Topics in Cohomology of Groups_Serge Lang.pdf

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K[G]-module F = A | B, with the natural injection i and projec-
tion 7r in the sequence
A ~--~A
with rri = 1A, and both i, 7r are K[G]-homomorphisms. Since F is
K[G]-free, it follows that 1F = Sa(v) for some K-homomorphism
v. Then by Lemma 2.4,
1A = 7clFi = 7rSe(v)i = Sa(Trvi),
whence A is K[G]-regular. Conversely, let A be K-projective and
K[G]-regular. Let
71"
F---+ A ---~ O
be an exact sequence in Mod(K, G), with F being K[G]-free. By
hypothesis, there exists a K-morphism if : A --~ F such that
7riK = 1A, and there exists a K-morphism v : A --+ A such that
1A = SG(v). We then find
~s~(i~v) = s~(~i~v) = Sc(v) = 1~,
which shows that SG(iKv) splits % whence A is a direct summand
of a free module, and is therefore K[G]-projective. This proves the
proposition.
With the same type of proof, taking the trace of a projection,
one also obtains the following result.
Proposition 2.14. In Mod(G), a direct summand of a G-
regular module is also G-regular. In particular, every projective
module in Mod(G) is also G-regular.
Proof. The second assertion is obvious for free modules, whence
it follows from the first assertion for projectives.
For finite groups we have a modification of the embedding rune-
for defined previously for arbitrary groups, and this modification
will enjoy stronger properties. We consider the following two exact
sequences:
(3)
(4)
o~ Ia ~ Z[G] & Z ~0
o--, Z--~ Z[G] ~ Ya ~ O.