\input fontmac \input mathmac \def\d{\,d} \def\bar{\overline} \def\graph{\mathop{\rm graph}\nolimits} \def\ran{\mathop{\rm ran}\nolimits} \def\Stab{\mathop{\rm Stab}\nolimits} \def\Out{\mathop{\rm Out}\nolimits} \def\Aut{\mathop{\rm Aut}\nolimits} \def\Inn{\mathop{\rm Inn}\nolimits} \def\matr#1#2#3#4{ \big({#1\atop #3}{#2\atop #4}\big) } \def\tto{\longrightarrow} \def\m{\frak m} \def\inj{\hookrightarrow} \def\surj{\rightarrow\!\!\!\!\!\rightarrow} \input cyracc.def \font\tencyr=wncyr10 \font\tencyri=wncyi10 \def\cyr{\tencyr\cyracc} \def\cyri{\tencyri\cyracc} \maketitle{The Burnside Problem}{by}{Marcel K. Goh}{15 April 2020} \section 1. Introduction \jankscsp ONE\ OF\ THE oldest and most studied problems in group theory began life in a 1902 paper by William Burnside [3]. He wrote: A still undecided point in the theory of discontinuous groups is whether the order of a group may not be finite while the order of every operation it contains is finite.'' This question is now known as the {\it general} Burnside problem and it was solved in 1964 by E. S. Golod and I. R. Shafarevich [4], who found an example of an infinite finitely-generated group all of whose elements have finite order. A more specialised version of the problem, which Burnside himself tackled in his original paper, requires that the order of every element of the group be bounded by a {\it fixed} integer $n$. The problem of whether such groups are necessarily finite is called the Burnside problem and to solve it, we consider a specific class of groups. The {\it free Burnside group with $r$ generators of order $n$}, denoted $B(r,n)$, is the group with $r$ generators $x_1, \ldots, x_r$ such that for every $s\in B(r,n)$, $s^n = e$. It can be regarded as the quotient of the free group $F_r$ by the normal subgroup $\{s^n : s\in F_r\}$. Any group with $r$ generators and all orders dividing $n$ is the image of a homomorphism from $B(r,n)$, so the Burnside problem boils down to determining if $B(r,n)$ is finite for all $r$ and $n$. \section 2. Small Cases The simplest case occurs when $n=2$. \parenproclaim Theorem A (Burnside, {\rm 1902}). The order of $B(r,2)$ is $2^r$. \proof For every $s,t\in B(r,2)$, $s^2 = t^2 = (st)^2 = e$, since every element is an involution. Then from $stst = e$ we can multiply by $s$ on the left and $t$ on the right to obtain $ts = st$. Thus $B(r,2)$ is abelian and, having $r$ generators as a basis, is isomorphic to $(\Z/2\Z)^r$, with order $2^r$.\slug Burnside also proved that the order of $B(2,3)$ is $27$, that $B(2,4)$ has order at most $2^{12}$, and that $B(r,3)$ is finite for all $r$. Using a convoluted method that spans two and a half pages, Burnside proved that $|B(r,3)| \leq 3^{2^r-1}$, but finiteness of $|B(r,3)|$ can be shown more easily if we relax the bound. The following argument was presented by Marshall Hall, Jr.\ in his 1959 textbook {\sl The Theory of Groups} [6]. \proclaim Theorem B. For $r\geq 1$, the order of $B(r,3)$ is of the form $3^{m(r)}$ for some integer $m(r) \leq 3^{r-1}$. \proof Because every non-identity element in $B(r,3)$ has order $3$, the group is a $3$-group and has order a power of $3$. We will perform induction on $r$. When $r=1$, the group is cyclic and its order is 3, so setting $m(1) = 0$ satisfies the inequality with equality. Now suppose that the theorem holds for some integer $k$; i.e.\ $|B(k,3)| = 3^{m(k)}$ and $m(k) \leq 3^{k-1}$. Note that for any Burnside group of exponent 3, the relation $(st)^3 = e$ implies that $$tst = s^{-1}t^{-1}s^{-1} \oldno 1$$ for any $s,t$ in the group. We form $B(k+1,3)$ by adding a new generator $g$ to the generating set of $B(k,3)$. Let $s\in B(k+1,3)$ It can be expressed as a finite product of the form $$s = s_1g^{\pm 1}s_2g^{\pm 1} \cdots g^{\pm 1}s_n, \oldno 2$$ where $n$ is some integer and the $s_i$ belong to $B(k,3)$. If, in this product, any two consecutive $g$'s have the same sign in the exponent, we can use the identity ({\oldstyle 1}) with $g = t$ to perform the following replacements, each reducing the total number of $g^{\pm 1}$ terms by one: $$gs_ig = {s_i}^{-1}g^{-1}{s_i}^{-1} \qquad \hbox{and}\qquad g^{-1}s_ig^{-1} = {s_i}^{-1}g{s_i}^{-1} \oldno 3$$ In this way, $s$ can be reexpressed in the form ({\oldstyle 2}) but with the exponents of $g$ alternating in sign. But we can make further reductions. Since $g^3 = e$, we have $g^{-1} = g^2$ and we can perform surgery on $s$ as follows: \eqalign{ s &= s_1\cdots s_i g s_{i+1} g^{-1} s_{i+2} g \cdots s_n\cr &= s_1\cdots s_i g s_{i+1} g\cdot g s_{i+2} g \cdots s_n\cr &= s_1\cdots (s_i {s_{i+1}}^{-1}) g^{-1} ({s_{i+1}}^{-1}{s_{i+2}}^{-1}) g^{-1} {s_{i+2}}^{-1} \cdots s_n,\cr } reducing the total instances of $g^{\pm 1}$ by one \big(the third line is due to ({\oldstyle 3})\big). An analogous reduction can be done when $s$ contains a $s_i g^{-1} s_{i+1} g s_{i+2} g^{-1}$, and in fact, in both cases we now have two consecutive matching exponents so we can immediately apply one of the identities in ({\oldstyle 3}) again. After doing this repeatedly, we find that any $s\in B(k+1,3)$ is of the form ({\oldstyle 2}) with at most two $g^{\pm 1}$s. As a final refinement, using $g = g^{-1}g^{-1}$ we can write $$s_1g^{-1}s_2gs_3 = s_1g^{-1}s_2g^{-1}g^{-1}s_3 = (s_1{s_2}^{-1})g{s_2}^{-1}g^{-1}s_3,$$ so any $s\in B(k+1,3)$ can be written in one of the forms $$s_1,\quad s_1gs_2, \quad s_1g^{-1}s_2, \quad s_1gs_2g^{-1}s_3,$$ where $s_1$, $s_2$, and $s_3$ are in $B(k,3)$. There are $3^{m(k)}$ possibilities for the first form, $3^{2m(k)}$ possibilites for each of the second and third forms, and $3^{3m(k)}$ for the fourth. Thus $$|B(k+1,3)| = 3^{m(k)} + 2\cdot3^{2m(k)} + 3^{3m(k)} < 3^{3m(k) + 1},$$ so $m(k+1) \leq 3m(k)$ and $|B(k+1,3)| \leq 3^{3^{r-1}}$.\slug In 1933, F. W. Levi and B. L. van der Waerden [8] determined the exact value of the exponent to be $$m(r) = {r\choose 3} + {r\choose 2} + r.$$ \section 3. Burnside Groups of Exponent Four For $n$ even just slightly larger than 2 or 3, much less is known about $B(r,n)$. For example, the exact orders of the groups $B(r,4)$ are only known for $r\leq 5$. It is known, however, that $B(r,4)$ is finite for all $r$. This result was proved by I. N. Sanov in 1940 [9], but since his paper is written in Russian, we present a treatment due to M. Hall [6]. The proof has been reorganised for the sake of clarity. We begin with a lemma. \proclaim Lemma C. Let $G$ be a finite group of order $M$, all of whose elements have orders dividing 4. Then if we adjoin to $G$ a new generator $g\notin G$ such that $g^2\in G$, the resulting group $G' = G \cup \langle g\rangle$ is finite, when divided by the normal subgroup $\{s^n : s\in G'\}$. \proof Since $g^{-1} = g^3$ and $(gs)^4 = e$ for any $s\in G'$, we have \eqalign{ gsg &= s^{-1}g^{-1}s^{-1}g^{-1}s^{-1} \cr &= s^{-1}g (g^2 s^{-1} g^2) gs^{-1};\cr } since $g^2\in G$, we have, for any $s\in G'$, $$gsg = s^{-1}gs'gs^{-1} \oldno 4$$ for some $s'\in G$. Armed with this identity, we proceed to fix some $s\in G'$, which, since $g = g^{-1}$, can be expressed as $$s = s_1gs_2g\cdots gs_n,$$ where each $s_i\in G$. We will say that this word has length $n$'' because it consists of $n$ elements of $G$. Applying ({\oldstyle 4}) to some $s_i$ we get $$s = s_1g \cdots g (s_{i-1}{s_i}^{-1})g {s_i}' g ({s_i}^{-1}s_{i+1}) g \cdots gs_n,$$ and this word has not increased in length. Note that $s_{i-1}$ has been replaced by $s_{i-1}{s_i}^{-1}$. If some $s_i = e$, then its two neighbouring $g$ terms combine to form an element of $G$, so the length of the word decreases by one. Repeatedly applying ({\oldstyle 4}), we can replace $s_{i-1}$ with $s_{i-1}{s_i}^{-1}$, then replace $s_{i-2}$ with $s_{i-2}(s_{i-1}{s_i}^{-1})^{-1} = s_{i-2}s_i{s_{i-1}}^{-1}$, etc. So, in particular, we can replace $s_2$ with any one of $$s_2, \quad s_2{s_3}^{-1}, \quad s_2s_4{s_3}^{-1},\quad s_2s_4{s_5}^{-1}{s_3}^{-1},\quad \ldots\ \oldno 5$$ (we repeat this $n-2$ times). Observe that the there is a pattern to the way new $s_i$'s appear. When $i$ is even, $s_i$ appears to the left of $s_{i-1}$ and when $i$ is odd, it appears to the right of $s_{i-1}$ and it is inverted. If any one of these products is the identity, we can reduce the length of the word. Suppose that $n\geq M+2$. Then in the list ({\oldstyle 5}) of $M$ possible replacements for $s_2$, either one of them is the identity or there is a repeated element, say $$s_2\cdots s_{2j}{s_{2j+1}}^{-1}\cdots {s_3}^{-1} = s_2 \cdots s_{2k}{s_{2k+1}}^{-1} \cdots {s_3}^{-1},$$ where \$j