By Janet E. Aisbett

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**Additional info for On K * Z/n) and K (Fq[t]/(t2)**

**Example text**

4 this leads to the following. 7: Theorem. H (SL Z/9; Z) = Z/8 © Z/9. K3Z/9 = Z/8 © Z/9. 3-4 (App): Lemma. If p = 2 and n is large and odd: (i) H ^ G L ^ ; Kj 8 M^) = 0. (ii) If U = cokerfSq1:^ + H2(M ; Z/2)), then for i = 0 or 1, H^GL Z/2; U) = Z/2. Proof: Consider the Bockstein sequences associated with the coefficient sequences: (a) M n >^SU HZ(Mn; Z/2) — » U , (b) U >-^> M # 9 M # — • H2(M ; Z/2), K w J n n n and where a (a b) = a 0 b + b 8 a. 2 to break the sequence from (a) into: 0 = [M*] GLnZ/2 > — [H2(Mn; Z/2)]GLnZ/2 -> [U] GL n Z/2 — 0, and 0 + H1(GL Z/2;U) -• H 2 (GL Z/2;M*) = Z/2 -> H 2 (GL Z/2; H2(M ;Z/2)) -> .

U.. if i / j. f. 1). 36 JANET E. AISBETT 2 whereas (1 + 2-2 ) E 1 mod 8. Hence I 2 eP. = u. , p odd 0 , p = 2. , j = r + i, i = s; 0 , else. 3) gives [f] = 7 at. % u. ,t, . 0 ii. . t.. ) 0 u. w.. ) 0 u. . 1:L li ij lj 22 iJ iJ iJ JJ ij 2 ~ ~ The restriction of [f] to H (M ; Z/p) 0 M is therefore seen to be I {6w 0 v } + I {(I w w 13 1J i,j i,j k l k kJ ) 0 v 1J : (i,j) f (1,1)}. 1). '1 v.. = -6w.. ,w, . 4)! v. = {Sq1 w. w. } ij = n J ik kji = j Wij When J. -w.. is a summand in the image of v..

10 injects 4 k 4 k £ 0 in H (G ; Z/p). And (App) has T\*I2 / 0 i n H 4 ( G 3 ; Z / 2 ) . o z n 2 i n t o H (G ;Z/2) f o r k > 3 ( 1 1 1 , 2 . 6 ) , TT*03 o ** _ When p = 2, K Z ^ 0. Since TT*H4(M ;Z/2) o And 6x, n i s i n ke r TT* K . K+J. while TT*03? is not, so 6x, and fr*a)9 are linearly independent in H (G ;Z/2) . Their p^-preimages are also linearly independent. 56 JANET E. 4: Proposition. 4 k (i) For k > 3 and p odd, Tor(H (G ;Z), Z/p) has SL Z/p- invariants Z/p © Z/p, generated by $x, and TT*6C .