Structure theorem of symplectic modules My question comes from the content presented on slide 31 of the following presentation given by Jean-Pierre Tignol (unfortunately, I do not have access to the main reference on the subject, that is Tignol and Amitsur's paper on symplectic modules).
A symplectic module is a finite abelian group $M$ with a nondegenerate alternating bilinear pairing $b:M\times M \rightarrow \mathbb Q/\mathbb Z$. 
On the slide mentioned above, the following structure theorem is refered to as "De Rham theorem, 1931".

Every symplectic module $M$ has a symplectic basis 
  $$M\cong (\mathbb Z/n_1\mathbb Z)^2\times\ldots\times (\mathbb Z/n_s\mathbb Z)^2$$
  with generators $e_1,f_1,\ldots ,e_s,f_s$ such that $b(e_i,f_i)=\frac{1}{n_i}+\mathbb Z$ and $b(e_i,f_j)=0$ for $i\not = j$ and $b(e_i,e_j)=b(f_i,f_j)=0$ for any $i,j$.  

I have two questions in regards to this.
First, can we assume furthermore that $n_1|n_2|\ldots |n_s$ ?
Second, what does this statement have to do with the renowned De Rham theorem of 1931, expressing that for a smooth manifold $M$, we have an isomorphism between De Rham cohomology groups $H_{dR}^p(M)$ and singular cohomology groups $H^p(M;\mathbb R)$ ?
 A: First if you don't have academic access to the papers, you can definitely ask people for them, including the authors, especially when they are super nice people like Jean-Pierre Tignol ! (There are also other less legal ways to get articles, which I can of course not publicly endorse but which I heard work pretty well).
For your second question, it turns out that in his 1931 paper, De Rham also introduces symplectic modules (under a different terminology) and derives their basic properties. So there is no direct link with his famous theorem about equivalence of cohomologies, it just happens to be in the same paper.
For your first question, the answer is yes, and follows directly from the following lemma: if $M\simeq (\mathbb{Z}/n\mathbb{Z})^2\times (\mathbb{Z}/m\mathbb{Z})^2$ with $n$ and $m$ coprime, and $(e_1,f_1,e_2,f_2)$ is a symplectic basis, then there is a symplectic basis $(e_3,f_3)$ for the natural decomposition $M\simeq (\mathbb{Z}/nm\mathbb{Z})^2$.
It is not hard to find, I'll let you fill in the details: just take $u\in \mathbb{Z}$ such that $u(a+b)\equiv 1$ mod $ab$, and define $e_3=u(e_1+e_2)$ and $f_3=f_1+f_2$.
