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I'm looking for a proof of a theorem of Swan [1, Theorem 3]:

If $G$ is a finite group and $P$ a finitely generated projective $\mathbb{Z}G$-module, then $\mathbb{Q}\otimes_\mathbb{Z}P$ is a free $\mathbb{Q}G$-module.

The cited paper merely says: This can be proved by computing the character of $G$ on $\mathbb{Q}\otimes_\mathbb{Z}P$.

Unfortunately, I don't see how this hint shows freeness.

(1) Swan, Richard G. Projective modules over finite groups. Bull. Amer. Math. Soc. 65 (1959), no. 6, 365--367

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    $\begingroup$ It is also exercise 16.4 in Serre's "Linear representations of finite groups". $\endgroup$
    – Chris Wuthrich
    Commented May 1, 2016 at 8:37
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    $\begingroup$ Tensor the $\mathbb{Z}G$-module with $\mathbb{Z}_{p}$ for any prime $p$. Then it is a projective $\mathbb{Z}_{p}G$-module, so its character vanishes on each element of order divisible by $p$. Since $p$ is arbitrary, the character afforded by the $\mathbb{Q}G$-module vanishes on all non-identity elements of $G$, hence its character is an integer multiple of the regular character. By Maschke, it is free. $\endgroup$
    – Geoff Robinson
    Commented May 1, 2016 at 10:35
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    $\begingroup$ This argument shows that if the order of an element $g\in G$ is $n=p_1^{a_1}\cdots p_r^{a_r}$ then $\phi(g)$ is divisible by $p_1\cdots p_r$. How do you deduce that it is zero? $\endgroup$
    – Ehud Meir
    Commented May 1, 2016 at 22:55
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    $\begingroup$ @EhudMeir : Here $\mathbb{Z}_{p}$ denotes the $p$-adics, (not the field with $p$ elements ) and characters afforded by projective $\mathbb{Z}_{p}G$-modules do genuinely vanish on all $p$-singular elements of $G$ ( see Curtis and Reiner (1962 or later two part edition, for example)). I retyped this comment. $\endgroup$
    – Geoff Robinson
    Commented May 2, 2016 at 18:08

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