Theorem 10.55 from Rotman's Homological algebra book.
If $l$ is a prime and $\pi$ is an elementary abelian group of order $l^n$, then $H_2(\pi, Z)$ is elementary abelian of order $l^{n(n-1)/2}$.
In the proof, Rotman first showed that $lH_2(\pi,Z) = {0}$ and then went on using induction: For the inductive step, he chose a subgroup $N$ of $\pi$ of index $l$ so $\pi/N$ is cyclic of order $l$. Then we got a spectral sequence:
$E_{p,q}^2 = H_p(\pi/N, H_q(N,\mathbb{Z})) \Longrightarrow_p H_n(\pi, \mathbb{Z})$
Then he went on stating that $E_{2,0}^2 ={0}$ which I understand. Then $E_{2,0}^\infty = {0}$ and the filtration of $H_2(\pi, \mathbb{Z})$ has only two steps. Then he argued that "The ususl bidegree argument shows that $E_{0,2}^\infty = E_{0,2}^2; E_{1,1}^\infty = E_{1,1}^2$".
This last conclusion I cannot get. I can see that the first equality holds when r starts from 4 and the second holds when r starts from 3.
Any help would be appreciated!