In this note I show that for the cohomology of a group with coefficients in Z/pZ, the transfer homomorphism commutes with the Steenrod reduced power operations. Suppose that G is a group and H a subgroup of finite index. Let P denote the cyclic group of prime order p considered as a subgroup of Sp, the symmetric group on p symbols. As in [2 ], we denote by S 5 G, the semi direct product of the permutation group S with the group GP where the former acts on the latter by permuting the factors. Then P X G, for example, may be considered a subgroup of P S G by imbedding G in GP via the p-fold diagonal map. If this is done, the basic Steenrod construction may be described simply: Given a E Hq(G, Z/pZ), form the element 1 5 a E Hvq(P S G, Z/pZ) as follows. Let X be a G-projective resolution of Z and suppose f cHomG(Xq, Z/pZ) represents a. Let W be a P-projective resolution of Z with augmentation e. Then W0XP becomes a P 5 G projective resolution of Z and e0fP is a PS G homomorphism of W0XP into Z 0 (Z/p)P zZ/pZ which is in fact a cocycle whose cohomology classdenoted by 1 5 a-depends only on a. (More generally, we may replace P by any subgroup S of Sp, but then it is necessary to include a sign in the action of S on ZO (Z/pZ)P.) Denote by P(a) =PG(a) CHPq(P X G, Z/pZ) the restriction of 1 5 a to the subgroup P X G. P(a) is the basic object from which the reduced powers are constructed. Suppose next that j-Hq (H, Z/pZ). Let T= fr} be a left transversal of H in G. Since H is a subgroup of G, X is also an H-projective resolution of Z and if gCHomH(Xq, Z/pZ) represents 3, then 17ET 'rg represents trH G(O). (See [1, Chapter XII, ?8].) We wish to study P(tr /) =res(1 5 (tr /)). As above, 1 5 tr / is represented by
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