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Let $F$ be a field and $G$ be a finite group. Let $V,W$ be two $G$-representations of the same dimension. I am interested to find all modules $U$ with the property that $V\otimes U$ and $W\otimes U$ are always isomorphic, whenever $V,W$ have the same dimension as an $F$ -vector space.

Free modules have this property, since then both $V\otimes U$ and $W\otimes U$ are also free. That seems like a rather specific property of free modules. Are any other representations with this property?

More precisely, let $I$ denote the Kernel of the ring homomorphism $K_0(FG)\to K_0(F)$. Is the annihilator ideal $\mbox{Ann}(I)=\{a\in K_0(FG)\mid ai=0 \; \forall i \in I\}$ generated by $[FG]$ ?

Let us have a look at one example: For $F_2C_3$ we have $K_0(F_2C_3)\cong F_2[V]/(V-2)(V+1)$ where V is the nontrivial 2-dimensional representation, $I=(V-2)$ and $\mbox{Ann}(I)=(V+1)$. And $V+1$ is exactly the free $FG$-module.

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    $\begingroup$ Could you state the property? One possibility, free and isomorphic, is answered by Jeremy. Another possibility, just isomorphic, may have a different answer. $\endgroup$
    – Bugs Bunny
    Commented Oct 16, 2023 at 6:03
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    $\begingroup$ I edited the question to make it more clear. But I think Jeremy's answer applies to both variants. $\endgroup$
    – HenrikRüping
    Commented Oct 16, 2023 at 6:23

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Suppose $U$ has this property. Let $V=FG$ and let $W$ be the direct sum of $|G|$ copies of the trivial representation.

Then $V\otimes U$ is free of rank $\dim(U)$, $W\otimes U$ is a direct sum of $|G|$ copies of $U$, and $V\otimes U\cong W\otimes U$, so by the Krull-Schmidt theorem $U$ must be free.

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  • $\begingroup$ I dont understand the last step. For a general ring $R$ it could happen that $V\oplus V$ is free, although $V$ is not. For example, for the ring $M_2(F_2)$ and $V = F_2^2$ this is the case. Krull-Schmidt (e.g. any module can be written uniquely as a sum of indecomposables) should still hold for $M_2(F_2)$. Maybe there is a reason that this cannot happen for group rings, and I just dont see it yet. $\endgroup$
    – HenrikRüping
    Commented Oct 11, 2023 at 8:42
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    $\begingroup$ @HenrikRüping For the $K$-theory version of the question, the $K$-theory group is freely generated by irreducibles or indecomposables (depending on the definition) so in particular is torsion-free, and this argument shows that the class of $U$ is a scalar multiple of the class of $FG$. If $F$ has characteristic zero then the trivial representation occurs in $FG$ with multiplicity one, showing that the class of $U$ is an integer multiple of the class of $FG$. I guess in characteristic $p$ there is the problem of making the tensor product in $K$-theory wel-defined. $\endgroup$
    – Will Sawin
    Commented Oct 11, 2023 at 9:57
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    $\begingroup$ Alternatively, and this works in any characteristic, for $F$ the trivial module $\dim\operatorname{Hom}(F,FG)=1$, so if $[V]=\alpha[FG]$ is a rational multiple of $[FG]$ in the representation ring, then $\dim\operatorname{Hom}(F,V)=\alpha$ is an integer. $\endgroup$
    – Jeremy Rickard
    Commented Oct 11, 2023 at 10:12

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