Generalized Burnside Problem: Suppose that $G$ is a finitely generated group where every element has finite order. Is $G$ a finite group?
Burnside Problem: Suppose that $G$ is a finitely generated group, such that there exists an $N$ for which $g^N=1$ for all $g\in G$. Is $G$ finite?
It turns out that the answer to both these questions is no, and in this blog post we will explore how using a detour through the theory of algebras will help us settle this question!
In 1941, Kurosh formulated the following question. Suppose that $F$ is a field, $A_F$ is an associative $F$ algebra (not assuming a unit) and that $A$ is finitely generated. An element $a\in A$ is said to be nilpotent if $\exists n\geq 1 \,\,a^n=0$, and an algebra is called a nilalgebra if all of its elements are nilpotent. The Kurosh problem is
Kurosh nilalgebra problem: Is every finitely generated nilalgebra finite dimensional?
Kurosh noticed that if there is a counterexample to this question for a field of positive characteristic, then there is a counterexample to the Generalized Burnside Problem. Here is the argument for that.
Theorem: If there is a counterexample to the Kurosh problem, there is counterxample to the generalized Burnside problem.
[Proof]
Suppose that there is some finitely generated infinite-dimensional nilalgebra over $F$ where $\mathrm{char}(F)=p>0$. Let $A=\langle a_1,\ldots, a_m\rangle$ be such an example. Let $\hat{A}=A\oplus F\cdot 1$ be the unital hull of $A$: the algebra you get by adding a unit. (Since $A$ was nil, it couldn't possibly have had a unit to begin with!) Now notice that for all $a\in A$, $1+a$ is invertible since $1-a+a^2-a^3+\cdots$ is a finite sum and an inverse. Thus $1+a\in G(\hat A)$, the group of units of $\hat A$. Consider the subgroup of the group of units
$$G=\langle 1+a_1,\ldots,1+a_m\rangle\subset 1+A\subset G(\hat{A}).$$
Now every element in $1+A$ is torsion: we know there is some $n(a)$ where $a^{n(a)}=0$. Choose $k$ so that $p^k\geq n(a)$. Then $$(1+a)^{p^k}=1+a^{p^k}=1$$ since we are in characteristic $p$. Therefore $G$ is a finitely generated torsion group. If we can show it is infinite, we have a counterexample to the Generalized Burnside Problem.
Suppose to the contrary that $|G|=d$. Let $g_1\ldots g_d$ be a product of elements of $G$. Then $\{1,g_1,g_1g_2,\ldots,g_1\ldots g_d\}$ is a set of $d+1$ elements, and by the pigeonhole principle two of the elements must coincide: say $g_1\ldots g_i=g_1\ldots g_j$ $(j>i)$. Then $g_{i+1}\ldots g_j=1$. So we can remove a segment from the original word $g_1\ldots g_d$ without changing the group element it represents. In particular any element of $G$ is a product of $<d$ generators (we don't need inverses because the group is torsion).
In other words,
$$\forall g\in G\,\, g=(1+a_{i_1})\cdots (1+a_{i_k})\,\,\,\,(k<d)$$
But now consider an arbitrary product of $d$ generators of $A$: $a_{i_1}\ldots a_{i_d}$. Then
$$(1+a_{i_1})\cdots (1+a_{i_d})=(1+a_{j_1})\cdots(1+a_{j_k})\,\,\,\, k<d.$$
Now expanding this out, we see that $a_{i_1}\ldots a_{i_d}$ is a linear combination of products of generators of length $<d$. This implies $A$ is finite dimensional, which is a contradiction. $\Box$
So it now suffices to find a counterexample to the Kurosh problem. We begin with some graded algebra conventions. Let $F\langle X\rangle$ be the free associative algebra on a finite alphabet $X=\{x_1,\ldots ,x_m\}$. This is graded by length of words:
$$F\langle X\rangle =F\cdot 1+F\langle X\rangle_1+F\langle X\rangle_2+\cdots$$.
We say a subspace $H\subseteq F\langle X\rangle$ is graded if $H=\sum_i(H\cap F\langle X\rangle_i)$. If $R$ is a graded subspace, we can form the quotient algebra $\langle X\,|\, R=(0)\rangle$. Then since the ideal generated by $R$, $I=\mathrm{id}(R)$, is also graded, the factor algebra $$A=F\langle X\rangle/I=F\cdot 1+A_1+A_2+\cdots$$ is graded as well. (Note we assume all relations have degree $\geq 2$.)
Definition: Given a graded vector space $V=\oplus_{i\geq 0} V_i$ in which each graded summand is finite dimensional, define the Hilbert series $H_V(t)=\sum_{i\geq 0} (\mathrm{dim} V_i)t^i$. Note this is a formal power series.
We can order formal power series lexicographically by coefficients.
Proposition (Golod-Shafarevich): In the situation above, $$H_A(t)(1-mt+H_R(t))\geq 1,$$ where the comparison is of formal power series.
We will come back to the proof of this proposition at the end of the post. According to Zelmanov, it was given to Golod as a homework problem by Shafarevich!
Now, suppose that the proposition is true, and moreover suppose that we can find a $t_0$ with $0<t_0<1$ such that
- $H_R(t_0)$ converges.
- $1-mt_0+H_R(t_0)<0$.
The construction of a finitely generated nilalgebra which is infinite dimensional:
Let $F$ be a countable field. Then the free algebra on $X$ without identity is also countable:
$$F_0\langle X\rangle =\{f_1,f_2,\ldots,\}.$$
Let $d_1=2$. Let $d_2>\mathrm{deg}(f_1^{d_1})$, and $d_3>\mathrm{deg}(f_2^{d_2})$ etc. Let $R$ be the set of homogeneous components of $f_i^{d_i}$. By choice of the $d_i$'s, every degree occurs no more than once.
Therefore $H_R(t)\leq t^2+t^3+\cdots$, and so $H_R(t)$ converges when $0<t<1$. In order to satisfy the criterion 2. above, we need $1-mt_0+H_R(t_0)<0$, which will be satisfied if
$1-mt_0+t_0^2+t_0^3+\cdots<0$ for some $0<t_0<1$. This is an easy precalculus exercise!
So the algebra $A=\langle F_0\langle X\rangle\,|\, R=(0)\rangle$ is infinite dimensional. Moreover it is nil by construction. Every element of $F_0\langle X\rangle$ has a power which is trivial.
Returning to the proof of the Golod-Shafarevich inequality:
Let $r_i=\mathrm{dim}(R_i), \mathrm{dim}(A_i)=a_i$ and, letting $I=\mathrm{id}(R)$, $b_i=\dim(I_i)$. Note that $a_i+b_i=\mathrm{dim} F_0\langle X\rangle_i=m^i$. The Golod-Shafarevich inequality can now be rewritten as
$$
a_n-ma_{n-1}+\sum_{i+j= n} a_ir_j\geq 0
$$
We use the following basic observation: $I\subseteq XI+RF\langle X\rangle$. This follows since a general element of $I$ is a linear combination of elements of the form $u_iRv_i$. If $R$is not at the left of the word, then $u_iRv_i\in XI$. Otherwise, it is in $RF\langle X\rangle$. Now choose a complementary graded subspace $S$ so that $F\langle X\rangle=I+S$. Then $\mathrm{dim}(S_i)=a_i$. Moreover, we can now say
$$I\subseteq XI+RS$$
Thus $b_n\leq mb_{n-1}+\sum_{i+j=n}r_ia_j$, which is $$(m^n-a_n)\leq m(m^{n-1}-a_{n-1})+\sum_{i+j=n}r_ia_j,$$
which is equivalent to the desired inequality.
No comments:
Post a Comment