\documentclass[12pt]{article} \usepackage{amsmath,amsthm,amssymb, amstext, amsopn, amsxtra} \setlength{\textheight}{8.8in} \setlength{\headheight}{-0.5in} \setlength{\oddsidemargin}{0.3in} \setlength{\textwidth}{6.3in} \begin{document} \newcommand{\nt}{\noindent} \newcommand{\bs}{\bigskip} \newcommand{\ms}{\medskip} \newcommand{\mk}{\medskip} \newcommand{\sk}{\smallskip} \[ \] \begin{center} \Large{\bf Finite Groups Embeddable in Division Rings} \bigskip \large{\bf T. Y. Lam } \\ \end{center} \bigskip \begin{center} {\bf \S1. Introduction} \end{center} \mk Finite groups that are embeddable in the multiplicative groups of division rings $\,K\,$ were completely determined by S.~A.~Amitsur [Am] in 1955. In case $\,K\,$ has characteristic $\,p>0$, the only possible finite subgroups of $\,K^*\,$ are cyclic groups, according to a theorem of I.~N.~Herstein [He]. Thus, the only interesting case is when $\,K\,$ has characteristic $\,0$; that is, when $\,K\supseteq {\Bbb Q}\,$. \bs In [He], Herstein conjectured that odd-order subgroups of division rings $\,K\,$ were cyclic, and he proved this to be the case when $\,K\,$ is the division ring of the real quaternions. Herstein's conjecture was settled negatively in [Am]. As part of his complete classification of finite groups in division rings, Amitsur showed that the smallest noncyclic odd-order group that can be embedded in a division ring is one of order $63$ (and this group is unique). \bs Amitsur's paper is daunting to read as it is long and technically complicated. In lecturing to a graduate class on division rings, I tried to find a simple reason for the ``first exceptional odd order'' $63$ (to Herstein's conjecture). After some work, I did come up with a reason that was simple enough to be explained to my class, without having to go through {\it any part\/} of Amitsur's paper. Furthermore, the method I used led easily to the {\it second\/} exceptional odd order, $117$ (which was not mentioned in Amitsur's paper). Since this line of reasoning did not seem to have appeared in the literature before, I record it in this short note. To better motivate the results discussed here, I have also included a quick exposition on the beginning part of the theory of finite subgroups of division rings. As a result, this paper can be read independently of [Am]. \sk \bs\nt \begin{center} {\bf \S2. Background on Amitsur Groups} \end{center} \mk Let us say that a finite group $\,G\,$ is {\it Amitsur\/} if $\,G\,$ can be embedded in the multiplicative group $\,K^*\,$ of some division ring $\,K$. We start with some background information on such Amitsur groups. First the following easy observation. \bs\nt {\bf (2.1) Theorem.} {\it Any abelian Amitsur group $\,G\,$ is cyclic.} \bs\nt {\bf Proof.} Say $\,G\subseteq K^*$, where $\,K\,$ is a division ring. If $\,F\,$ is the center of $\,K$, then the $\,F$-span $\,L\,$ of $\,G\,$ is a finite-dimensional $\,F$-algebra. Since $\,L\,$ is a commutative domain, it must be a field. From field theory (see, e.g., [Ja\,:\,p.\,132]), it then follows that $\,G\subseteq L^*\,$ is cyclic. \ \ \ \ \ \ \ \ {\bf QED} \bs Next, we classify Amitsur $\,p$-groups. (Here, $\,p\,$ denotes any prime number, unrelated to the characteristic of our division rings.) This is possible thanks to some standard results in $\,p$-group theory. \bs\nt {\bf (2.2) Theorem.} {\it Let $\,G\,$ be an Amitsur $\,p$-group. Then $\,G\,$ is cyclic if $\,p\,$ is odd, and $\,G\,$ is cyclic or dicyclic if $\,p=2$.} \bs Recall that a group is {\it dicyclic\/} (of order $\,4n$) if it is generated by two elements $\,x,\,y\,$ subject to the relations $\,x^{2n}=1,\;y^2=x^n$, and $\,yxy^{-1}=x^{-1}$, where $\,n\geq 2$. Dicyclic groups are also known as {\it generalized quaternion groups\/}; they are $\,2$-groups iff $\,n\,$ is a power of $\,2$. It is easy to see that any dicyclic group $\,\langle x,\,y \rangle\,$ as above is embeddable in the division ring of real quaternions, so the conclusion in (2.2) is indeed the best possible. \bs\nt {\bf Proof of (2.2).} In the theory of $\,p$-groups, there is a well-known result which guarantees that, if a $\,p$-group has a unique subgroup of order $\,p$, then it satisfies the conclusion of the present theorem; see Th.~12.5.2 in [Ha]. Thus, it suffices for us to verify that, if $\,G\neq \{1\}$, $\,G\,$ has a unique subgroup of order $\,p$. Say $\,G\subseteq K^*$, where $\,K\,$ is a division ring, with center $\,F$. \sk By elementary group theory, we know that $\,G\,$ has a nontrivial center. Fix a central element $\,x\in G\,$ of order $\,p$, so $\,|\langle x\rangle |=p$. If $\,\langle y\rangle \subseteq G\,$ is any subgroup of order $\,p$, then $\,L:=F(x,y)\,$ is a field. In this field, the equation $\,t^p-1=0\,$ has at most $\,p\,$ solutions, so we must have $\,\langle y\rangle = \langle x\rangle $, as desired. \ \ \ \ \ \ \ \ {\bf QED} \bs In order to get some general results for non $\,p$-groups, we introduce the following class of finite groups, where $\,m,\,n,\,r\,$ are natural numbers: $$ G_{m,n,r}=\langle a,\,b\,: \;\;a^m=b^n=1,\;\;bab^{-1}=a^r \rangle \,. \leqno (2.3) $$ Whenever we use this notation, it will be assumed that $\,r^n\equiv 1\; (\mbox{mod}\;m)$. This assumption guarantees that $\,G_{m,n,r}\,$ is a semidirect direct product of $\,\langle a\rangle\,$ and $\,\langle b\rangle\,$, with $\,|\langle a\rangle|=m\,$ and $\,|\langle b\rangle|=n$. In particular, $\,|G_{m,n,r}|=mn$. If necessary, we may assume $\,0\leq r1\,$ and} $\,o\,(\overline{r})\,$ ({\it the order of $\,\overline{r}\,$ in the unit group} $\,\mbox{U}({\Bbb Z}_m)\,$) {\it is exactly $\,n$. Then $\,G\,$ is not Amitsur.} \bs\nt {\bf Proof.} Assume, instead, that $\,G\subseteq K^*$, where $\,K\,$ is a division ring. We use the presentation (2.3) for $\,G$. From $\,bab^{-1}=a^r$, we get $\,b^iab^{-i}=a^{r^i}$. Thus, $\,b^ia=a^{r^i}b$, and $\,a^{-1}b^ia=a^{r^i-1}b^i$. From $\,b^n=1$, we get $\,b^{n-1}+\cdots+b+1=0\,$ in $\,K$. Conjugating this equation by $\,a$, we get $$ a^{r^{n-1}-1}b^{n-1}+a^{r^{n-2}-1}b^{n-2}+\cdots +a^{r-1}b+1=0. \leqno (3.2)$$ Subtracting the last two equations and cancelling $\,b\,$ (from the right) gives $$ \bigl(a^{r^{n-1}-1}-1\bigr)b^{n-2}+\bigl(a^{r^{n-2}-1}-1\bigr)b^{n-3}+ \cdots +(a^{r-1}-1)=0. \leqno (3.3)$$ Conjugating this by $\,a\,$ again and subtracting (3.3) from the resulting equation, we get after another cancellation of $\,b$: $$ \bigl(a^{r^{n-1}-1}-1\bigr)\bigl(a^{r^{n-2}-1}-1\bigr)b^{n-3}+\cdots +(a^{r^2-1}-1)(a^{r-1}-1)=0.$$ Carrying this process to the bitter end produces the equation $$ \bigl(a^{r^{n-1}-1}-1\bigr)\bigl(a^{r^{n-2}-1}-1\bigr)\cdots (a^{r^2-1}-1)(a^{r-1}-1)=0 \in K. \leqno (3.4)$$ Therefore, $\,a^{r^i-1}=1\,$ for some $\,i\in [1,\,n-1]$. But then $\,r^i\equiv 1\,(\mbox{mod}\;m)$, which contradicts $\,o\,(\overline{r})=n$. \,\,\,\,\,{\bf QED} \bs In my lectures, I referred to the above as the ``triple $(-1)$-proof'', because of the three remarkable layers of $(-1)\,$'s in the first factor of the key equation (3.4)! Note that this proof actually showed that, under the assumption that $\,o\,(\overline{r})=n>1$, $\,G_{m,n,r}\,$ cannot be multiplicatively embedded in any domain (that is, a nonzero ring without 0-divisors). This, however, is not a real improvement, since it is easily seen that, once a finite (multiplicative) group $\,G\,$ is embeddable in a domain, then $\,G\,$ is in fact embeddable in a division ring. \bs We now record some consequences of (3.1). \bs\nt {\bf (3.5) Corollary.} {\it If $\,G=G_{m,n,r}\,$} ({\it generated by $\,a,\,b\,$ as in} $(2.3)$) {\it is Amitsur, then the largest square-free subgroup of $\,\langle b\rangle\,$ must be central in $\,G$.} {\it In particular, if $\,G=G_{m,n,r}\,$ is Amitsur and $\,n\,$ is square-free, then $\,G\,$ is cyclic.} \bs\nt {\bf Proof.} For any prime $\,\ell\,$ dividing $\,n=o(b)$, let $\,\langle c \rangle\,$ be the $\,\ell$-Sylow group of $\,\langle b \rangle\,$, say of order $\,\ell ^t$. Then $\,cac^{-1}=a^s\,$ for some $\,s$, and $\,\langle a\rangle \langle c\rangle \cong G_{m,\ell ^t,s}\,$ is still Amitsur. By (3.1), we must have $\,o\,(\overline{s})< \ell ^t\,$ in $\,\mbox{U}({\Bbb Z}_m)$. This means that $\,c^{\ell ^{t-1}}\,$ acts trivially on $\,a\,$ (by conjugation), and hence $\,c^{\ell ^{t-1}}\,$ is central in $\,G$. From this, the first conclusion in the Corollary follows. If $\,n\,$ happens to be square-free, this conclusion means that $\,b\,$ is central in $\,G$. Thus, $\,G\,$ is abelian, and hence cyclic by (2.1). \ \ \ \ \ \ \ \ \ {\bf QED} \bs\nt {\bf (3.6) Corollary.} {\it Let $\,G\,$ be an Amitsur group of order $\,p^{t}q_1\cdots q_k\,$, where $\,k\geq 1$, and $\,p>q_1>\cdots >q_k\,$ are primes. Then $\,G\,$ is cyclic.} \bs\nt {\bf Proof.} Since $\,k\geq 1$, $\,p\,$ is an odd prime. Thus, $\,G\,$ is Sylow-cyclic, and hence solvable by (2.4). We proceed by induction on $\,k$. Note that the solvability of $\,G\,$ implies, by Philip Hall's Theorem, that it has a subgroup $\,H\,$ of order $\,m:=p^tq_1\cdots q_{k-1}$ (see [Ha:~Th.~9.3.1]). This Hall subgroup is still Amitsur, and hence cyclic by the inductive hypothesis. Also, since $\,[G:H]=q_k\,$ is the smallest prime divisor of $\,|G|$, $\,H\,$ must be normal in $\,G$. Thus, by taking a generator of $\,H\,$ and an element of order $\,q_k\,$ in $\,G$, we can represent $\,G\,$ as $\,G_{m,q_k,r}\,$ for some $\,r$. Since $\,q_k\,$ is a prime, (3.5) implies that $\,G\,$ is cyclic. Note that this argument is also sufficient to get the induction started, so the proof is complete. \ \ \ \ \ \ \ \ {\bf QED} \bs A special case of (3.6) is that {\it any square-free Amitsur group is cyclic.} This recaptures Corollary 5 on p.\,384 of [Am]. \bs Note that, in (3.1) and (3.5), the integers $\,m,\,n\,$ and $\,r\,$ were arbitrary. To get some explicit numerical results, let us specialize these results to odd-order groups. \bs\nt {\bf (3.7) Theorem.} {\it If $\,G\,$ is an Amitsur group of odd order $< 171$, then $\,G\,$ is cyclic except possibly when} $\,|G|=63\;\,or \;\,117.$ \bs\nt {\bf Proof.} If $\,|G|\,$ is either a prime power, or square-free, or of the form $\,p^{t}q\,$ where $\,p>q\,$ are primes, the foregoing results imply that $\,G\,$ is cyclic. Among odd integers from $1$ to $169$, the only ones {\it not\/} of any of the above forms are $$ 5\!\cdot\! 3^2=45, \;\;\, 7\!\cdot\! 3^2=63, \;\;\, 11\!\cdot \!3^2=99, \;\;\, 13\!\cdot\! 3^2=117, \;\;\, 5\!\cdot\! 3^3=135 \;\;\, \mbox{and} \;\;\, 17\!\cdot\! 3^2=153.$$ If $|G|=45$ or $99$, $\,G\,$ is easily seen to be abelian (by Sylow theory), and hence cyclic (by (2.1)). Next, consider $\,|G|=135$. Since $\,G\,$ is Amitsur, we can represent it (thanks to (2.5)) in the form $\,G_{m,n,r}\,$ with $\,(m,\,n)=1$. We may assume $\,m,\,n>1$, so we have $\,m=5\,$ or $\,3^3$. But $\,\mbox{U}({\Bbb Z}_5)\,$ has no element of order $\,3$, and $\,\mbox{U}({\Bbb Z}_{3^3})\,$ has no element of order $\,5$. Therefore, $\,G\,$ must be abelian and hence cyclic. The case $\,|G|=153\,$ can be treated similarly (with exactly the same conclusion). Thus, $\,G\,$ is cyclic in all cases, except possibly when $\,|G|=63\,$ or $\,117$. \ \ \ \ \ \ \ \ \ {\bf QED} \bs What about $\,|G|=63$? There are, up to isomorphism, four groups of order 63; they are: $$ {\Bbb Z}_{63},\;\;\;\;\; {\Bbb Z}_3 \times {\Bbb Z}_{21}, \;\;\;\;\; {\Bbb Z}_3\times G_{7,3,2}\,, \;\;\;\;\mbox{and} \;\;\;\; G_{7,9,2}.$$ The first group is cyclic, and hence Amitsur. The second and third ones are not Sylow-cyclic, and hence not Amitsur. The remaining question is whether the non-cyclic (but Sylow-cyclic) $\,G_{7,9,2}\,$ is Amitsur. Note that, since $\,o\,(\overline{2})=3<9\,$ in $\,\mbox{U}({\Bbb Z}_7)$, (3.1) does not apply to this group. Thus, there {\it is\/} a chance that $\,G_{7,9,2}\,$ is Amitsur. Indeed, the embeddability of this group was proved by Amitsur, and subsequent expositions on this have been given by C.~Ford [Fo] and J.~Dauns [Da]. For the sake of completeness, we give a quick sketch for the embeddability of $\,G_{7,9,2}\,$ below. \bs Let $\,L={\Bbb Q}\,(\zeta)\,$ where $\,\zeta\,$ is a primitive 21st root of unity, and let $\,\sigma\in \mbox{Gal}\,(L/{\Bbb Q}\,)\,$ be defined by $\,\sigma(\zeta) =\zeta^{16}$. Then $\,o\,(\sigma)=3$, so $\,L\,$ is a cubic extension of the fixed field $\,F:=L^{\sigma}$. Note that $\,F\,$ contains the primitive cubic root of unity $\,\omega:=\zeta^7$, since $$\,\sigma(\omega)=\zeta ^{7\cdot 16}=\zeta^7=\omega.$$ On the other hand, for the primitive 7th root of unity $\,a:=\zeta^3$, we have $$\,\sigma(a)=\zeta^{3\cdot 16}=\zeta^{6}=a^2.$$ Now introduce a new symbol $\,b$, and form the cyclic $\,F$-algebra $$ K:= L\oplus L\,b \oplus L\,b^2, \;\;\;\; \mbox{with}\;\;b^3=\omega, \;\;\;\mbox{and} \;\;\; b\,\ell=\sigma(\ell)\,b \;\;\; (\forall \;\,\ell \in L),$$ which has center $\,F$. We have $\,\mbox{dim}\,_{\Bbb Q}\,F=\varphi(21)/3=4$, so $\,\mbox{dim}\,_{\Bbb Q}\,K=3^2\!\cdot\!4=36.$ In the algebra $\,K$, the elements $\,a,\,b\,$ satisfy the equations $$ a^7=1,\;\;\; b^3=\omega,\;\;\;b^9=\omega^3=1, \;\;\;\mbox{and}\;\;\; ba=\sigma(a)b=a^2b. $$ Therefore, $\,a,\,$ and $\,b\,$ generate a group $\,G\cong G_{7,9,2}\,$ in the group of units $\,\mbox{U}(K)$. {\it It can be checked that $\,K\,$ is a division $\,F$-algebra} (see, for instance, [Fo]), so this establishes the noncyclic $\, G_{7,9,2}\,$ as an Amitsur group (of order $63$). The center $\,F\,$ of the division ring $\,K\,$ is easily seen to be the biquadratic extension $\,{\Bbb Q}\bigl(\sqrt{-3},\,\sqrt{-7}\,\bigr)\,$ of the rationals. \bs As for the case $\,|G|=117$, we have a noncyclic candidate $\,G_{13,9,9}\,$ (noting that $\,o\,(\overline{9})=3\,$ in $\,\mbox{U}({\Bbb Z}_{13})$). This group can be shown to be Amitsur in the same way as $\,G_{7,9,2}\,$ was. \bs A substantial part of the argument in showing that a certain group is Amitsur is to establish that a certain $\,{\Bbb Q}\,$-algebra is a division algebra. This is by no means routine. In Amitsur's paper, this part of the work is handled by number-theoretic tools, such as the Hasse Norm Theorem and the Hasse-Brauer-Noether Theorem. Thus, to fully understand Amitsur's work would require considerable preparation in algebraic number theory. \bs In closing, we should mention that it is also of interest to {\it fix\/} a field $\,F\,$ (of characteristic zero), and to determine the finite groups $\,G\,$ that are embeddable in $\,K^*\,$ for some finite-dimensional {\it central\/} $\,F$-division algebra $\,K$. This problem has been studied by B.~Fein and M.~Schacher [FS], who called such $\,G\,$ ``$F$-adequate groups''. The embedding constructed for $\,G=G_{7,9,2}\,$ above showed that $\,G\,$ is $\,{\Bbb Q}\bigl(\sqrt{-3},\,\sqrt{-7}\,\bigr)$-adequate, but Fein and Schacher showed that $\,G\,$ is already $\,{\Bbb Q}\bigl(\sqrt{-3}\,\bigr)$-adequate. This is remarkable, since Fein and Schacher have also proved the following interesting positive result on Herstein's conjecture: {\it If} $\,[F:{\Bbb Q}\,]\leq 2$, {\it then any $\,F$-adequate odd-order group is cyclic, except when} $\,F={\Bbb Q}\bigl(\sqrt{-3}\,\bigr)\,$! \begin{thebibliography}{ABC} \vspace{.2cm} \bibitem[Am]{Am} S.~A.~Amitsur: Finite subgroups of division rings, Trans. Amer. Math. Soc. {\bf 80}(1955), 361-386. \bibitem[Da]{Da} J.~Dauns: A Concrete Approach to Division Rings, Heldermann Verlag, Berlin, 1982. \bibitem[FS]{FS} B.~Fein and M.~Schacher: Embedding finite groups in rational division algebras, I, II, J. Algebra {\bf 17}(1971), 412-428, and {\bf 19}(1971), 131-139. \bibitem[Fo]{Fo} C.~Ford: Finite groups and division algebras, L'Enseig.~Math. {\bf 19}(1973), 313-327. \bibitem[Ha]{Ha} M.~Hall: The Theory of Groups, Macmillan Company, New York, 1959. \bibitem[He]{He} I.~N.~Herstein: Finite multiplicative subgroups of division rings, Pacific J. Math. {\bf 3}(1953), 121-126. \bibitem[Ja]{Ja} N.~Jacobson: Basic Algebra, I, Second Edition, W.~H.~Freeman and Co., New York, 1985. \end{thebibliography} \mk\nt University of California \\ Berkeley, Ca 94720 \sk\nt lam@math.berkeley.edu \end{document}