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.

No, this is not true: Take the canonical dot product, with $k=2$, $a=p=(0,1)$, $b=m=(1,0)$.
