Symmetrical bilinear form $G\times G\to \mathbb{Z}_{2}$ Lets $G$ is finite abelian group (such that for any $x\in G$ $x+x=0$, i.e. $G=\mathbb{Z}_{2}^{\oplus k}$ for some $k\in\mathbb{N}$) and $(\cdot,\cdot):G\times G\to \mathbb{Z}_{2}$ is symmetrical bilinear form.
Know that:
$$(a, m)=0,$$
$$(a, p)=1,$$
$$(b, m)=1,$$
$$(b, p)=0.$$
Is it true that
$$(a, b) = 1?$$
Thanks.

It can go either way. Consider $k=3$ and the bilinear form defined by the usual inner product
$$(x,y)=x_1y_1+x_2y_2+x_3y_3$$
Then we can arrange $a,b,p,m$ so that $(a,b)$ is either $0$ or $1$ to our liking:
$$\begin{array}{|c|c|c|}\hline  a &(1,0,0) & (1,0,1) \\ b & (0,1,0) & (0,1,1) \\ p & (1,0,0) & (1,0,0) \\ m & (0,1,0) & (0,1,0) \\ \hline (a,b) & 0 & 1 \\ \hline \end{array} $$
