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Setting: $G$ is a finite abelian group and any bicharacter on $G$, where a bi-character on $G$ is a map $b:G \times G \to \mathbb{Q}/\mathbb{Z}$ such that $$b(x+y,z)=b(x,z)+b(y,z),b(x,z+y)=b(x,z)+b(x,y).$$ My question assumption is: given two positive integers both greater than $2$, ie $p,q \geq 2$ and $p,q$ are coprime, and suppose we know that bi-characters $b(a_1,b_1)=\frac{x}{p}$ and $b(a_2,b_2)=\frac{y}{q}$. $x$ and $p$ are coprime and $y, q$ are coprime for some $(a_1, a_2, b_1, b_2) \in G$.

Now my question is:

Does there exist $c,d \in G$ such that $b(c,d)$ can be written as the fraction like $b(c,d)=\frac{z}{pq}$ such that $z$ is coprime to $pq$?

I can prove this when $G$ is cyclic. If $G$ is a finite cyclic group $\mathbb{Z}_N=<\mathbb{I}>$. We use addition notation for the finite abelian group. We only need to choose $b(\mathbb{I},\mathbb{I}) \in \mathbb{R}/ \mathbb{Z}$, then the bi-character on all other values are fixed since \begin{equation} b(m\mathbb{I},n\mathbb{I})=(mn)b(\mathbb{I},\mathbb{I}) \end{equation} Remember that our condition becomes that $$b(a_1\mathbb{I},b_1\mathbb{I})=\frac{x}{p}=(a_1b_1)b(\mathbb{I},\mathbb{I}),b(a_2\mathbb{I},b_2\mathbb{I})=\frac{y}{q}=(a_2b_2)b(\mathbb{I},\mathbb{I}) , (x,p)=(y,q)=1.$$

It is obviously that $b(\mathbb{I},\mathbb{I})$ must be form of $\frac{z}{pq}$ and $z$ and $pq$ are coprime to each other.

Now I am going to prove for the general case by reducing the finite abelian group into a direct sum of the cyclic group and use some elementary number theory argument. However, I am stuck. I am wondering is there easy way to prove it or there is a counterexample here.

Any help would be very appreciated.

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  • $\begingroup$ Is this post and its answers helping? $\endgroup$ Nov 1, 2020 at 19:59
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    $\begingroup$ In the "question assumption" you talk about two bi-characters, but then the equations are about the same bi-character $b$. Also, in your argument for cyclic groups, the relation of the bi-characters $b$ and $c$ is unclear. Please clarify. $\endgroup$
    – GH from MO
    Nov 1, 2020 at 21:10
  • $\begingroup$ Sorry, I typed them wrong, the c should be b. I had edited the question $\endgroup$
    – enjuikuo
    Nov 1, 2020 at 22:02
  • $\begingroup$ I asked this since I think this statement is correct for the general finite abelian group but I can not prove it. However, I think to prove them should be easy. $\endgroup$
    – enjuikuo
    Nov 1, 2020 at 22:06
  • $\begingroup$ Sorry about that, I have edited. Basically, I assume we know one single bi-character has two values. One is x/p, another is y/q. I should write it more carefully. $\endgroup$
    – enjuikuo
    Nov 2, 2020 at 5:32

1 Answer 1

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You can choose integers $m_1,m_2,n_1,n_2$ so that $m_1$ and $n_1$ are coprime to $p$ and $m_2$ and $n_2$ are coprime to $q$, and such that $n_1m_2b(a_1,b_2)=0=n_2m_1b(a_2,b_1)$.

Then $$b(n_1a_1+n_2a_2,m_1b_1+m_2b_2)=\frac{n_1m_1x}{p}+\frac{n_2m_2y}{q}=\frac{n_1m_1xq+n_2m_2yp}{pq},$$ whose numerator is coprime to $pq$.

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  • $\begingroup$ Thank you very much, let me check them. $\endgroup$
    – enjuikuo
    Nov 2, 2020 at 15:34

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