3 p-Groups

For our purposes a p-group is a finite group such that it has p^a 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 p^a 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, G_i = [G, G_{i - 1}]

Recall that G is nilpotent if G_i is the trivial group of order 1 for some i. Now if G is a p-group then the abelian groups G_i/G_{i + 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 andG_{i + 1} is the subgroup generated by[G, G_i] and the setG_i^p

By the above discussion it follows that G_i becomes trivial for large enough i; in addition, each G_i/G_{i + 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) = $$\oplus_{i}^{}$$ G_i/G_{i + 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

G_i/G_{i + 1}×G_j/G_{j + 1}$$\to$$G_{i + j}/G_{i + 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

$$\sum_{\sigma\in S_{p^n-1}}^{}$$ad(a_$\sigma$(1))oad(a_$\sigma$(2))o ^... oad(a_{$\sigma$(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 x_i be the elements of G₀/G₁ $\subset$ L(G) corresponding to the finitely many generators g_i of G. Then for any $\rho$ a commutator on the x_i 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 x_i 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].