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3 p-Groups

For our purposes a p-group is a finite group such that it has pa elements for some non-negative integer a. The Restricted Burnside problem for such groups can be stated as follows.

Problem 4 (Restricted Burnside Problem for p-groups)   Let G be a finite group with exponent pn which is generated by k elements. Then G has pa elements for some integer a. Is there a uniform bound a(k, n) for a?

To study p-groups we first note that these are nilpotent. We define the Central series for G

G1 = G and by induction on i, Gi = [G, Gi - 1]

Recall that G is nilpotent if Gi is the trivial group of order 1 for some i. Now if G is a p-group then the abelian groups Gi/Gi + 1 have order a power of p. Thus we can construct a finer series called the p-Central series for G a p-group

G1 = G andGi + 1 is the subgroup generated by[G, Gi] and the setGip

By the above discussion it follows that Gi becomes trivial for large enough i; in addition, each Gi/Gi + 1 is a vector space over $ \Bbb$Z/p$ \Bbb$Z for all smaller i. The $ \Bbb$Z/p$ \Bbb$Z-vector space L(G) is defined as

L(G) = $\displaystyle \oplus_{i}^{}$ Gi/Gi + 1

The non-commutative structure of G can be caught by a Lie algebra structure on L(G). We recall the definition of a Lie algebra.

Definition 1   A vector space L over a field k is said to be a lie algebra if there is a pairing [,] : L×L$ \to$L with the following properties

[x, y] = - [y, x] and [x,[y, z]] + [z,[x, y]] + [y,[z, x]] = 0

The Lie algebra structure on L(G) is given by

Gi/Gi + 1×Gj/Gj + 1$\displaystyle \to$Gi + j/Gi + j + 1

where the map is ($ \overline{x}$,$ \overline{y}$) $ \mapsto$ $ \overline{xyx^{-1}y^{-1}}$ (Check that this is well-defined!).

The above lie algebra has some additional structure. First is an identity proved by Higman [7]. If G has exponent pn then

$\displaystyle \sum_{\sigma\in S_{p^n-1}}^{}$ad(a$\scriptstyle \sigma$(1))oad(a$\scriptstyle \sigma$(2))o ... oad(a$\scriptstyle \sigma$(pn - 1)) = 0

as a map L(G)$ \to$L(G); here ad(a) : L$ \to$L for any element a in a Lie algebra is the map b $ \mapsto$ [a, b].

The second identity is proved by Sanov [11]. Let xi be the elements of G0/G1 $ \subset$ L(G) corresponding to the finitely many generators gi of G. Then for any $ \rho$ a commutator on the xi we have ad($ \rho$)pn = 0.

The main result of Zelmanov can formulated as follows.

Theorem 1 (Zelmanov)   Let L be any Lie algebra over $ \Bbb$Z/p$ \Bbb$Z which is generated as a Lie algebra by k elements xi such that we have the Higman and Sanov identities.Then L is nilpotent as a Lie algebra.

The interested reader can find this proof outlined in [13].

next up previous
Next: 4 The Proof of Up: The Work of Efim Previous: 2 Groups
Kapil Hari Paranjape 2002-11-22