 square-free group.

Let $G$ be a square-free group. Then by Theorem \[meta\] and the result of Propostion \[det1\], the determinant of matrix of the bilinear form $(-,-)_{\phi} : \mu_{\mathbb{Z}}(G)\times \mu_{\mathbb{Z}}(G)\to \mathbb{Z}$ is $\pm 1$. There exist a non degenerate bilinear associative symmetric form for $\mu_{\mathbb{Z}}(G)$, so this algebra is symmetric. Conversely, let $G$ be a finite group and $p$ be a prime number such that $p^2 \mid |G|$, then $G$ has a $p$-subgroup $P$ of order $p^2$. We prove that all the associative symmetric bilinear form ${<}-,-{>}$ on $RB(P)$ are degenerate.

-   Suppose that $P = C_{p^2}$, let $B$ be the Burnside functors of $Mack_{\mathbb{Z}}(P)$, then there are $a,b,c\in \mathbb{Z}$ such that the matrix $M$ of ${<}-,-{>}$ in the usual basis of $B(G)$ is: $$M=\left(\begin{array}{ccc}a & b & c \\b & pb & pc \\c & pc & p^2c\end{array}\right),$$ If we reduce modulo $p$ this matrix, it is clear that the two last columns are proportional. So the $det(M)$ is divisible by $p$, so $B$ is not isomorphic to its $\mathbb{Z}$-linear dual $B^{*}$.

-   Suppose that $P=C_{p} \times C_{p}$. Let $B$ be the Burnside functors of $Mack_{\mathbb{Z}}(P)$. There are elements $a,b_1,b_2,\cdots,b_{p+1},c\in \mathbb{Z}$ such that the matrix $M$ of ${<}-,-{>}$ in the usual basis of $B(G)$ is: $$M:=\left(\begin{array}{cccccc}a & b_1 & \cdots & b_{p} & b_{p+1} & c \\b_{1} & pb_{1} & c & \cdots & c & pc \\\vdots & c & pb_{2} & \ddots & \vdots &  \\b_{p} & \vdots &  \ddots & \ddots & c & \vdots \\b_{p+1} & c & \cdots &  c & pb_{p+1} & pc \\c & pc & \cdots &  pc & pc & p^2c\end{array}\right)$$ By reducing this matrix modulo $p$ it is enough to look at the following $(p+1)\times (p+1)$ matrix: $$\left(\begin{array}{cccc}0 & c & \cdots & c \\c & 0 & \ddots & \vdots \\\vdots & \ddots & \ddots & c \\c & \cdots & c & 0\end{array}\right)$$ the sum of the lines is zero modulo $p$, so $det(M)$ is divisible by $p$.

Let $G$ be a finite group and let $p$ be a prime number such that $p^2 \mid |G|$. The proof