What is the Probability an Auto Morph Ism Fixes a Group Element - Gary Sherman

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What is the Probability an Automorphism Fixes a Group Element?Author(s): Gary ShermanReviewed work(s):Source: The American Mathematical Monthly, Vol. 82, No. 3 (Mar., 1975), pp. 261-264Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/2319852 .

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1975] MATHEMATICAL NOTES 261Xr = Ur(O) and Ur(O) -+ X - 2Rs r -1 1. The harmonic unction an be recoveredfrom ts boundaryvalue by Poisson's formula (re'0)= X * Pr(O), wherePr(O) isthe Poisson kernel.A necessary nd sufficientondition hat an operatorbe theboundaryvalue of a harmonic unctions that

limsup Icn(X) |/jnj 4 oo2. For any operator X, Cn(X(k)) = (in)kCn(x) and in view of the Riemann-Lebesgue emma necessarynd sufficientondition hat n operatorbe a derivativeof someorder f a continuous unctions thatcn(X) = O(j n Ik)as jn -+ oofor omek. These operators re exactly he distributions. lso, since0 e C' ifand only fc(4 = (| n1-k) s n -+ oo for very , thedistributionsreexactly hoseoperatorssuch that X *4 E C for every4 E C'.3. Simpleexamples fapplications o partialdifferentialquations an be madeby applying hetheoryn the case of severalvariables.For example, f g(O, b) s aCO functionf 0 and 4which s periodicwithperiod2t in each of0 and q, an oper-atorX = X(0, 4) whichsatisfies henon-homogeneous aplace equationAX = gmust be a C' functionnd thus mustsatisfyheequation n theusual sense. Inparticular ny twicecontinuously ifferentiableunctionwhich satisfies hisequa-tion must n fact be infinitelyifferentiable.The firstuthorwassupportednpartbyNSF grantGP-34558.References1. A. Erdelyi, Operational Calculus and Generalized Function, Holt, Rinehartand Winston,New York, 1962.2. J. Mikusinski,Operational Calculus, Pergamon Press,New York, 1959.DEPARTMENT OF MATHEMATICS, UNIVERSITY OF CALIFORNIA, SANTA BARBARA, CA 93106.

WHAT IS THE PROBABILITYAN AUTOMORPHISM FIXES A GROUP ELEMENT?GARY HERMAN

Introduction.etthe initeroup operaten the initeon-emptyetX (i.e.,Gis representeds a group fpermutationsfX). Weask: Whats theprobabilityan elementhosen trandomrom fixes n elementhosen trandom rom ?If thisprobabilitys denoted yPG (X), then1 (g,x)|gx = x for 9EG andxEX}I!PG X) = xGIi IX

Setting .,= {g e-GIgx = x} we find hat


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262 GARY SHERMAN [Marchk{(g,x) |gx = x forgE G and xEX} = G., = E [G:GxJGIt = kI G|xeX i=1

where x I *,xk is a set of representativesf thedistinct rbits of X under G.Thus,PG(X) = k/|X ; theratio of thenumber forbits n X underG to the orderof X.Of special interest,nd the motivation or thepreceding emarks,s whenGoperates n tself yconjugation.n this aseoneobtains heprobabilityf wogroupelementsommutingsee [1]). It is easyto showthisprobabilityo be at most5/8fornonabeliangroups.Recently, ustafson2] has shown hebound5/8 o be validwhenG is an infinite onabeliancompactgroup of course the countingmeasureis replaced by thenormal eftHaar measureon G).The purposeofthisnote s to consider A (G) whereG is a finitebeliangroupandA is itsgroupofautomorphisms. e show:(i) PA(G) = lif and onlyif G = Z2,(ii) PA(G) ? 3/4 f G 7 Z2,(iii) there s onlya finite umber fgroupswith givenprobability,(iV) PA(G) -O as fG -oo .

Finite abeliangroups. It is trivial hatPA (Z2) = . Conversely,fPA (G) = 1,thenG is an elementarybelian 2-group ince the automorphism -- -x mustbe the dentitymapping.ViewingG as a Z2 space itfollows hat nytwonontrivialelements f G are in thesameorbit.Thus,2/2i 1 wheref f-2i'. This impliesj = 1; i.e., G = Z2.Fromthe ntroductoryemarkswehave jG F j + I 1 j + + j0? whereF is thesubgroupof trivial rbits nd 01, , ?r are thenontrivial rbits.Hence,(|G| - fF|)/2 r since Ioif > 2 for = 1,2,** ,r. Fromk = r+ fFfwe obtaink ? (IGI + fFf)/2. If G = Z2, then F #A and therefore G: F] ? 2; i.e.,|FJ G1/2. hus k ? (3/4) Gf, O PA(G) ? 3/4 proving ii). Observe thatPA(Z4) = 3/4.If G is decomposable, ay(*) G DGi,i=lthen t is routine o verify(**) PA(G) < Fl PA (Gi),where, or = 1,2, --,s, Ai is thegroupofautomorphismsfGi equalityprevailsin **) ifG is decomposednto a direct umof tsSylow ubgroups).n viewofthisinequality,o obtain a bound forPA(G) it sufficeso obtain a boundfor heSylowsubgroups.To thisend, assumeff pn forsome primep. If G is elementaryabelian,thenPA(G) = 2/pn.This follows ince G is a Zp-space.For cyclicG thereis at least one element f orderpm for m = 0, 1, .., n As elementsn the same


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1975] MATHEMATICALOTES 263orbitmusthave equal orders,G has at least n + 1 orbits. inceelements f the sameordercan be written s powersof elementswith ordersprimeto p, the elementsof a particular rderform n orbit.ThusPA(G) = (n + 1)/p"when G is cyclic.Thefollowingemma s helpfuln establishing generalbound forPA(G) when G is ap-group.

LEMMA.Let n be a positive ntegergreater than 1. The maximumvalue of]7J=1ni,for 1 ni= n, is

i= 4 if n 1 (mod3)3(n-i)/3 * 2l/2, where i = 2 if n 2 (mod3)

i = O if n-O (mod3).Proof.The maximum ccurswhenno n, = 1. Further,m-2) *2 > m if, andonly f, m > 4. Thus each ni> 4 can be replaced by (ni - 2) + 2 in the partitionand the associated productwillbe increased. f someni = 4, replacingt by 2 + 2leaves the productunchanged. f 2 + 2 + 2 occurs n the partition, eplacingt by3 + 3 increases he product.Hence the maximum ccurswhen each niis a two ora three. The conclusionof the lemmafollows mmediately.We observe thatthe maximum roduct ssociated withthe partitions f n issmaller han the correspondingroduct btainedfromm if n < m.PROPOSITION1. If IG = pn, thenPA(G) < 2 * 3/p2)n/2.Proof. Supposethe nvariants f G are Mi1,, iMk,b , , 1 where XI'=l mi= mand i < j impliesmi_ mj. Let Ge denotethe summands f G of orderp and G'denotethe summands f G of order at least p2. Thus G G' C Ge. From (**)we get

PA (G) ? PA (G') PA Ge) _ (in m ) (P2)k(*) _~~~~~~~~~2/p') r (mi 1).i=1

Since I 1(mi + 1) = m+ k, maximizingk maximizes the sum. The largestvalue fork occurswhenthemi's are smallestall twos,exceptfor one three fmis odd). Takingk' to be the nteger f {m/2,m 1)/2}and applying he emmato() w6 havePA(G) _ 2 3(m+k')/3 < 2 3(m+*m)/3


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264 L. CARLITZ [MarchPROPOSITION 2. If 0 < p < 1, then here s onlya finitenumber ffinitebeliangroupsG with A(G) ? p.Proof.We maychoose a positive ntegerN and a primeq bothso large that

2 * 3/4)N/2 < p and 2 * 3/q2)1/2 p. If PA (G) ? p and pJdividesIG j, wherep isa prime, hen < N and p < q. This conditionmposes n upperboundon theorderof G. The result s clear.PROPOSITION 3. If {Gn} s a sequenceoffinitebelian groupsforwhichIGn oOas n -+ oo, then A G,,) 0 as n -+ ooProof. Since {PA (G,)} is boundedabove by 1, the limitsuperior f{PA(Gn)}is finite.ndeed, im up A(Ge) = 0, otherwisewe contradict roposition . ThusO liminfPA(Gn) < limSUPPA(Gn) 0.A problem.As a finalremarkwe pose thefollowing roblem.Suppose G is afinitenot necessarily belian) groupand S is its set of subgroups.Let G operateon S byconjugation nd considerPG(S). It is clearthatPG(S) = 1 if,and only f,each subgroup fG is normal.This s equivalent o G being belianorHamiltonian.The problemhenstodeterminef here xists omerealnumber , where < p < 1,forwhichPG(S) ? p whenG is neither belian nor Hamiltonian.The author on-jecturesp = 2/3. If thisconjectures true, he boundis sharpsincePS3(S) = 2/3.

References1. P. Erdos and P. Turan,On someproblemsf a statisticalroup-theory,V, ActaMath.Acad. Sci. Hung.,19 (1968) 413-435.2. W.H. Gustafson, hat stheprobabilityhat wogroup lementsommute?hisMONTHLY,80 (1973) 1031-1034.3. S. Lang, Algebra,Addison-Wesley,eading,Mass. 1965.DEPARTMENT OF MATHEMATICS, ROSE-HULMAN INSTITUTE OF TECHNOLOGY, TERRE HAUTE,IN 47803.

SOME SUMS INVOLVING FRACTIONAL PARTSL. CARLITZ

1. Put(.- [x] - I (x : integer)

0 (x = integer),where is real and [x] denotes hegreatestnteger x. Berndt 1] has proved hefollowing esult:

What is the Probability an Auto Morph Ism Fixes a Group Element - Gary Sherman

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