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Atiyah in his famous paper , Characters and cohomology of finite groups, after proving completion of representation ring in augmentation ideal is the same as $ K(BG)$, gives bunch of corollaries of this main theorem. One of them that catches my interest is: For any finite non-trivial group $G$ there exists arbitrary large integer $n$ such that $H^n(G,\mathbb{Z})\neq 0 $. I just wonder if anyone can prove this without this powerful theorem.

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Could you clarify a qualifier in your statement. Is the theorem that for any non-trivial finite group there are infinitely many $n$ such that $H^n(G,\mathbb Z)$ is non-trivial? Or is there only one $n$, and it depends on the group $G$? –  Ryan Budney May 11 '11 at 20:29
    
No, I meant $H^n(G,\mathbb{Z})$ cannot vanish for all sufficiently large $n$. –  Sam Nariman May 11 '11 at 20:45
    
BTW: Swan also showed the stronger result: If $H^n(G,\mathbb{Z}) = 0$ for all sufficiently large odd $n$ then $G$ is trivial or cyclic. –  Ralph May 11 '11 at 23:57
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@Ralph: I do not believe this is literally true. The cohomology of the binary icosahedral group is concentrated in degrees divisible by $4$. –  Johannes Ebert May 12 '11 at 8:23
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@Johannes: You're right. The correct statement is that the p-Sylow subgroups are cyclic or (generalized) quaternion. –  Ralph May 12 '11 at 18:33

4 Answers 4

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. 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 $Z/p$ is

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

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 $k\geq 1$.

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And should that first term in the total Chern class be (1+x) ? –  Chris Gerig May 11 '11 at 22:12
    
thanks for spotting the typo. –  Johannes Ebert May 12 '11 at 8:19
    
Could you, Johannes, elaborate more? How did you compute chern class of $L_i$? –  Sam Nariman May 14 '11 at 20:00

Yes, Richard Swan is the first to prove it, The Nontriviality of the Restriction Map in the Cohomology of Groups (1959):

The restriction $res^G_H:H^i(G)\rightarrow H^i(H)$ is nonzero for an infinite number of values of $i>0$.

As a corollary, for any prime $p$ dividing $|G|$, the $p$-primary component $H^i(G)_{(p)}$ is nonzero for an infinite number of values of $i>0$.

He actually proves this in a more general case, where $G$ is a compact Lie group and $H$ is a closed subgroup (defining group cohomology with the classifying spaces $BG$ and $BH$). The proof uses basic cohomological and Lie group principles.

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I did not know that paper. Swan's argument begins similarly as the argument in my answer, but then it becomes more involved since a Lie group does not have a regular representation. Swan has to use a spectral sequence. –  Johannes Ebert May 12 '11 at 18:01
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Yea; both arguments are awesome. –  Chris Gerig Jan 10 '12 at 23:01

The first purely algebraic proof of this fact seems to be from Leonard Evens:

  • A Generalization of the Transfer Map in the Cohomology of Groups, Trans. Amer. Math. Soc. 108(1963), 54-65 [Theorem 3]

where he proves the result with help of his norm map. After having established the basic properties of the norm map, the proof is rather elementary: Let $C$ be a cyclic subgroup of prime order of $G$ and let $x$ be a generator of $H^2(C,\mathbb{Z})$. Then the powers of $y = N^G_C(x)$ yield non-trivial cohomology classes of $H^*(G,\mathbb{Z})$ in degrees divisible by $(G:C)$.

From a historical point of view the norm map already occured in disguise in Evens´ paper

  • The Cohomology Ring of a Finite Group, Trans. Amer. Math. Soc. 101(1961), 224-239

where he proves finte generation of the cohomology ring.

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One way is to use another powerful theorem :)

For example, one knows from a result of Dan Quillen that the Krull dimension of the cohomology ring with coefficients in a field with bad characteristic is the same thing as the $p$-rank of the group, that is, the maximal rank of an elementary abelian $p$-subgroup. This is nicely explained in Dave Benson's Representations and cohomology, vol. II, and is proved in [Quillen, Daniel. The spectrum of an equivariant cohomology ring. I, II. Ann. of Math. (2) 94 (1971), 549--572; ibid. (2) 94 (1971), 573--602. MR0298694 (45 #7743)]

It follows that for some prime $p$, the Krull dimension of $H^\bullet(G,\mathbb F_p)$ is positive, so it cannot vanish for $\bullet\gg0$, and then the same thing holds with integer coefficients. Indeed, this shows there are infinitely many non-zero degrees, answering Ryan's question in the comments to the main question.

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I realized we need not to use that powerful theorem of Quillen to show that $H^*(G,\mathbb{Z}_p)$ is nonzero for arbitrary large dimension. Actually by embedding $G$ in $U(n)$ for some $n$ we can show that $H^*(G,\mathbb{Z}_p)$ is finitely generated $\mathbb{Z}_p$-algebra. Using this observation, it is easy to see $H^*(\mathbb{Z}_p,\mathbb{Z}_p)$ is finitely generated $H^*(G,\mathbb{Z}_p)$-module. So we are done. –  Sam Nariman Mar 10 '12 at 2:24

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