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 nontrivial 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.

$\begingroup$ Could you clarify a qualifier in your statement. Is the theorem that for any nontrivial finite group there are infinitely many $n$ such that $H^n(G,\mathbb Z)$ is nontrivial? Or is there only one $n$, and it depends on the group $G$? $\endgroup$ – Ryan Budney May 11 '11 at 20:29

$\begingroup$ No, I meant $H^n(G,\mathbb{Z})$ cannot vanish for all sufficiently large $n$. $\endgroup$ – Sam Nariman May 11 '11 at 20:45

$\begingroup$ 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. $\endgroup$ – Ralph May 11 '11 at 23:57

3$\begingroup$ @Ralph: I do not believe this is literally true. The cohomology of the binary icosahedral group is concentrated in degrees divisible by $4$. $\endgroup$ – Johannes Ebert May 12 '11 at 8:23

3$\begingroup$ @Johannes: You're right. The correct statement is that the pSylow subgroups are cyclic or (generalized) quaternion. $\endgroup$ – Ralph May 12 '11 at 18:33
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
$$L_0 \oplus L_1 \ldots \oplus L_{p1}.$$
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
$$((1+x)(1+2x) \ldots 1+(p1)x))^m.$$
In particular, the $m(p1)$st Chern class is
$$z=\prod_{k=1}^{p1} k^m x^{m(p1)} \neq 0,$$
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 $H^{2 m (p1)k } (G) \neq 0$ for all $k\geq 1$.
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.

1$\begingroup$ 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. $\endgroup$ – Johannes Ebert May 12 '11 at 18:01

1
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), 5465 [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 nontrivial 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), 224239
where he proves finte generation of the cohomology ring.
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), 549572; ibid. (2) 94 (1971), 573602. 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 nonzero degrees, answering Ryan's question in the comments to the main question.

$\begingroup$ 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. $\endgroup$ – Sam Nariman Mar 10 '12 at 2:24