3 added 136 characters in body

Let $G \to GL_N (C)$ be the regular representation. Pick a nontrivial subgroup $Z/p \subset G$ for a prime $p$ and consider the composition $Z/p \to GL_N(C)$, inducing $BZ/p \to BGL_N (C)$. If we can show that this map is nonzero in arbitrarily high cohomological degree, the theorem is proven. Let $L_k$ be the $1$-dim representation of $Z/p$ with the generator acting by $e^{2 \pi i k/ p}$. The restriction of the regular representation of $G$ to $Z/p$ is a multiple (say $m$ times) of the sum

$$L_0 \oplus L_1 \ldots \oplus L_{p-1}.$$

Let $x \in H^2 (Z/p)$ be the first Chern class of $L_1$; this is a generator. The Since $L_i$ is the $i$th tensor power of $L_1$, the total Chern class of $L_i$ is $1+ix$. Therefore, the total Chern class of the regular representation on $Z/p$ isthen

$$((1+x)(1+2x) \ldots 1+(p-1)x))^m.$$

In particular, the $m(p-1)$st Chern class is

$$z=\prod_{k=1}^{p-1} k^m x^{m(p-1)} \neq 0,$$

the latter because $p$ is a prime. Because $H^{\ast} (Z/p; Z) = Z[x]/(px)$, all powers of $z$ are nonzero. Therefore: write $|G|=pm$, $p$ prime. Then $H^{2 m (p-1)k } (G) \neq 0$ for all $k\geq 1$.

2 added 97 characters in body

Let $G \to GL_N (C)$ be the regular representation. Pick a nontrivial subgroup $Z/p \subset G$ for a prime $p$ and consider the composition $Z/p \to GL_N(C)$, inducing $BZ/p \to BGL_N (C)$. If we can show that it this map is nonzero in arbitrarily high cohomological degree, the theorem is proven. Let $L_k$ be the $1$-dim representation of $Z/p$ with the generator acting by $e^{2 \pi i k/ p}$. The restriction of the regular representation of $G$ to $Z/p$ is a multiple (say $m$ times) of the sum

$$L_0 \oplus L_1 \ldots \oplus L_{p-1}.$$

Let $x \in H^2 (Z/p)$ be a generator. The total Chern class of the regular representation on $Z/p$ is then

$$((1-x)(1+2x) ((1+x)(1+2x) \ldots 1+(p-1)x))^m.$$

In particular, the $m(p-1)$st Chern class is

$$z=\prod_{k=1}^{p-1} k^m x^{m(p-1)} \neq 0,$$

the latter because $p$ is a prime. Because $H^{\ast} (Z/p; Z) = Z[x]/(px)$, all powers of $z$ are nonzero. Therefore: write $|G|=pm$, $p$ prime. Then $H^{2 m (p-1)k } (G) \neq 0$ for all $k\geq 1$.

1

Let $G \to GL_N (C)$ be the regular representation. Pick a nontrivial subgroup $Z/p \subset G$ for a prime $p$ and consider the composition $Z/p \to GL_N(C)$, inducing $BZ/p \to BGL_N (C)$. If we can show that it is nonzero in arbitrarily high cohomological degree, the theorem is proven. Let $L_k$ be the $1$-dim representation of $Z/p$ with the generator acting by $e^{2 \pi i k/ p}$. The restriction of the regular representation of $G$ to $Z/p$ is a multiple (say $m$ times) of the sum

$$L_0 \oplus L_1 \ldots \oplus L_{p-1}.$$

Let $x \in H^2 (Z/p)$ be a generator. The total Chern class of the regular representation on $Z/p$ is then

$$((1-x)(1+2x) \ldots 1+(p-1)x))^m.$$

In particular, the $m(p-1)$st Chern class is

$$z=\prod_{k=1}^{p-1} k^m x^{m(p-1)} \neq 0,$$

the latter because $p$ is a prime. Because $H^{\ast} (Z/p; Z) = Z[x]/(px)$, all powers of $z$ are nonzero.