The Dimension Problem for Groups and Lie Rings
Thomas Sicking
University of Göttingen
[email protected]
Abstract
For a group G and its integral group ring ZG, one can define its n-th dimension subgroup by Dn(G) = G ∩ (1 +
∆n(G)), where ∆(G) denotes the augmentation ideal of ZG, i.e. the kernel of the map ZG → Z sending each
group element to 1. The identification of the dimension subgroups is known as the dimension subgroup problem.
For a Lie ring L one can similarly define the n-th Lie dimension subring by Dn(L) = L ∩ ∆n(L), where in this
setting ∆(L) ⊆ UL, the universal enveloping algebra of L, is the kernel of the map UL → Z sending each u ∈ L
to zero. Again, the identification of Lie dimension subrings is an interesting problem.
Introduction
The same for Lie Rings
For a group G one can define two descending series of normal subgroups. Firstly, we have the lower central
series {γn(G)}n≥1 defined by γ1(G) = G and γn+1(G) = [G, γn(G)], where [H, K] denotes the subgroup of
−1k −1hk with h ∈ H, k ∈ K. Secondly, there is the dimension
G generated by all commutators [h, k] =
h
P
series defined as follows. Let ZG = { g∈G ag g : ag ∈ Z} be the integral group ring of G. Then there is
a ring homomorphism ZG → Z mapping each g ∈ G to 1 ∈ Z. We denote its kernel by ∆(G), and define
Dn(G) = G ∩ (1 + ∆n(G)). It is obvious, that D1(G) = G = γ1(G) for every group G, and it can also be
shown that D2(G) = γ2(G) and D3(G) = γ3(G) for every group G. Generally, the equation
Recall that a Lie ring L is a Z-module with a bilinear operation [·, ·] satisfying
1. [x, x] = 0 and
2. [[x, y], z] + [[y, z], x] + [[z, x], y] = 0
for all x, y, z ∈ L. Every associative ring becomes a Lie ring with [x, y] := xy − yx. Furthermore, every
Lie ring L can be embedded into a unital associative ring, the universal enveloping algebra UL. Then every
Lie ring morphism L → L0 can be extended to a morphism of associative rings UL → UL0. In particular, the morphism L → 0 can be extended to a morphism ε : UL → U0 = Z. We define ∆(L) = ker(ε).
Define Dn(L) = L ∩ ∆n(L). In L, we again have the lower central series defined by γ1(L) = L and
γn+1(L) = [L, γn(L)]. By the equation
[x, y] = xy − yx
[g, h] − 1 = g −1h−1((g − 1)(h − 1) − (h − 1)(g − 1))
shows by induction on n, that γn(G) ⊆ Dn(G) for every group G and every n ∈ N. However, for n ≥ 4,
equality does not always hold, as first shown by Rips [6].
in UL, we can easily see that γn(L) ⊆ Dn(L) for every n ∈ N.
Some Results
Some Results
Using the embedding of a free group F on a set X into the unit group of the ring of non-commutative formal
power series in X with integer coefficients given by x 7→ x + 1 for x ∈ X, Magnus was able to prove that
Dn(F ) = γn(F ) for every n ≥ 1. Taking an arbitrary group G and write it as G = F/R with a free group
F , the dimension subgroup problem boils down to identifying the subgroup F ∩ (1 + ∆n(F ) + r), where
r = ZF (R − 1), and comparing it with γn(F )R. Using a descending series of intermediate groups, Sjogren
was able to prove the following, which is the best known general bound on dimension quotients.
Theorem 1 (Sjogren’s Theorem). Define
bi = lcm(1, . . . , i).
Then for every group G and every n ∈ N, the exponent of the group Dn(G)/γn(G) divides
cn =
n−2
Y
It turns out that the theories of groups and Lie rings run remarkably parallel. First of all, one can easily see
that the universal enveloping algebra of a free Lie ring F on a set X is the free associative algebra ZhXi on
X. Then ∆n(F ) is the ideal of all non-commutative polynomials involving no monomial of degree less than
n. Thus, an element u ∈ F ∩ ∆n(F ) has no monomials of degree less than n, and is also a Lie element, and
therefore it is in γn(F ). Thus we get
Theorem 6. Let F be a free Lie ring. Then Dn(F ) = γn(F ) for all n ≥ 1.
Again, for a given Lie ring L we can write it as L = F/R. Determining the dimension quotients Dn(L)/γn(L)
is then equivalent to determining (F ∩ (∆n(F ) + r))/(γn(F ) + R), where r = UF · R.
Theorem 7. For every Lie ring L and every n ∈ N, the exponent of the Lie ring Dn(L)/γn(L) viewed as an
abelian group divides
n−2
Y (n−2)
cn =
bi i ,
i=1
(n−2
i )
bi
.
with the same bi as in Theorem 1.
i=1
The sequence {cn} grows rapidly, e.g. c4 = 2, c5 = 48, c6 = 995328, . . .. But cn is only divisible by primes
up to n − 2, therefore one has the immediate corollary.
Corollary 2. Let G be a p-group for some prime p. Then Dn(G) = γn(G) for all n ≤ p + 1.
The dimension quotients D4(G)/γ4(G) and D5(G)/γ5(G) were characterized by Tahara [7]. With this
(very complicated) description of the fifth dimension quotient, he was able to prove that the exponent of
D5(G)/γ5(G) is actually a divisor of 6. This is much sharper than the bound arising from Sjogren’s Theorem,
but however it still has prime divisors 2 and 3.
The metabelian case
In the world of Lie rings, a stronger bound than the one known for metabelian groups has been established in
[5]. As in groups, for L = F/R one defines S = R + γ2(F ). Then F/S ' L/L0 is a finitely generated abelian
Lie ring. identifying certain ideals in UF , we were able to show the sharpest possible bound on the exponent
of Dn(L)/γn(L).
Theorem 8. Let L be a metabelian Lie ring. Then, for every n ∈ N, we have
2Dn(L) ⊆ γn(L).
However, it has not yet been possible to prove an equivalent of Theorem 4.
The metabelian case
For metabelian groups G, i.e. groups with abelian commutator subgroup, or eqivalently G00 =
[γ2(G), γ2(G)] = 1, stronger results than Sjogren’s result are known.
Take a finitely generated metabelian group G and write G = F/R with a free group F . As G is metabelian,
we have F 00 ⊆ R, and therefore, defining S = Rγ2(F ) ⊆ F , we see that G/G0 ' F/S, and is a finitely
generated abelian group. The group ring Z(F/S) is therefore a commutative noetherian ring. Also note that
[S, S] ⊆ R. Then one can prove
Theorem 3 (Gupta [2]). For every metabelian group G, the exponent of Dn(G)/γn(G) divides
2c∗n = 2
n−2
Y
bi ,
i=1
with bi as in Theorem 1.
This is again a much sharper bound than the one given by Sjogren, but has the same set of prime divisors.
Using the Artin-Rees Theorem, Gupta, Hales and Passi [4] were able to show
Theorem 4. If G is a finitely generated metabelian group, then there exists an integer n0 depending only on
the abelianization G/γ2(G) of G, such that Dn(G) = γn(G) for all n ≥ n0.
Counter Examples
Counter Examples
The examples for the group case by Rips and Gupta can be transferred to Lie rings as well.
Example 9. Let X = {r, a, b, c} and let F be the free Lie ring generated by X. Set
x0 = y0 = z0 = r,
xi = [xi−1, a], yi = [yi−1, b], zi = [zi−1, c] for i ≥ 1.
For n ≥ 4, let R(n) be the Lie ideal of F generated by the following elements:
22n−1r,
2n+2a − 4yn−3 − 2zn−3, 2nb + 4xn−3 − zn−3, 2n−2c + 2xn−3 + yn−3,
zn−2 − 4yn−2, yn−2 − 4xn−2,
xn−1, yn−1, zn−1,
[a, b, u], [a, c, u], [b, c, u] for all u ∈ F,
[xi, b], [xi, c], [yi, a], [yi, c], [zi, a], [zi, b], i ≥ 1,
[xi, xj ], [xi, yj ], [xi, zj ], [yi, yj ], [yi, zj ], [zi, zj ], i, j ≥ 0.
Then L(n) := F/R(n) is easily seen to be metabelian with γn(L(n)) = 0. Let
u := 22n−1[a, b] + 22n−2[a, c] + 22n−3[b, c].
Then u ∈ Dn(L(n)), but u 6= 0.
Open Problems
For every n ≥ 4, there is an example of a metabelian group G with Dn(G) 6= γn(G). However, in all known
examples, the exponent of Dn(G)/γn(G) is equal to 2. The first such example is due to Rips [6], and it has
been generalized by Gupta [3].
Example 5. Let X = {r, a, b, c} and let F be the free group generated by X. Set
x0 = y0 = z0 = r,
In general, it seems that working in Lie rings is easier. This is mainly because the Lie commutator is, contrary
to the group commutator, a bilinear operation. The main open problems are:
• Is there a finitely generated metabelian Lie ring with Dn(L) 6= γn(L) for infinitely many n?
• Is there a group G (or a Lie ring L) with γ4(G) = 1 (γ4(L) = 0) and D5(G) 6= 1 (D5(L) 6= 0)?
• Is there a group G (or a Lie ring L) such that Dn(G)/γn(G) (Dn(L)/γn(L)) has 3-torsion (or p-torsion for
some odd prime p)?
xi = [xi−1, a], yi = [yi−1, b], zi = [zi−1, c] for i ≥ 1.
For n ≥ 4, let R(n) be the normal subgroup of F generated by the following elements:
2n−1
2
r
,
n+2 −4
n
n−2
−2
−1
2
2
4
2
a yn−3zn−3, b xn−3zn−3, c x2n−3yn−3,
−4 , y
−4 ,
zn−2yn−2
x
n−2 n−2
xn−1, yn−1, zn−1,
[a, b, u], [a, c, u], [b, c, u] for all u ∈ F,
[xi, b], [xi, c], [yi, a], [yi, c], [zi, a], [zi, b], i ≥ 1,
[xi, xj ], [xi, yj ], [xi, zj ], [yi, yj ], [yi, zj ], [zi, zj ], i, j ≥ 0.
Then G(n) := F/R(n) is easily seen to be metabelian with γn(G(n)) = 1. Let
2n−1
2n−2
2n−3
2
2
2
g := [a, b]
[a, c]
[b, c]
.
Then g ∈ Dn(G(n)), but g 6= 1.
References
[1] L. Bartholdi and I. B. S. Passi: Lie dimension subrings, Int. J. Algebra Comput. 25, 1301–1325 (2015)
[2] N. Gupta: Sjogren’s Theorem for dimension subgroups - the metabelian case, Combinatorial Group Theory and Topology (Alta, Utah, 1984) 197–211, Ann. of Math. Stud., 111, Princeton University Press,
Princeton, NJ (1987)
[3] N. Gupta: The dimension subgroup conjecture, Bull. London Math. Soc., 22, 453–456 (1990).
[4] N. Gupta, A. Hales and I.B.S. Passi: Dimension subgroups of metabelian groups, J. reine und angew.
Math. 346, 194–198 (1984)
[5] I. B. S. Passi and T. Sicking: Dimension quotients of metabelian Lie rings, Int. J. Algebra Comput. 27,
251–258 (2017)
[6] E. Rips: On the fourth integer dimension subgroup, Israel J. Math. 12, 342–346 (1972)
[7] K. Tahara: The augmentation quotients of group rings and the fifth dimension subgroup, J. Algebra, 71,
141–173 (1981)