The problem is:
Let Z denote the set of all integers.
Consider Z/nZ = Zn as trivial G-module.
Show that there is a isomorphism between the First Homology Group H1(G, Zn) and the factor G/G'Gn;
Where G' is the commutator (derived) subgroup G'=[G,G] and Gn is the subgroup of G generated by all its n-th Powers.
- With this isomorphism proved, we want to conclude that if G is a finite p-Group, then H1(G,Fp) is a Fp-module (vector space) with dimension equal the number of least generators of G.
Useful answers to this question depend on what you can use. One might use the fact (Proposition 2.3 of Brown: Cohomology of Groups) that H^1(G,A) classifies the conjugacy classes of splittings of the extension 0 --> A --> semidirect product(A,G)-->A-->0. In your case, however, A is abelian and a trivial G-module. So the splittings should correspond to homomorphisms G-->Z_n. Such homomorphisms must map G' and G^n to zero.
The difficulty depends on your view of the (co)homology groups. Generally speaking, explicit formulas involving (co)cycles are not of great use. The best approach is to go back to the fundamental property of the (co)homology functor, as is done in Serre's "Local Fields", chap.VII, which is in my opinion the best introduction to the subject.
Let us look e.g. at homology. Given a group G and its group algebra A = Z[G], an exact sequence of Z[G]-modules o --> A --> B --> C --> 0 gives rise to an exact sequence of co-invariants ... --> AG --> BG --> CG --> 0, where H0 (G,M) := (M)G denotes the maximal quotient of M on which G acts as identity. But this sequence is not exact on the left, where it extends to an a priori infinite exact sequnce of homology groups ... --> H1(G,A) --> H1(G,B) --> H1(G,C) --> H0(G,A) --> ... The philosophy is that the homology functor is an automatic machinery which transforms a short exact sequence into an infinite exact sequence on the left. Think of the Taylor development of a function f at a point M, which you can write down without thinking, but the terms of the expansion give you valuable information on f itself in the neighbourhood of M.
1) Let us apply this to your specific problem. Denote by IG the augmentation ideal of A , i.e. the ideal generated by all the elements 1-g when g runs through G. Obviously H0(G, M) = M/IGM . Consider the natural exact sequence 0 --> IG --> A --> Z --> 0, where G acts by conjugation on G and as identity on Z, and the rightmost map consists in taking the sum of the coefficients of an element of the group algebra A. Taking homology gives ... --> H1(G, A) --> H1(G, Z) --> H0(G, IG) --> H0(G,A) -->... But A is a free A-module, hence H1(G, A) = 0, and the image of H0(G, IG) --> H0(G,A) is null practically by definition, so that H1(G, Z) = IG/ IG2 , and one can check "by hand" that the map g --> 1-g induces an isomorphism H1(G, Z) = G/G' (remember that G acts by conjugation on itself).
Consider now the exact sequence 0 --> Z --> Z --> Z/n --> 0, where the leftmost map is multiplication by n , hence is injective.Taking homology gives H1(G, Z) --> H1(G, Z) --> H1(G, Z/n) --> Z --> Z . Applying our preliminary result H1(G, Z) = G/G' , we get immediately that G/G'Gn = H1(G, Z/n) as desired.
2) Suppose now that G is a finite p-group, where p is a prime, and write Fp instead of Z/p. After 1), we know that G/G'Gp = H1(G, Fp). Both sides are finite abelian groups killed by p, so they can be viewed (when written additively) as Fp finite - dimensional vector spaces. Let d(G) denote the minimal number of generators of G. Passing to the quotient, we get obviously d(G) >= d(G/G'Gp ), and this last quantity is = dim G/G'Gp by elementary linear algebra.
The reverse inequality holds true, but its proof is somewhat independent from the previous homological considerations. Write G*=G/G'Gp. A theorem of M. Hall in group theory implies that a homomorphism f : G1 --> G2 is surjective iff the induced map f* : G1* --> G2* is surjective. The equality d(G) = dim G/G'Gp is known as the Burnside basis theorem.
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