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pag 1: \item{\dash}{We compute the ring structure of~$\TH*(G,\Zmod{p^a}\times\Zmod{p^b})$ for $p$~an odd prime.} \item{\dash}{We compute the structure of~$\TH*(G,\Zmod p\times\Zmod p)$ both as a ring and as an algebra over~$\TH*(G,\Z)$.} becomes \item{\dash}{We compute the ring structure of~$\TH*(\Zmod{p^a}\times\Zmod{p^b},\Z)$ for $p$~an odd prime.} \item{\dash}{We compute the structure of~$\TH*(\Zmod p\times\Zmod p,\Zmod p)$ both as a ring and as an algebra over~$\TH*(\Zmod p\times\Zmod p,\Z)$.} pag 5: An abelian group (also called {\defnt module}) is a $G$-module for $G=\{0\}$. becomes An abelian group (also called {\defnt module}) is a $G$-module for $G=\{e\}$. The element~$(e)$ is the ring unity. The ring is abelian if and only if~$G$ is. becomes The element~$(e)$ is the identity element of the ring. The ring is commutative if and only if~$G$ is. The element~$N=\sum_{g\in G}(g)$ of~$\Z[G]$ is called the {\defnt norm element}. becomes If $G$ is finite, then the element~$N=\sum_{g\in G}(g)$ of~$\Z[G]$ is called the {\defnt norm element}. pag 6: Given a diagram of $G$-modules and $G$-maps with exact bottom row, if $P$~is projective, then there exists a $G$-map~$f$ such that the following diagram commutes: This property actually characterizes projective modules.\par becomes Projective modules are characterized by the following property: given a diagram of $G$-modules and $G$-maps with $P$~projective and exact bottom row, then there exists a $G$-map~$f$ such that the following diagram commutes: with~$P_i$'s projective. becomes with~$P_i$'s projective and the trivial action of~$G$ on~$\Z$. pag 12: The cohomology class of the unity of~$R$ becomes the unity in the ring~$\TH*(G,R)$ and the class of an invertible element of~$R$ becomes invertible in~$\TH*(G,R)$, since the cup product on~$\TH0(G,R)\tensor\TH0(G,R)$ is induced by the product in~$R$.\par becomes The cohomology class of the identity element of~$R$ becomes the identity element in the ring~$\TH*(G,R)$ and the class of an invertible element of~$R$ becomes invertible in~$\TH*(G,R)$, since the cup product on~$\TH0(G,R)\tensor\TH0(G,R)$ is induced by the product in~$R$.\par pag 13: after Lemmma: Any finite cyclic group has periodic cohomology. In fact, if $g$~is a generator of~$G\isom\Zmod n$ and $N$~is the norm element in~$\Z[G]$, then $$\dots\arrow\Z[G]\map{(g)-(e)}\Z[G]\map{N}\Z[G]\map{(g)-(e)}\Z[G]\map{N}\Z[G]\arrow\dots$$ is a complete resolution and one has $$\TH i(G,M)\isom\TH{i+2}(G,M)$$ for any $G$-module~$M$ and any integer~$i$.\par pag 17: $$\TH i(G,\Z)\tensor\TH{-i}(G,\Z)\arrow\TH{-1}(G,\Z)$$ becomes $$\TH i(G,\Z)\tensor\TH{-i}(G,\Z)\arrow\TH0(G,\Z)$$ pag 18: As seen in \periodiclemma, integral duality implies that each element of the integral duality ring with integer coefficients of order the order of the group is a unit in the ring. becomes As seen in \periodiclemma, integral duality implies that each element of the Tate cohomology ring with integer coefficients of order the order of the group is a unit in the ring. pag 21: For at least one~$i$ we have~$l_i\ge d_i$. becomes For at least one~$i$ we have~$l_i\ge ld_i$. It is possible to decompose $a^Nb$ as a product of $a^{ll_i}\alpha_i^{l}$ times a remaining factor. But up to sign, we have $$a^{ll_i}\alpha_i^{l}=(a^{l_i}\alpha_i)^l.$$ Since~$a^{l_i}\alpha_i\in\ppart{\TH0(G,\Z)}$ is not a unit, it must be a multiple of~$p$ and so its~$l$-th power is divided by~$p^l$ and hence is zero in~$\ppart{\TH0(G,\Z)}$. becomes It is possible to decompose $a^Nb$ as a product of $a^{ld_i}\alpha_i^{ld}$ times a remaining factor. But up to sign, we have $$a^{ld_i}\alpha_i^{ld}=(a^{d_i}\alpha_i^d)^l.$$ Since~$a^{d_i}\alpha_i^d\in\ppart{\TH0(G,\Z)}$ is not a unit, it must be a multiple of~$p$ and so its~$l$-th power is divided by~$p^l$ and hence is zero in~$\ppart{\TH0(G,\Z)}$. \corollary{}{Suppose $G$ is a finite group with non $p$-periodic cohomology for a~$p$ dividing its order. Then any element~$a\in\dirsum_{i<0}\TH i(G,\Z)$ is nilpotent.} becomes \corollary{}{Suppose $G$ is a finite group with non $p$-periodic cohomology for a~$p$ dividing its order. Then any element~$a\in\dirsum_{i<0}\ppart{\TH i(G,\Z)}$ is nilpotent.} This is particularly easy if we have an immersion of~$\TH*(G,\Z)$, or at least of one of its $p$-parts. We first need the following easy lemma: becomes This is particularly easy under the equivalent conditions of the following easy lemma: pag 22: \proof{By the previous lemma, the hypothesis implies the existence of a submodule isomorphic to~$\TH*(G,\Z)$, which is not noetherian.} becomes \proof{By the previous lemma, the hypothesis implies the existence of a submodule isomorphic to~$\ppart{\TH*(G,\Z)}$, which is not noetherian.} pag 26: $$\max(0,d'-n+1)\le\prod_{i=0}^{l'}a_i\le d'.$$ becomes $$\max(0,d'-n+1)\le\prod_{i=0}^{l'}d_i\le d'.$$ pag 29: \proposition{}{Let $G$ be a finite group with non $p$-periodic cohomology such that the ordinary cohomology ring~$\H*(G,\Zmod p)$ is generated by elements of degree at most~$2$. becomes \proposition{}{Let~$p$ be an odd prime and let~$G$ be a finite group with non $p$-periodic cohomology such that the ordinary cohomology ring~$\H*(G,\Zmod p)$ is generated by elements of degree at most~$2$. $$0\arrow\Z\map p\Z\map\pi\R\arrow0$$ becomes $$0\arrow\Z\map p\Z\map\pi\Zmod p\arrow0$$ pag 30: Such a set consists of products~$c_1c_2$ with~$c_1$ of degree~$1$ and~$c_2$ of degree~$2$, as there is a set of multiplicative generators of~$\H*(G,\R)$ of degree~$1$ and~$2$. becomes Such a set consists of products~$c_1c_2$ with~$c_1$ of degree~$1$ and~$c_2$ of degree~$2$, as there is a set of multiplicative generators of~$\H*(G,\Zmod p)$ of degree~$1$ and~$2$. pag 39: after For abelian $p$-groups of higher rank, one can exploit Chapman's results on generators of the ordinary cohomology ring with coefficients in~$\Z$ to derive the Tate product structure almost everywhere.\par put Throughout this chapter, $p$~denotes an odd prime.\par pag 40: \theorem{}{Let~$G=\prod_{i=1}^l\Zmod{p^{c_i}}$. For all~$i>0$ let's call~$\dual{\alpha_l^i}\in\TH{-2i}(G,\Z)$ the dual of~$\alpha_l^i$ according to a basis of~$\TH{2i}(G,\Z)$ containing~$\alpha_l^i$. becomes \theorem{}{Let~$G=\prod_{i=1}^l\Zmod{p^{c_i}}$. For all~$i>0$, call~$\dual{\alpha_l^i}\in\TH{-2i}(G,\Z)$ the dual of~$\alpha_l^i$ according to a basis of~$\TH{2i}(G,\Z)$ containing~$\alpha_l^i$.pag 6: $$\matrix{ &&P\cr &&\downarrow&\searrow\cr A&\arrow&B&\arrow&C\cr }$$ becomes $$\matrix{ &&P\cr &&\downarrow&\semmap{}{0}\cr A&\arrow&B&\arrow&C\cr }$$ $$\matrix{ &&P\cr &\swmmap{f}{}&\downarrow&\searrow\cr A&\arrow&B&\arrow&C\cr }$$ becomes $$\matrix{ &&P\cr &\swmmap{f}{}&\downarrow&\semmap{}{0}\cr A&\arrow&B&\arrow&C\cr }$$ pag 18: \proposition{}{Let $p$ be a prime and let $p^{n_1}$ and $p^{n_2}$ be the orders of two different cyclic direct summands of~$\TH{i}(G,\Z)$ as an abelian group. Call $p^{n_p}$ the biggest power of~$p$ dividing~$\card G$. Then $n_1+n_2\le n_p$. If $i$~is odd, then both~$n_1$ and~$n_2$ are not greater than~$n_p/2$. } becomes \proposition{}{Let $p$ be a prime and let $p^{n_1}$ and $p^{n_2}$ be the orders of two different cyclic direct summands of~$\TH{i}(G,\Z)$ as an abelian group. Call $p^{n_p}$ the biggest power of~$p$ dividing~$\card G$. Then $n_1+n_2\le n_p$. If $i$~is odd and $p$~is an odd prime, then both~$n_1$ and~$n_2$ are not greater than~$n_p/2$. } In case $i$~is odd, suppose there is an element~$a$ of order~$p^{n_1}$ greater than~$\sqrt{p^{n_p}}$. Consider an element~$b\in\TH{-i}(G,\Z)$ such that~$ba={\card G/\ord a}$. Then, as above, $$abab={(\card G)^2\over(\ord a)^2},$$ which is not~$0$. But by the anticommutativity of the cup product, as~$i$ is odd, $a^2$ must be zero and then so must be the product~$abab$. becomes Let~$i$ be odd, let~$p$ be an odd prime and suppose there is an element~$a$ of order~$p^{n_1}$ greater than~$\sqrt{p^{n_p}}$. Consider an element~$b\in\TH{-i}(G,\Z)$ such that~$ba={\card G/\ord a}$. Then, as above, $$abab={(\card G)^2\over(\ord a)^2},$$ which is not~$0$. But by the anticommutativity of the cup product, as~$i$ is odd, $a^2=-a^2$ must be zero, since its order is odd, and then so must be the product~$abab$. pag 19: \corollary{}{There can be at most one cyclic direct summand of~$\TH i(G,\Z)$ of order greater than~$\sqrt{\strut\card G}$. If $i$~is odd, then all elements (and hence all cyclic summands) of~$\TH i(G,\Z)$ have order at most~$\sqrt{\strut\card G}$. } becomes \corollary{}{There can be at most one cyclic direct summand of~$\TH i(G,\Z)$ of order greater than~$\sqrt{\strut\card G}$. If $i$ and the order of~$G$ are odd, then all elements (and hence all cyclic summands) of~$\TH i(G,\Z)$ have order at most~$\sqrt{\strut\card G}$. } pag 24: $$\eqalign{ K_1&:=\{(i,j)\ |\ i+j=r\}\cr K_2&:=\{(i,j)\ |\ i,j\ge0\}\cr K_3&:=\{(i,j)\ |\ r-n+1