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, we know there are two 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.

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.

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, we know there are two 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 $$b(a_1\mathbb{I},b_1\mathbb{I})=\frac{x}{p}=(a_1b_1)b(\mathbb{I},\mathbb{I}),c(a_2\mathbb{I},b_2\mathbb{I})=\frac{y}{q}=(a_2b_2)b(\mathbb{I},\mathbb{I}) , (x,p)=(y,q)=1.$$

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.

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, we know there are two 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.

Setting: $G$ is a finite abelian group and any bicharacter on $G$（a bi-character on $G$ is a map $b:G \times G \to \mathbb{Q}/\mathbb{Z}$ such that $b(x+y,z)=b(x,z)+b(y,z),b(x,z+y)=b(x,z)+b(x,y).$） So now my question assumption is that: given two positive integers both greater than $2$, ie $p,q \geq 2.$ and $p,q$ are coprime and we know there are two 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:

Is there exists, $c,d \in G$ 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 $b(a_1\mathbb{I},b_1\mathbb{I})=\frac{x}{p}=(a_1b_1)b(\mathbb{I},\mathbb{I}),c(a_2\mathbb{I},b_2\mathbb{I})=\frac{y}{q}=(a_2b_2)b(\mathbb{I},\mathbb{I}) , (x,p)=(y,q)=1.$

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.

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, we know there are two 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 $$b(a_1\mathbb{I},b_1\mathbb{I})=\frac{x}{p}=(a_1b_1)b(\mathbb{I},\mathbb{I}),c(a_2\mathbb{I},b_2\mathbb{I})=\frac{y}{q}=(a_2b_2)b(\mathbb{I},\mathbb{I}) , (x,p)=(y,q)=1.$$

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.