Groups Generated by Two Operators of Order 3 Whose Commutator Is of Order 2

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6. Conformal Mapping.-Analogous to the author's previous work, the preceding theory gives for n = 2 the conformal mapping of any crescentshaped plane region-i.e., one bounded by two Jordan curves with a unique point in common, one curve enclosing the other-on a circular crescent, including the continuous attachment of the conformal map to a topological correspondence between the boundaries. 1"The Problem of Plateau for Two Contours," Jour. Math. Phys., 10, 315-359 (1931). This paper will be cited as "Two Contours." 2 "Solution of the Problem of Plateau," Trans. Amer. Math. Soc., 33, 263-321 (1931). This paper will be cited as "One Contour." 8 One Contour, p. 282.

GROUPS GENERA TED B Y TWO OPERA TORS OF ORDER 3 WHOSE COMMUTA TOR IS OF ORDER 2 By G. A. MILLER DEPARTMENT OF MATHEMATICS, UNIVERSITY OF ILLINOIS

Communicated December 12, 1932

Let s and t represent two operators of order 3 such that their commutator s2t2st is of order 2. It is known that then each of the four commutators whose elements are powers of s and t StS2t2 S2tSt2 st2s2t S2t2St

is of order 2 and hence there result the following identities s2t2st = t2s2ts, s2tst2 = ts2t2s, st2s2t = t2stS2, StS2t2 = tSt2S2. Since t2s2ts.s2tst2 = t2s2t2st2 and t2S2t2St2.t2S2t2St2 = t2S2t2StS2t2St2 = t2S2t2. tst2s2. st2 = 1, it follows that the first and the second of these four commutators are commutative. Similarly it follows from the identities s2t2st. t2StS2 = S2t2S2tS2 and S2t2S21S2.S2t2S2tS2 = S2t2s2tsl2s2ts2 = S2t2S2.sts2t2.ts2 = 1, that the first of these four commutators is also commutative with the third. It seems desirable to employ a different method to prove that the first and fourth of these commutators are commutative but a similar method wiU be employed to prove that the fourth of these commutators is also commutative with the second and third, as follows: From the identities s2tSt2. tSt2S2 = S2tS2t2S2, and s2ts2t2S2. S2tS2t2S2 = s2ts2t2sts2t2s2 = s2t. t2s2ts. s2t2s2 = 1, it results that the second and the fourth are commutative. Similarly, the fact that the third and the fourth are commutative results from the identities t2StS2 . stS2t2 = 12St2s2t2 and t2St2S2t2.t2St2S2t2 = t2St2S2tSt2S2t2 = t2s.S2t2St.t2S2t2 = l

It has not yet been proved that the first and last of these four commu-

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tators as well as the second and third, are commutative. Since these two pairs of operators are of order 2 they generate two dihedral groups. To prove that they are commutative it is only necessary to prove that the orders of these two dihedral groups cannot exceed 4. We shall first prove that these groups have the same order. The transform of the former of these groups by s is generated by the two operators st2sts = st2s2t. t2s2ts and ts2t2s. Since t2s2ts is commutative with each of these two generators it results that the dihedral group generated by them is simply isomorphic with the one generated by the second and the third of the given four commutators. That is, the two dihedral groups under consideration have a common order. To prove that this common order cannot exceed 4 it should be noted that the four given commutators generate the commutator subgroup H of the group G generated by s and t, and that the two given dihedral groups are invariant subgroups of H. The transform of the former of these dihedral groups, as noted in the preceding paragraph, cannot be invariant under H unless t2s2ts is thus invariant. This is, however, possible only when the common order of these dihedral groups does not exceed 4. Hence it has been proved that each of the four given commutators is commutative with each of the others and hence the order of H cannot exceed 16 and the order of G cannot exceed 144. In order to determine G it is therefore only necessary to construct a group of order 144 which is generated by two operators of order 3 whose commutator is of order 2. The direct product of two tetrahedral groups is obviously such a group and hence we have established the following theorem: If two operators of order 3 have a commutator of order 2 but are not otherwise restricted they generate the direct product of two tetrahedral groups. It is now easy to determine all the possible groups which are generated separately by two operators of order 3 whose commutator is of order 2 since all such groups must -be quotient groups of the direct product of two tetrahedral groups. Moreover, each of them must be non-abelian since two commutative operators of order 3 could not have a commutator of order 2. The smallest such group is therefore the tetrahedral group. This corresponds to the quotient group of G with respect to either of its two invariant subgroups of order 12. There are also two and only two invariant subgroups of order 4 in G and each of these gives rise to a quotient group which is simply isomorphic with the direct product of the tetrahedral group and the group of order 3. As G obviously contains no other invariant group besides the identity whose order is less than 12 it has been proved that there are three and only three groups which can be separately generated by two operators of order 3 whose commutator. is of order 2. These are the tetrahedral group, the direct product of the tetrahedral group and the group of order 3, and the direct product of two tetrahedral groups.

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It may be desirable to add that a simple method of constructing s and I so that they satisfy the given conditions and generate the direct product

of two tetrahedral groups is as follows: Represent two tetrahedral groups on distinct sets of letters and construct the operators s and t so that each of them has a constituent from each of these two tetrahedral groups. The two constituents of s and t from one of these two tetrahedral groups appear in the same co-set of this group with respect to its invariant subgroup of order 4 while those from the other appear in two different such co-sets. Hence it results that both st and st2 are of order 6 and that their squares are operators of order 3 in the two given tetrahedral groups.. Since the Sylow subgroups of order 9 in G are not invariant and do not involve any invariant subgroup of G besides the identity it is possible to represent G as a transitive permutation group of degree 16, and when it is thus represented its characteristic subgroup of order 16 will be regular. Hence G appears in the holomorph of the abelian group of order 16 and of type (1, 1, 1, 1), and corresponds to one of the 280 Sylow subgroups of order 9 in the group of isomorphisms of this abelian group. This holomorph was considered by the present writer in the American Journal of Mathematics, volume 20 (1898), page 229, and hence it results that we may take for s and t the following two permutations of this holomorph as it is there represented: agc. bin. doj.fkl . hmp, ajf . bhk. cdn. eog. ilp. When G is represented according to the method noted in the preceding paragraph the operators s and t can be represented as permutations of degree 6. It is well known that when two operators of order 3 have a product of order 2 they must generate the tetrahedral group and hence it is of interest to note that the tetrahedral group plays such a fundamental r6le when their commutator is of order 2. Whenever a group is determined by the orders of two operators and the order of their commutator it must have for a quotient group the abelian group which is the direct product of cyclic groups whose orders are the orders of these two operators. This results directly from the fact that to each of these two operators we can adjoin an operator of equal order which is commutative with the two operators without affecting their commutator. In particular, if a group is determined by two operators of order 2 and the order of their commutator it is dihedral and has the four group for a quotient group whenever this commutator is of even order and if it is determined by two operators which are restricted only by the facts that they are of order 3 and the order of their commutator is given it has the non-cyclic group of order 9 for a quotient group. If it is determined by two operators of orders 2 and 3, respectively, which are not further restricted except that the order of their commutator is given it

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has the cyclic group of order 6 for a quotient group. If the order of this commutator is 2 it is known that the corresponding group is the direct product of the tetrahedral group and the group of order 2. When G is determined by the two operators s and £ of orders 2 and 3, respectively, which are not otherwise restricted except that the order of their commutator is 3 then there is no upper limit for the order of G, but all such groups are solvable since the commutator subgroup of the commutator subgroup must always be abelian. In fact, it is easy to verify that the commutator of stst2 and t2sts is then (ts)6 while that of tst2s and st2st is (st)6 and the product of these two commutators in order is t(st2) 612. As (st)6 is commutative with this product it results that (ts)6 and (st)6 are commutative. This implies that the commutator subgroup of the commutator subgroup is abelian since it is obviously generated by these two sixth powers. If the order of st is assumed to divide 6 but no additional restrictions are imposed upon s and t then G is the group of order 54 obtained by extending the non-abelian group of order 27 which involves no operator whose order exceeds 3 by an operator s of order 2 under which it is invariant and which transforms into their inverses all the operators of one of its subgroups of order 9 and transforms an operator t of this group of order 27 but not found in this subgroup into itself multiplied by an operator contained therein but not invariant under this group of order 272.

It is easy to verify that (st)2 is commutative with (tS)2 in all the groups considered in the preceding paragraph and that these two operators generate an invariant subgroup of all of these groups. Hence this infinite system of groups may be constructed as follows: The direct product of two cyclic groups C1, C2 of the same arbitrary order is extended by an operator t of order 3 such that !2C2t = C1, I2C1C2t = C2-1, while s transforms Ci into C2, and t into Cl. t2. The group generated by C1, C2 and t involves only operators of order 3 in addition to those of the direct product Cl and C2. This may be regarded as a direct generalization of the tetrahedral group. The entire group G may be regarded as a direct generalization of the octahedral group and involves one and only one subgroup of index 2. When Ci is of order 2 there results the group of the cube and when Ci is of order 3 there results the group of order 54 noted in the preceding paragraph. 1 G. A. Miller, Proc. Nat. Acad. Sci., 18, 665 (1932). 2

Cf. H. R. Brahana, Am. Jour. Math., 50, 347 (1928).