# 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.

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The "i.e." is incorrect. Nothing before that excludes the possibility that $G$ is infinite. – joriki Mar 1 '12 at 9:03
@joriki: Lets $G$ is finite – Aspirin Mar 1 '12 at 9:05

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

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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:

$$\begin{array}{|c|c|c|}\hline a &(1,0,0) & (1,0,1) \\ b & (0,1,0) & (0,1,1) \\ p & (1,0,0) & (1,0,0) \\ m & (0,1,0) & (0,1,0) \\ \hline (a,b) & 0 & 1 \\ \hline \end{array}$$

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