a_{\mu,\nu}\right)\cdot a_{\mu,N+1}\right) \\
&\stackrel{(2)}= \prod_{\mu=1}^m\prod_{\nu=1}^{n+1} a_{\mu,\nu}.
\end{align}$$
