M.A. MATHEMATICS
GROUP PROPERI'IES WITH THE AID OF REPRESENTATION THEORY
A Thesis
by
Pushp Lata Jain
Submitted to the Faculty of Graduate Studies and Research in partial ful fil.Iœnt of the requirements for the degree of Master of Arts
McGill University, Montreal, Canada. July, 1960 ACKNOWLEDGEMENTS
The author expresses her sincere gratitude to Dr. M.D. Burrow for his guidance and encouragement during the course of this work. No other assistance was received. TABLE OF CONTENTS
1. Introduction
2. Representation Modules m • • • • • • • • • • • • • • • • • • • • 3 Realization of the Representation of a Ring into Linear Transformation of '1'7i. as Matrices • • • • • • • • • • • • • • • 4
Reducibility and Decomposability of 17!. • • • • • • • • • • • • • 5 3. The Regular Representation or a Special Representation Module and Its Properties • • • • • • • • • • • • • • • • • • • • 8
Form of the Regular Representation of a Semi-simple Ring • • • • • 21 Multiplicity of Principal Indecomposable Components in the Regular Representation • • • • • • • • • • • • • • • • • • • • 25
4. Applications to Groups • • • • • • • • • • • • • • • • • • • • • • 29 Group Algebra 1 and a Close Connection between the Representations of r and the Group • • • • • • • • • • • • • • 29
The Trace Function ) on r 1 Simplification of this Function when Restricted ~o the Group • • • • • • • • • • • • • • • • • 30 The Number of Distinct Irreducible Representations of r and G (group) ••••••••••••••••••••• • • • • 33
Character Relations • • • • • • • • • • • • • • • • • • • • • • • 36 The Properties of Groups by an Application of the Theory • • • • • • • • • • • • • • • • • • • • • • • • • • • • 42
5. Concluding Re marks • . . . . • • • • . . . .• • • • • • . . • • • 57
6. Bibliography • • • • • • • • • • • • • • • • • • • • • • • • • 59 GROUP PROPERI'IES WITH THE AID OF REPRESENTATION THEORY
Section l
Introduction
A representation of degree n of a group is a hororrorphisrn of the group into a group of nxn matrices 1dth coefficients in a given field K. The image group of matrices is necessarily a sub-group of
GL (K), the group of all non-singular ~~ matrices with coefficients in n
K. Two representations ~ and ~, necessarily of the same degree n, are equivalent if there is a matrix T € GLn (K) such that T 1(g)T;"~g) 1for every g f G. Equivalent representations are to be regarded as essentially the sarne.
The representation ~ is said to be reducible if ~ is equi- valent to ~ and the matrix ~(g) is of the form:
A(g) 0 yg c;G , ~(g) = * B(g) where A(g) is an IlDCIJl matrix, B(g) is an (n - rn) x (n - m) matrix while
0 stands for the ~-m) zero matrix and ~~ is sorne (n - m) x rn ma.trix. If
~ is reducible in this way it is clear that ~l and ~2 defined thus: ~l (g) = A(g), ~2 (g) = B(g) provide representations of G. Thus a reducible representation gives rise to representations of smaller degree.
A representation which is not reducible is called irreducible. An important objective of the theory of representations is the survey of all inequivalent irreducible representations of a given group.
By considering the traces of the matrices of a representation
€: , a numerical function fon the group is defined: f(g) =trace ~(g). Since trace (T ~(g)T-1 ) = trace ~(g) it is apparent that equivalent representations give rise to the same nurnerical functions on the group. Of special importance are the function J, the so-called character--3, derived from the irreducible representations: ~(g) = trace ~(g), ~ irreducible. As we will see, important properties of a group can often be deduced from a knowledge of its characters. Indeed there are results in group theory which have never been proved in any other way. On the other hand, the characters can be determined when relatively little is known about the group so that the representation theory is an invaluable aid to group theory.
The classi cal theory, the case where the given field K is the 1 field of complex numbers, is due to G. Frobenius • His most important 2 results were proved independently by Burnside who gave a number of important applications as well as its classical treatrnent. Fundamental work in this field was also done by I. S chu~. An important treatment of the theory can be given which is closely related to the representation theory of hypercornplex numbers4 considered by E. Noether. 5"
In this t hesis we atternpt to give an account of the theory of representations from a consistently algebraic point of view. \~ile the results are weil known, many of the proofs involved in t his approach appear only in the journals or in lecture notes, notably of R. Brauer, 3
but so far as the writer is aware are not to be found in any published textbook.
Section 2
Representation Hodules
Let 1J be a ring and K be a field. By a representation module
rrc.. we mean a (K, '0' )-rodule )'i( , that is an additive abelian group nt.. having the elements of K and 1J as left and right operators respectively.
This means that 'd u (;- rrt., V a ~ K, V at "()" there are unique eleJœnts au ~nt and ua (;;rn and the following rules hold:
1) a(u + v) au + av
2) (a + 13)u ;: au + !3u
3) (a.j3)u ;: a(13u)
4) (u + v)a = ua+ va
5) u(a + b) ;: ua+ ub
6) u(ab) = (ua)b
7) (au)a = a.( ua)
From now on we will asswne that '0"' is a finite algebra over K with the identity 1 and that furthermore 8) a(ua) = u(aa) u(aa).
For every a t '0' we can define a mapping ~(a): m -~ rn thus:
'V u é-llt , u ~ (a) = ua.
· · (u + v) ~ (a ) = (u + v)a = ua + va = u ~ (a) + v~ (a)
and (ku) ~ (a) = (ku) a = k( ua) = k( u 0::: (a)). 4
We see that ~(a) is a linear transformation in -,re. regarded as a K-vector space. Moreover since u ~(a+b) .. u(a + b) = ua + ub = u~(a) + u~(b) =~(a)+ ~(b))
and u ~(ab) = u(ab) = (ua)b ~ (ua) c (b) = u(_Ma) ~(b))
We have
~(a + b) =~(a) +~ (b) and
a:;: (ab) = ~(a) ~ (b).
Hence the mapping a-P da) is a ring-homomorphism. In this way each representation module ~ives rise in a natural way to a hommorphism !:'of the ring ~into the ring of linear transformations in
-rrt regarded as a vector space.
Now if ~has a finite K-basis (u , u , ••• un), then there 1 2 are uniquely determined elements aiJ ~ K such that n (i = l, 2, ••• , n) and we arrive at a realization of c- (a) as an (nxn) matrix (a J). It 1 is easi.ly sean that the correspondence ~(a) - /A = (aiJ) is an isom:>rph- ism of the ring !::'"'("'0"') onto a ring of nm matrices. Moreover if v , ••• , v 1 0 is another basis of ~connected with the first basis by the non-singular -1 ..... ) matrix T = (t1J); T ~ (tiJ ; n v. = ~ t.JUJ (i = 1, 2, ••• , n) ~ J==l ~ n n then since vi ~(a) = via= z tiJuJa '"' ~ tirjKu.k J=:l J,k==l 5
n ~ J,k,S=l
the new correspondance is
~(a) -~ TAT-l
To sum up: each representation module n1. provides a represent- ation ~ of 1)' into linear transformation of nt which can in turn be realized as matrices determined uniquely up-to equivalence. More generally it is easily seen that operator-isomorphic(a) representation modules lead to equivalent representations.
A representation module )')"t is reducible if n
If mis reducible and we take a basis of '17t adapted to n, i.e. a ba sis \ ul' u , ••• , of 'h"t such that u , u , ••• , uT'J is a basis 2 un~ ~ 1 2 of )lt then the matrix corresponding to o_:::(a) is of the form:
A(a) 0 ) o:_(a) = { * B(a)
where A(a) is the rxr matrix corresponding to~(a) and B(a) is the Ln-~-~ matrix corresponding to ~2 (a).
These results lead us back to the similar remarks of Section l. 6
)1(. is decomp:>sable if Tit = 1'r! + nt • 1 2
In this case the corresp:>nding representation 0::: of mis said to be
decomposable into the representations o_: and o.::, called comp:ments, 1 2 corresp:>nding to the representation .m:>dules and m respectively. 11'\ 2 \'lith a suitable basis the corresponding matri.xa:::(a) has the form
0:::. (a) = ( A(a) 0 ) 0 B(a)
It is to be noted that ~may be indecomposable, that is not decomposableJ without being irreducible.
Since l'Y't is a group with operators, we have from group theory the following results.
Il Theorem 2-1: (Jordan Holder) For any composition series of subrodules of
111. ; (l) m. the number k and the factor modules --~- , i = 1, 2, ••• , k, are uniquely rrt. . 1 ~- determined, the latter up to operator-isomorphism.
Theorem 2-2: (Remak-Krull-Schmidt) For any decomposition of ~ into indecomposable submodules: . (2) + m,r
the number r and the indecomposable submodules nti are uniquely deter
mined, the l at ter up to oper ator-isomorphi sm. 7
Theorem 2-3: (Fittings Lemma,or rather a special case of it) If '"'is indecomposable and both chain conditions hold, then any operator endo- m:>rphism of nt is either nilpotent or an isomorphism. In terms of representation theory these results give:
I' _ Every representation ~ of a ring -& has a unique number of unique irreducible constituents o:J; i = 1, 2, ••• , k. Also if a suitable basis of ~ is chosen such that it is adapted to the chain of submodules in (1) ~(a) will be of the form
At a) ~(a): *
Vfuere A. (a) is the representation corresponding to a suitable basis of J.
~~ ;r(.l-l' namely 111.. i-l + U J ( ~ U.~' basis of ~ ) 1 where UJ runs through the basis elements wh~ch belong to 1'"r( . but not to rtt . • J. J.-1 0 is a set of rectangular matrices with zero entries and * is a set of rectangular matrices.
Il' - Every representation !:"of a ring -o- has a unique nwnber of unique indecomposable components~ .• -J.
In this case the corresponding matrix O::(a) will take the form
0 o::Ca) =
0 ~a) 8
Section 3
The Regular Representation
The algebra "'0' with its own additive structure as a mdule can also be regarded as a representation module, for its product satis fies the rules 2.1-2-8. The corresponding representation is called the regular representation and plays an important part in the theory. We note that the representation submodules of 1Y are then right ideals in c--.
Theorem 3-1: Every irreducible representation !:: :f 0 of a ring ~ appears as a constituent of the regular representation t•
Proof: Let --m,be the irreducible representation module of~·
Since [::" 0 au f. ~ such that ua 0 for sorne :f 1 T a 1 te-
Now consider the mapping I_: ay= ua, v a € lY.
·: 71'(. is irreducible this mapping is ont_ç (otherwise the image
of -(! would be a proper submdule of "')"T'{.) besides:
(a + b)! = u (a + b) = ua + ub a~ + b _E.
(ab)!= u(ab) (ua)b = (a!)b
(ka)~= u(ka) = (ku)a = k(ua) = k(a~) so that ~ i s an operator-homomorphism of the module of ïr onto the module
)T(... If fi is the kernel of this homomrphism then we know from group theory that
But as the r epresentation from 1ijf.i i s a constituent of L we have the theorem. 9
In greater detail, as theorem 3-2 below shows, every irreducible representation occurs as the unique bottom constituent of some principal indecomposable representation. It will be expedient to establish first the following:
LemmajA If f.. is an operator homorrorphism of a component oc of the ring "0' into -rt,..and ~ is an operator homomorphism of 'l?'(onto -n , then 3 an operai:.or hom:>morphism ~ : "'- ~ ne. such that l = ~~·
Proof: Let ~ be the decomposition operator (an operator homomorphism) which maps -o- -onto the component ()(... Note that a~ = a, 'ta~ «..
Let (l.~)L. = c ~ 1t
• ,. @_ is onto 3 b e; m. b ~ = C
Let 1 be the mapping ot ~l'l'(. given by a;t= ba, v a e: ot.
·.-(a + a') 1. = b(a + at) = ba + ba' = aJ. + a'.J
and (aa')! = b(aa•) = (ba)ar = (a~)a•,
~ is operator hommorphism.
Also a(l ~) = (ax)~ = (ba)§ = (b!)a = aa
= (l.~)ta = (l~~)a
= (l.a)~ = (a1)L = a& ~ a ~ ot
Corollary 3-A-1 : If at. is a component of -c- and f.. is an operator homororphism of ot into tr7" l'C- ( 7't is a (K, -&') subrodule of -rrc.,the representation module of ~ )
The n 3 :! : 01.. ~ l'Y"(,. 10
~: Let ~be the natural hommorphism of ntonto 'r1 )"'(...•
Then the lemma determines ~·
Corollary 3-A-2: If in Cor. 3-A-1, "')"(.. is the unigue maximal sub- module of rn ( that is lt'C. m, ?è f "nt. ~ 1'1! ~ )'(.. ) then ~ thus obtained is -onto ~: If Dt '{. =t m. then
ot!_C: rt.. ·: c:t.1 is a submdule
and O<.ll rrc~!!!.) re. or oc. t_ = 0 (contradiction fort;_ is not zero map of 0(..) Corollary 3-A-.3: If ~ is an operator hommorphism of a repres- entation module )'l'( onto a comJ;lCment oc. of -r:r. Then 17e..= ffl:J.+ and 77( ~, = 0 such that )'Xl "'=' rx 2 ~ • Proof: Let E.. = f be the identity map of ot- _, 0(, • From Lemma JA, :1 operator homomorphism ::E ~ en.~ l1'(.. auch that1:_ ~ = ±_. Let oc.!_= n:t_1 ~ = ~ ue:-~ 1u!] = 0 f If u (;- m1 n ~~ then :3 a +- 0, a ~ ot. , a l.. = u a~ = Ul) • 0 = a which is impossible. -·. u = o. However v u ~ rrc., :1 a ~O'C. ; u JJ = a Then ·.- u = u - a "! + a ~ and ( u - a :rh!. = a - a ~ 0 u - a 'I 6 ~ il whereas a~ é "l by definition .:. bt = -ml ..j.. )7t2 }1oreover·:a.r = 0~ al' l = 0 =9 a = 0, !_ is an isomorphism Ot. '::;; nt 1 Theorem .3-2: Let tf = ot + ••• oc. indecomposable 'Vi. Then 1 ~ (a) ::1 unique maximal right ideals h .(1 at., ~ . f= 0t . 1 (b) Moreover ~ ~ ~ ~ ot.; ex.. Dt.~ ot. ~-.l. ~ ~J and every irreducible representation of lr ~ J t;,. ~ J is provided by Oti for sorne i. bi Note: It could be that l1. = 0 for ail or sorne i, in which case ~ oti is irreducible as weil; in the former case -(J is said to be completely reducible. Proof: a) ·: "0" is finite over K, the maximal condition holds and =.:t a right ideal h .a. OC.., ~ • '/= Dt., i ma.xima.l in ~ ~ ~ ~ A7 ~. Let h . 1 f= /.:; . be another right ideal E: ot.. and k. 1 ~ ~ ~ ~ Then 0t. = l-; . + /-; . t. For ~.; is maximal in f'rf •• ~ ~ ~ .l. .... ~ 1 -& = ./; i n .Oï f 0 1 ·:IX i is indecomposable. Let f_ be the ma.pp:i.ng of oc i into l>-i/ ~ defined thus: at_ b mod ..S if a i' b (:; .t;. i' b 1 c;. i 1 and a "" b + b]_ = 1 ~ ex 1 1 1:r 1 e_ is weil defined, ·: if a= b + b •, b - b = b 1 - b 1c: {) 2 2 1 2 1 2 whence b b rnod and b ' b ' mod 1 = 2 ~ 1 = 2 --s ·: !Ti/i> Cl Oti/~ we have the following scheme of ma.ppings -->1 ot·~ oci/~ oc. ex. where l) is the natural ~ ~> ~~~ homomorphism oftt. onto ot ·;, ~- ~'\). 12 Hence by Cor~-A-1 -3 :r : ~. ~ ot. and f ;::: 'l' JJ. ~ --, ~ - -- Now 'V bE. J::s i and b 4 ~ as b = b + 0, b f ;::: b (rod ~ ) So b t. ~ ;::: b (JWd ~ ) But (b !) ~ ;::: b "L (nod ~ ) ~-. b 'l:_ ==. b (mod -t ) ,•. 'Vi b 'Z i =b ( m:>d ~ ) Hence 'r is not nilpotent and .. by Fitting's Lemma (Theorem 2-3) .1 is an isomorphism of ex . ~ ot.. .• ~ - ~ naps CXi Hence r_ ;::: r ~TOti onto T and this is a contradiction ..' t maps oc.i onto .ô i /rb C i 0tJ we show that ot. ~ b) If 0t. i/ I:J . i ~ 1 ~J The following scheme of mappings is evident: ot· OtJ oc.. v ~ i ~ ~~ J,. ..:a ~ ~J ot.J Ot. ~ J hJ By corollary J -A-2 with t_ = ~ !, 3 a '1. : Ot.i ---?JYO(J Then by corollary 3-A-3, ·: 0(... is indecomposable Ot;. ":f: OtJ' ~ On the other hand if 1: is the isomorphism oc i '::;;' OC.J then .e, • '"'( ;::: ~J which leads to an isomorphism oti/ ~- hi ot.J/ IrJ Definition 3-1: An operator homon:orphism of a (!{, tr) roodule nt. into a (K, t1) m:>dule n is called an intertwining_ of n-t and x • 13 Let ~ ( m., re.) = l 0::. ) ~ an intertwining of 7r(. and n} It is easy to check that -;; ( ~, n) is a K-module, and that its K-dimension is < mn when m and n are the K-dimensions of -ne and rc respectively. Let i ( m , n ) = K-dimension of ":1 ( m., 1"C.) • Ive will also write i~ ~) = i( rrc, ~ ) where).:!, v are the corresponding representations provided by m and rt respective1y. Theorem 3-3: If "j ( ot' rtf. ) 1 "j ( 0(.. , l"L ) ~ J ( at ' rrt;'ft. ) Proof: If a:_ E:. "j ( ot , irl't) and ")) is the natura1 homororphism of m~1 n. the n (_ = ~ ~ ~ ":1 ( oc. , TYj -rt ) • Converse1y by 1emma. 3-A if t_ t:: ":1 ( o<., -,..rt/rt ), there exists a o- t "J ( ot,-rn.) andr:.._ = ~ ~. Hence the ma.pping );: : ~ _:;;. ~ ~ = L is easi1y seen to be a K- hom:nwrphism of ":J ( crc. , m) onto "j ( ot., ~- Corollary 3-3-1: i( ot., m) = i( oc., rrl 1-rt ) + i( ot.., n). Corollary 3-3-2: If -& is an al gebra over K and n-<..is sorne (K, '""0"' ) m::> dule then a) ~( ~, r1'<) :;;;' m as K-IrPdule b) ~( ml + "'n'l2' re. ) ~ ., ( n')_' rt..) + ~ ( m.2, ~ ), where 1'! is a (K,--& ) n:odule too. Proof: a) ' ( tJ, rn) : 1 .C.j [_ operator hoirPmorphism of tT~n-c.J 'V t_, I -7 I.f.. t -rn Le..t' ~ : f. -;> I.f. Now (tl +l2)o::::= I.(e_l +.f2) = I.E_l + I.f2 = !.1~+ E..2!!: Also (k.P)o--.= I.(kf) = k(fov), so'~ is k-homorrorphism of the - iiiodule -- - intertwining of "tl and h't with the (K, -CJ ) module "'O't. 11oreover E_ ~ = 0 ~ I.!_ = 0 or (I._t_)a = 0 or a.f = 0 , i.e. f is 0 map. _.. 0::: is an isornorphism or "":r ~,v ~),,.. = dimens;on..... --..... b) Every ma.ppingo:= of m m - ':> 6"i gives rise to 1 + 2 a pair ~l' o;::2 ) where o_:::i is the restriction of q: to ~' i = 1, 2. Similarly every pair (o.::. , o::_ ) gives rise tao:::::- 1 2 If l ~ ( 1 1, }'2) ' 1 E- 'j ( 1'T<.l + ~2' M.) then ~ + :f e; ) C:-1 + - ~ , ~2 + ~2 ) :. The correspondence ~ c~ (~1 , ~2 ) is an isornorphism of 'J( ~ + nt , 1't) onto AJ( ~~ N.) + ) ( nt , tM.). 2 2 15 Definition 3-2: The ring of operator endomrphism of a (K, ~) nodule '17'( is called the conmuting ring of ~ denoted by "(; ( rrc.) • If mis irreducible (9 ( m) becoœs a s-field over K. In this case if r denotes the K-rank of ~ (rn) we have from definition 3-1 1"\'(. i( 1 rrt) ... r Theorem 3-4: If f is principal indecomposable representation oftr, and .!!:' is any arbitrary representation of -e-. Then i(f , ~) = mr where m is the multiplicity of the unique irreducible bottom constituent of ~ in ~ and r is the rank of the cornmuting ring of the representation module of this bottom constituent. Proof: Let mbe the representation rood.ule of~ and n'(.l ~ Trf~ be the representation module of~'">, where "l'l'(.. is a (K, -&) submdule of ~. -..:;. 0 Now by corollary 3-3-1 i(f..., ~) = i(tJ ~1) + i(f., ~2) ~l is representation provided by )"'1"(.. o· So if~ contains (the unique irreducible bottom constituent t 1 of !) m:t. tines and 1(:'2 contains it ~ tires and if we assure the result for ~l and ~2 • Then i(f.., d = ~r + ~r = (~ + ~)r. However !: now contains t 1 (Il)_ + ~) tilœs. This shows that if the theorem is true for a representation of lower degree than ~then it is true for ~also. So by induction the theorem will be true if it is true in the case when ~ is irreducible. 16 the representation Let t be~corresponding to the principal indecomposable component eX of -(J • And let !: E: 'J( (f(. , m) and '!::' =f 0. Since 7t'L is irreducible, .: , o.::. is onto and induces an unique isomrphism f.: otj~ ~ 7Jrt, where .(;. = the kernel of fJ'-'. But n'( is irreducible so h must be rrrud.mal and hence the unique maximal right ideal in ot. ( theorem 3-2). In this way -v ~ 1:. ':1 ( at, rrt.), ~ provides the unique &(:- "j ( o<.;J, , nt) and conversely Def inition 3-.3: The radical N of a ring ~ is the set of all elements which are mapped onto 0 by every irreducible representation of -&. or N = ta J .fi (a) = 0 'V' ft irreducible } ~ve list the following properties of N , assuming that tr has a left identity and that both t he chain conditions on right ideals hold t rue in 1r • 1) ri is a tw sided ideal of ,_ • Proof: Let a and b E: N ·: F. (a + b) = F . (a) + Fi ( b ) = 0 .•:. a + b G N -~ -~ -· Also ar and ra é N , a t: IY , s 2) N is nilpotent . That is 3 s suc h that N = 0 a Proof : Let Then oc n c 0(. independent1y of n. m Because ~ is representation module for some irreducib1e l"'\ representation f, wn=O , nE:-N or an E:Ot.m Take O't..= "'0' -&V L "'0'" '0" , -tf maximal right ideal in "0- • 1 c ~ ~ And by doing the same thing to "'0' , ~ We have • • • • • and so on. Or -cr v v • • • v (.. -& C. • • • C. ""&. 1 2 s m s-1 -; Mininru.m condition ho1ds, ~ s such that ~ = 0 ms ·: 1. f -o- ~ N6 = 0 as choice of the v. is arbitrary. l. 18 3) If Y é l:'f and Oti y c_ oc." . 'V i_, where ot. are principal ~ ~ 1\. indecomposable components of --o and ot. is the maximal right ideal in oc.. ~ ~ then r E- N. oc. . Proof: ~ . . Ot. r C. dt. ---:- r = 0 ' • ~ ~ r·~ But r.01- i gives an irreducible r e presentation of "'D'". Horeover ail irreducible rêpresentations of 1Y are obtained in this way. f!Cr) = 0 v i , or r ~ N • 4) If lf= . " Th en N = ~0(.. J where ot.." is the unique maximal right ideal C::: ot .• ~ ~ ~ Proof: Let n (::- N 1\. Then oc. n C:: 01. ~ ~ .: ·iot.. n ( ~ Bt . ~ ~ • 1\. or ï)"n C ~ 01.. ~ or l.n = n E -· · N ~ ~&i Conversely, consider the rna.pping ~ Ot_j_ -~ 0(.J l t fot.J and given by a -::Y ta fi.xed. It can be easily seen that '1. is operator homomorphism of ex.. ~ 19 But &c..i.j -7 O!.J is not onto. For if so 1\ ~a <::oc. , 3 IX.. such that a 1:: = a' 'Y l l - ,1,1. or (a- a').J= 0, or a- at f kernel X· But Kernel !. C: Ôti , St_i being maximal in Ot..i. a' necessarily belonging to & . l which is not true. 1\ (\ t. '' Ot.l ' OtJ 'Vi 1\ ,... or Ot. oc. . But & . is an arbitrary one l :, ~. ot." (! N l . (\ He nee N = t Ot.. l 5) If Otis a right ideal (1J and ot is nilpotent, then 0t.. C. N Proof: Suppose ex ~ rr then ;~ a C::- oc.., a 4 N 1\ and ::l i such that OC.. i .a 0<... 4- l ~. ot.. (1) O<.i .ot l From (1) ot. • ()(.2= Ot l i But ... Oi.. is nilpotent, 3 r such that r o-s_.oc = o or Ot..i = o. But oti t= 0 f\. ex.. . (){. c: ot.. . -. l l Vi 20 and hence 6) 1f 1r< is a semi-simple ring. Proof: i) tr1 t< is a ring. If the relation fis defined on 1r, such that ata', if-fa- ar~~ Then it can easily be checked that! is true equivalence. Let L§J denote the class of -o- (rood l'l'"') containing a. "'the swn of two classes defined as ~ + lQJ = La+ bj is obviously well defined. The product is also \'/ell defined . as: ~ . L.hl .. L§&) For if a= ar , b = b' (1) We Show ab : a 1b 1 a - a ' <.- !Y' also b - b 1 " ~ ab - a'b as i-rell as alb - albi E: rY' ( re being two sided ideal of -a-) or ab - a' b + a'. b - a 1 b ' é f'( or ab= albi ii) 151rr is semi-simple. Let F be an irreducible representation of if. De fine "'F of 111 \"(as ( ;:-) f (LaJ) = F (a) 21 1\ F is an irreducible representation of ~~~ otherwise it will 1\ contradict the irreducibility of f. F is well defined. For if a, b E, LaJ a-bE:: T"( ,·, f (a - b) = 0 or F (a) = F (b). 1\. Thus F (L§J) is the same for every element in this class. Conversely if f is an irreducible representation of -c /r't' the relation ( *) defines an irreducible representation f of ë:f • Noreover the correspondence f ~ ~ is clearly 1-1. If LêJ f n where 7!. is the radical of tf/ rc ,, Then F (W) ;:::: 0 ;:::: f (a), V irreducible representations of 11. or UlJ \.Q.J -a; rf is a serni.-simple ring. Corollary 1 1) and -tf/f( have· the same irreducible representations. Theorem 3-5: i) The regular representation of a serni-simple ring is completely reducible. 22 ii) Any representation of a semi-simple ring (if it does not contain the zero representation as a component) is completely reducible. Proof: i) Let -D' = at ot otn 1 + 2 + •••• + ( ot . are indecomposable right ideals in tJ' ) ~ By property 4, each ot , v i , have only 0 as a proper sub i ideal. ..:, ali the Oi . ts are irreducible • l Proof ii): Let nt be any representation rrodule of the semi-simple ring tf and let . + • • • • + rn. indecomposable n'''t ~ '\'le wish to show that the )'f'(. • are also irreducible. l Now by asswnption m . -& f 0 'Vi l Let nt. 1 t rr<.. be a maximal subrodule of '"'f'l'<.. , • l l ~ Th en }?"(. i is an irreducible subrwdule. 7rt,. ' ~ Yf"(., . If ~ does not provide t he zero representat i on, then it provides '"fr( 1 an irreducible ~epresentation of 'tr. )'Y'{. • i ~ "'1i[T ':::::: ot.J , f or sorne 'J • i Let)) O'ti --?) Î)'(.i (?? is t he natural horwmorphism of TY'{. ~ """-i) "i=lt ~ 1re. ' i l Ir(.. 1'Yi (1) "nt. (2) -·. By corollary3-A-3 with E. = 1.! = + i i ~ 7rl. (1) (2 ) """'= en. and 11"{. = nt.. ' i J i ~ 23 ·.·~ is indecomposable. »< (1) (:Ù (2) 1 ' either a) nt. = orb) or-e.. = nt. = nt. - ~ i ~ ~ "'i ~ b) is not true ' - . "rrt.. is irreducible for arbitrary i. ~ That rn i cannat provide the zero representation of -cr may be seen as '11'{. 1 follows:~ Let 11 be the representation of -cr corresponding to "l'K •• /:=. ~ Consider the ma.pping/.!(1): nti _ 7 nti defined by x._l!(l) = x.l It is easy to see that~(l) is an operator endomorphism. So by Fittingls lemm.a either }:!(1) is nilpotent or an isomorphism. If E(l) is nilpotent Then :'J r:l:!_(l)r = 0 or E_(l) = 0 But then ~(a) = E_(l.a) = E_(l) .E_(a) = 0 va é -rf or~ is the zero representation i.e., -mi provides 0 represent ation of "fr (contradiction) ., ~(1) is an isomorphism (l) yr-t. ·; Now if 1 provides the zero representation, ~ n-t.· ~ then it contradicts (l) Hence the t heorem. 24 Corollary 3-5-1: If the ring 'tf has a faithful completely reducible representation then 11 is semi-simple. Proof: Let ~ be the given representation i.e. M = F + • • • • + F -- 1 -s Let n ~ N (N the radical of Ir ) Now ~(n) = K1 (n) + •••• + ~lt)= 0 --Y n = 0 Since ~ is a faithful representation of lr. Corollary 3-5-2: If the ring tr has a faithful irreducible representation then it is simple, i.e., its two sided ideals are 0 and 1r only. Proof: -tr is semi-simple by Corollary 1 • So let -rr = Dt. 1 + • • • • .:,. 0(.. k" Suppose Then ot. 1 0t J '1= 0 't:f J Now :f : OC J ~ot1 defined by aJ-----" a1 · " Now let h be a t'WO sided ideal of 1:r. Then either OC.. f;- = 0 ~ = 0 Or Dt. Ot.- C.b-. f"= 1 But then OC.. J~ 'i= · 0, otherwise 0 = ot. f>- i J - i Ot.. ~ 01.- J 1" '=' ot i:r 0 Or Ot-J! 1J I = '. 01.- Ji:, = ot J cr. ,t:; 'rJ J ~·. N- '=' 1f' . 25 Theorem 3-6: The multiplicity h. of the principal indecomposable components l. Ot. in the regular representation of tris n.;/rank C(F. ), where n. is the l. ... -l. l. degree of the principal indecomposable representation corresponding to ~· l. and C(F.) is the commuting ring of the unique irreducible bottom constituent -l. OC..l..• -l.F. of Proof: Let "&= ( O'L (l) + oc. (1) + + ex. (1)) + 1 2 • • • • h ( Ot. (2) + O'L (2) (2) . 1 1 2 + ••• +OC.h )+ •••• + + (ot (k) + 2 + ~ (k)) ( otl(k)) + •••• 1 • • • • hk where oc (f.L) ~ ()t ())) i J if-f 1-L ""~ k D-t. (1)) ht ( 0(. (f.L) 0(.. (1)) i ( -o-, = i z , = 1 l.=f.L J-1 J 1 h)J.. k i( oc... (f.L) , ()(. (1)) i i l. 1 J=l ~1 (By Cor. 3-3-2b) = h1 rank C(f1) for i Ot.. (l)) = rank C(F ) if f.L = 1 (()(.~I-L)l. , 1 -1 =0 iff.L+l K-dimension ~l(l) or h 1 ""' (By Cor. 3-3-2a) rank C (f1) k f 2 Corollary 3-6-1: If 11 is of rank n then n ~ ~ _L. where ri i•l r. • rank of C (Fl..), and f. is degree of an irreducible representation F .• - l. -l, Equality if-f -(J is semi-simple. 26 k Proof: n• i h.n. i=l ~ ~ k 2 ~ n. (From Theorem J-6) ""' ~ i=l r. ~ However if -o-is semi-simple fi == ni 'Vi and we have equality. Theorem J-7: (Generalised Burnside Theorem). If F is an irreducible algebra of matrices of rank n and degree f. Then n = f 2/r where r is the rank of C(F). Moreover F automatically has an identity element. Also F ~ to a complete matrix algebra of degree f/r over a field of degree r over the ground field K. Proof: F ~If 1 Ex.tend F to F* :; F and a. If, a.(: K F* is of rank (n 4"- 1) by definition. It is irreducible also otherwise it will contradict the irreduc- ibility of F. F* is faithful irreducible representation of degree f. So F* is simple (Cor. J-5-2) But F is its proper two sided ideal. :, F* = F or F -:; If. F is simple of rank n and degree f 27 (By Cor. 3-6-1) ··or.. ""' cx \1' i as seen in the Now F = ~ + cx.2 + • • . i 1 proof of Cor. 3-5-2 .'. k = 1 and so n = ~ giving the first part of the theorem. Moreover h = f/r (By Theorem 3-6). The fact that F ~ to a complete matrix algebra follows from the structure of the endomorphism ring of a completely reducible module - (Van Der Waerden8). Theorem3-8: If K is an algebraically closed field and C(f) is the colllllUting ring of an irreducible representation f then r == rank of C(f) = 1. Proof: Let ~be an irreducible representation module correspond- ing to !:_. Then C(!:_) = {o::} ~ operator endororphism of4 and C(f) is a s-field. ~ .. ua ("-I- ~) = 1-.ua- u ~a= u (A.I- cr::- )a, u~~~ a el)". ~·. /... I - q:: f C(f), 'V !(. tK. Now since K is algebraically closed 3f...f;K, 3 u f ~ u 1 0 such that u ( 1-..I - ~) = 0 (*) Otherwise u ( A- I - ( t-. I - • • • • ( "- I -(!>-) 0 1 ~) 2 ~) s - + "Where the /.... i nay be taken as roots of the polynomial p (x) ... 0 and p ( ~ ) is the polynomial ex:pressing the linear dependence •••• that must subsist between u, u !:', u J- 1 If ( A. I - ~) f 0 then ~: C (f) is a s-field ( A. I - ~) -l 'f 0 and 28 from ( *) u "" 0 a contradiction •·. o.:: - ~r -·. C(f) is of diJœnsion 1 or r = 1. Corollary 3-8-1: If -o-is semi-simple and K is algebraically closed then formula T 3-6 becomes k n ... ~ i=l Theorem (Frobenius Schur) : If ,.. is a ring and , • • • • • • , 3-9: !:1 Kk are the irreducible representations; If bl" •••• , bk E -& are any arbitrary elements then =:~ a e:- tr such tha.t F. (a) = F. (b.) i = l, 2, • • • • ' k. -~ -~ ~ Proof: It will suffice to asswœ that tris semi.-simple for if we find lâ1 f- -f11 't(' ( tf-radical of "tr ) such tha t F. ( 1 a' ) = F. (\b.\) for the semi-simple ring -(!1 rr. -~ ~ -~ 1JJ then: F. (a) = ri l( a )j -~ And ri < ~ ) "" ri (bi) -·. F. (a) = F. (b.) -l. -~ ~ Let lr = ot.l + • • • • Where each ot. is a two sided ideal of ir and F. is the uninue ~ ~ ~ bottom constituent of ~i which is one of the isomorphic minimal right ideals of ot.. i. 29 = b (1) + b (1) + + b (1) Let bl 1 2 • • • • k • • • • :0 b (k) + + b (k) bk l • • • • • • k Forma=- b (l) + b (2) + •••• + bk(k) 1 2 k Nowf~ (a)= E~ ( ~ bk(k)) =- F. (b~) == F. (b.) ...... J=l --l. ~ -~ ~ ~1. For .El (bl) : lv l (1) -~ h 1 (1) , where ~ k . (l) =- Ot.. and i::.l ~ 1 [)' (l) is the minimal right 1 ideal for f • 1 k 1 1 o 0{.. : t.f (a)= t.a = t b}J) = tb (l) = t,E (b ) t tt~ c 1 1 ~ 1 J=l 1 1 1 tb (J) = 0 if J 1= 1 for tb (J) <:: OC (\ ot J J 1 ;;• Or in general F. (a) = F. (b.) -~ -~ ~ Section 4 Application to Groups The group algebra. Let f be the set of ail forma.l sums: c g + 1 1 + c g formed from the elements C. of a field K and the elements g. • • • • n~ ~ ~ .30 of a group G of order n. If we define the swn in r by n n n <. c.g. + i d.g. = ~ (c. + d.) g. i=l l l i=l l l. i=l l l l and the product by: n (< c.g.) = z_ ll < ~ (cidJ) gigJ i=l i J then r becomes a linear associative algebra with the elements of G as a basis. This is called the group algebra of G over K. Let ~ be any representation of r by linear transformations in a K-vector space. By restriction of 1-1 to the elements of g. \: G we get a - l 1'\ representation~ of G by linear transformations in the K-vector space: 1\ 1\ A H (g.g.J = M (g.) l1 (g ) - l ... - l - conversely if a is a representation of G we are led to the representation H of r by defining n = t i=l Hence the representations of the group can be studied through the representations of the group algebra. This is our procedure here and we will use the sarr~ symbols for both representations. If "YTt.is the representation module for the representation t!_ of 1 , uf is a basis of Îl"'"(. then for a t r and u1, ••. f ui ~ (a) = uia = ~ J=l 31 In this way we arrive at a matri.x C = (ciJ). The trace J of the representation~ is the function on r to K defined by f Î' (a) = trace C = ~ c ;t i=l ii It is easy to see that the choice of another basis for~ leads to an equivalent matri.x TCT-l which therefore has the sa~ trace as C. More generally, operator isomorphic representation modules have the same trace fWlctions. From matrix theory we know that if A and B are matrices of the same degree then 1) trace (A + B) = trace A + trace B 2) Also if M is the matrix arbitrary square natrices we have trace M = trace A1 + trace ~· The first relation shows that if J is the trace fWlction for some representation then and the second relation shows that the trace function of any representation. equals the sum of the trace functions of the constituants of that represent ation. Thus if , , •••• , f.k are the distinct irreducible constituants ,E1 ,E2 of!! each with nn.ù.tiplicity ~~ •••• , m krespectively and , ••• • , 11' J2 fk' X the corresponding trace fWlctions: f (a) = lil J1 (a) + • • • + ~ J k (a), Y a 3) Again if A is a matrix and An = If, for some integer n 32 Th en 3 a rnatri.x T such that TAT-l == ( t_l . .. 0 ) j 0 ~f O.Ye where f is the degree of A, ~- nth roots of unity. f ~" ·:trace A == trace A1 == ~ ~ i i==l -1 f and trace A == ~ 1:i , ('ëë.i is conjugate of ~i) i==l -1 :. trace A == trace A This shovm ~hat if g <:: G then Theorem 4-1: If ~ is the trace of the regular representation ~ of G of arder n then for g f: G ~(g) == n , g = 1 =O,gfl Proof: Here the representation module is r itself and using the basis g •• g : 1 • • n g .R(g) == g.g == g if-f g = 1 ~- ~ i from v1hich the result follows. Theorem 4-2: If the characteristic p of the field K does not divide n, the group arder of G, then r is semi-simple. Proof: Let )\be the radical of r and ~ be its regular represent- ation. 33 Let v E- ty .·.v = clgl + • • • • + en~ trace !! (v) = trace !! ( ~ c JgJ) = ~ But (v) 0, :. c 0 !! = 1 = -1 Let v• = g k v v• E: l"( · · r'l is a tw:> sided ideal. 1 1 But v• c g + • • • 0 = 2 2 • + ckgl + • • • • + n~ :. ck = 0 'V k in this way • . ·.v = o, V V (: I'C" Hence rt = o. Theorem 4-3: The number of irreducible representations k ofr (also of G) over a field K of characteristic zero or characteristic p, Pl n, is equal to the rank of the centre ofr 1 and also to the number of classes of conjugate elements in G. Proof: Let a~z.( r) 1z.( r ) = centre of r ) Then ab = ba , v b ~ r :. F. (a) F. (b) = F. (b) F. (a) yi, where F., i = 1, 2 . . -~ -~ -~ -~ -~ k are irreducible representations ofr. -. fi (a) = c .li' (By theorem 3-8) ~ i f. is degree ofF. ~ -~ We now have a correspondence a ---~ (c , c , •••• , ck) 1 2 at:Z.(r) 34 Conversely if (c , c , •••• , ck) is any arbitrary set of 1 2 ele~œnts given in the field, then we can find ~. <:- r such that F. ( 1. ) = ~ -~ ~ c I (By T 3-7) but by (T 3-9) we can find an element bt-r such that i f. 1 fi (li) = fi (b) or (c , c , •••• , ck) ---?> b. 1 2 b is unique, for if (cl, • • • • , ck)-7 bt also:t F. (b) = F. (bt) , v i -~ -~ or F. (b-bt)=O , i -~ " ~'. b - bi ~ 11, fYis the radical of r. or b = b t , •• • r is semi-simp1e. Also b f- 2. (r) : v ae:-r, F. (ab) = F. (a) F. (b) = F. (a) c. If "" -~ -~ -~ -~ ~ . ~ F. (a) = F. (b) F. (a) = F. (ba), ciIf. -~ -~ ~ -~ ~ ·t, F. (ab - ba) = 0, -~ or ab = ba. As the 1-1 correspondence is an isomorphism bet,reen z(r) and a k-diJœnsional vector space we have the theorem. Now it will be shown that rank k of z. (r) is also equal to the number of classes of conjugate elements in G. Let a(:- z. ( r) Then a a g + , = 1 1 -1 Also a = g ag _ ( ·: g E- r and g -1 exists) 5 5 6 6 • • • + a Il = a g g g·~ ·r • • n""ll 1 s 1 s · • • • + a g g g-l. n s 1 s • 35 + an8u = alg(sl) + a2g(s2) + • • • • + ag( ) by taking g g.g~l = n sn s ~ s ',' gl = 1) If gu and ~ are conjugate: -l. g g g g for sorne s. v = s u s · By equating coefficients of ~ on both sides a =a v u Hence all conjugates have the same coefficient + a c , . . n n c. = ~ g. g. (;- C. a class of ~ ~, ~ l. conjugate element a . ~ K. l. Conversely if a = ~a. ë. the n a ~ L ( 1 ) since l. l. Theorem 4-4: Two representations ~ and nr of a group G have the same irreducible constituents each with the same multiplicity if-f ' Proof: If ~1 and 1( have the sarne constituents with the same multiplicity then ~' \ 1( (g) = Conversely let /h(g) = }!. rf-g) , V' g <::: G. 36 Let ~1 (g) = ~ai ÀÇL (g) , (~i are irreducible constituents -~ of~) and f('(g) ~b. Îl,.,..., (g) ' ( rr. are irreducible constituents J ~ ;t... -~ -~ of!!:_) Now by (the Frobenius-Schur Theorem 3-9) 3c \ti ~1. (c) = (1) = f. JH. ~ ' -~ -~ and ~ - (c) = fH (c) = /M (o) 0 Jr i ..':..J -J -J ' Now ~ (c) = aifi and Î (c) =b.f. _[f ~ ~ But 4 (c) = X3'c) , hence a. = b. \>' i. ~ ~ J Remark a) - The values of the numerical functions f = J mentioned in section 1 determine fully the distinction or sameness of two representations of G. Character relations - If G is of arder n and Ka field of character C~ istic o and there are i = 1 ~ ••• k, distinct classes~of conjugate elements then the follo-v.fing t"JO relations hold true. k lst: ~ S . J-1 rY ( g . ) , ( ~ . J-1 = 1 if C. and t-=1 ~ ~ ~ ~ CJ are inverse classes, o otherwiseJ 2nd: ~Yhere {'('( g i) is the arder of the normalizer of gi' the gi are r epr esentat ives of Ci and the 1 are irreducible characters. 37 the Proof:lst: Let f 1, f 2, •••• , !k be distinct irreducible representations degree f , f , •••• , fk respectively. ot 1 2 Let cl, • • • ' ok contain hl , • • • , hk elements respect- ively. If c1 = ~ gi g. e C. ~ ~ Then -.· FJ Cc.) KT (a) = FJ (a) FJ (ë:) , 'V' a - ~ ~ -- ~ By theorem 3-8 ~ (ëi) = wi If v.>~ CS k.. · J J 1 By taking traces: yJ- i i 1 ryJ- fL (c) = w J fJ or \..1) J - tt (c.;) fJ ... integers) ~.J-1 h. , (For either no element of O. hasan inverse ~ ~ ~ in 0 J or all have their inverses in 0 J) k And ft (ci) ft (cJ~ = "z:'~ giJ'rFl (c1') . k ~ J 't or wl If v:> ~f = ~ giJ'! wl If 1 1 1 ?:: =1 1 Taking traces: k i J u::,.t wl ft = ~1 giJ"tlJ)i î r By the relation (*) this becomes . l 1 k l hihJ f (gi) f (gJ) = ~f. ~J . ft ht f (g't) [, =1 38 giJ 'è'h cr trace of B: (g 1:) ' (B: is the regu1ar representation of G) n.h. ~. J-1 ~ ~ He nee 2nd: Let ~ '(g1) l' (g2) • • • • • • J' (gk) '-"\ l = Î 2(g1) ;2(g2) . . • • • • )2(gk) • • . . . • • . .• . . . . . \ ! k(gl) . . • • • . • • . . 1- k(gk) '-"' V\ Let T = ~ 'X where 1' is the transpose of ).· Then k T = ~ j i(g ) ) i (g ) = ~ -1 n J»~ i=1 r ;- r ~ h ,/'l- Now T is non-singular, in fact ( T-~ = (ho-- ~fav-1) for: \. /p~ n = = I 1 if-f cr = f - and l'= i3"t or Cl = ~ ·',1··, 39 = 0 otherwise. :. T T-l-= 1 t-l rT-l~l = I Hence ~ f· (ge) ~ Sst-1 /Y (g~l) = ~JJ.Y s, 1' n / = By similar consideration of T and T-1 also assuming lst or a special case of ~ character relation. Nagao Hirosi7 has proved a n:ore general orthoganality relation < a. (u) (t) ak1(v) ( -1) 8 c c n t ~ G ~J t -= jlY o i:f d Jk ~ where t = (a (f)(t)) (n = 1, 2, •••• 1 k) is an absolut ely iJ r- irreducible r epresentat i on of degree f of G over a f iel d of char acteristic p o, n is the order of G. Remark b ) - The conjugate ~ of a character J is again a character . Proof : Let F be a representation of t he group G corresponding t o 1· Let f (a) = l-1 1 f (b) = 1Y wher e l-1, rf' are na triees, a, b (; G. 1 Set up a mapping ~: f (a ) -> (H- ) t 40 1 1 1 1 1 .[ (ab)Ol-' = ((H rY)- ) = (ry-lH- )r = (J:.C )f ( nr- )1 = F (a) ~ F (b) q- ~·. .[ tl"-" provides a representation of G also. But JN-1 = fg-1 1 = ~ c) - If / and are characters of a group G then is a 1 f 2 Î -.l-2 character too. Proof: Let 'ln = x \ and -rt = ~ y , • • • • y 1. 1~, .••• ' m , 1 n 1 be the representation IWdules of the representations of which and Î 1 1 2 are characters. Construct U If (xiyJ) a is defined as (xia) (yTa), 'fi aG- G then it is easy to see that l1 becomes a representation module for sorre representation 1-Jhose character is Î t • 1 2 Theorem 4-5: The degree of the irreducible representation of rdivides n, the order of the group G. The proof of this theorem depends on the following lemma: Lernma 4A: The linear representations(of degree 1) of z ( r) are algebraic integers. k Proof: Let cia ::: ~ qi"tcY, ~"(é K. ~1 then (qit) is the matrix ~(a) provided by the regular represent- ation R of -z.( r ). 41 Let~ be a representation of degree 1 of Z (r) then k e (ë.) ~ (a) "" ~'t ~ Cet: ) , e (a) , e (~J) t- K. - l. ~ 't""1 k k Hence "'C~ ~ <ë-r ) ~ (a)§ iL:"" z q.'t ~ (ë't ) 1: ""1 l. Cl) i ::: 1, 2, • • • ' k. Now if ~ (a) = t, then (1) shows that t satisfies tk - trace 1 (qi7.. )tk- + •••••• + (det (qi't )) = 0 or t is an algebraic integer. Proof of theorem: Using the same notations as in the character relations: . ~ ::: (g.-1) we have h. f~ (g.) l~ (gi-1) = n n or ~ wi !J..L (gi-1) ::: - i ~ fi! u:>i is the representation of degree 1 of z. ( 1 ) so by above ~ ~ -1 1e.nma. an algebraic integer. Al.so f (gi ) being sum of roots of wlity is an algebraic integer • .: . ..E...- is an a1gebraic integer. f~ and n/f is rational :. n/f is a rational integer ~ ~ -·. f~\ n For an application of this theory we require the following 1emma and corollaries. 42 Lemma. 4B: If = f. then F. (g) = g E G such that lJi (g)l J. -J. where f. is the degree of the representation F. whose character J. ' -J. is Ji• Proof: • . ~i (g) aa '[ 'è.J , t.J , nth roots of unity. J=l And we have equality if-f E. c· J = ~ sa~ If then J):i(g)) = f., .3T: J. T (F. (g)) T-l = c. I~ -J. J. .L • J. -·. F. (g) = c. If -J. J. . J. Corollary 4-B-1: 1F (g) = f ~_E(g) = I If F. is irreducible anu Corollary 4-B-2: -J. = f., ~g&G. J. Then F. is of the first degree. -J. But F. is irreducible representation of G. -J. ~. c is not repeated. or fi= 1. Information on the structure of a group can be obtained from a know1edge of its characters as the following theorem shows. 43 ~ Theo rem 4-6: G is not simple if-f ~ gi, and an irreducible ~ , ~ f g. + l, where f is the degree of F giving 1 x. ~J ~ -~ Proof: Let 'Y~ (g.*) = f for sorne g.* :fi. f 1 ~ 1 By cor. 4-B-1 !.~ (~* )=If ~ But N = ~gi } _E~ (gi) = If f is the kernel of the mapping 4. ~ ·:t- So N is a non-trivial normal sub-group of G, for clearly g 6 N. i Conversely - suppose G is not simple and so let N be its normal sub- group. Then F the representation of G/N is also a representation of G -~ in which ail eleœnts of N are ma.pped onto If ~ ~ So J (g. ) = f for R.. E: N. 1 ~ '"'1. Theorem 4-7: The number of linear characters of an abelian group G is equal to the number of its elements. Proof: Each element of an abelian group forms a class by itself. ~.the number of k distinct irreducible representations is equal to the number of its elements. Secondly using Theorem 3- 8-1 n = ~ f.z (n is the group order) J:.J.. 1 f. = 1 = > 1 Theorem 4-8: The number of linear characters of a group G is equal to the index G ~ C of the -commuta tor sub- group C of G. Proof: Observe that a linear character J is also a represent- ation and hence 44 'X(aba-~-l) = ~(a)~(b)l(a-1 )):(b-1 ) =,Y(a))'(a-1 ))'(b)J(b-1 ) 1 then C is fkernel of ~ and so }' indues a linear character } of G/C: Il. ) (~) = )"(a), where @1 is the class of a modulo c. Conversely ~ the character J of G/C gives rise to 'J' of G. :, the number of linear characters of G and G/C are the same. ·: G/C is abelian ail its characters are linear and their number = the order of G/C = index G:C. Theorem 4-9: Every p-group is solvable. s Proof: Let order n of a group G be p • .., n = ps = f 1 2 + f2 2 + • • • • . . f = 1 (,[ is the identity representation of 61) 1 1 5 But every fJ/p > f J = 1 . : . Number of linear characters) 2. Or the index of the commutator group ~ 2, so if G is not abelian its corrmutator sub-group is a proper sub-group, hence again a p-group of smaller order. : . G) Gl .J Gl 1 •••••• ).l Definition 4-1: The scalar product ( ',{ , ;t 1) of two characters À and ::(' 1 of a group G of order n is given by If J = Jf.L irreducible char- 45 acters of G, then ( ) Jl) 1 ~a b , Il Il Il .. 1 (g.) = . n ~ ;l ]. g.<':. G ]. Property 1: J is irreducible if-f ( J ,Y) = l 2 Proof: ( ;l ~)' ) = ~ai = l if-f 3 f..L: afl = 1, a L = 0 'V if=!-l = ~afl ). fl = l· )' Il = J Il, an irreducible character. Property 2: as often as contains ',(~ ~: Let m be the multiplicity of lv-- in ')' fJ.)'tf fJ. JI Then rn= l ~ lll~ -n g .€: G ]. Pro perty 3: If A is irreducible then rJ contains the ident i ty representation once and only once. ~: Follo•-Ts from Property 2; contains )' = ) (Identity character) as often as ) )' = l' J 1 1 1 ;r contains ) , which is obvious ly once and only once. 46 Hence the theorem. Theorem 4-10: The natural representation of a doubly transitive group over the field of complex numbers consists of identity representation and an irreducible representation. Corollary: Every symmetric group Sn has an irreducible represent ation of degree n-1. Proof: Let )'T(' = x , •••• , xnJ be the representation module 1 of G (g , •• , gni and let(l, 2, •• , n) be the symbols = 1 pernruted by G. Let x. J..g (1) Find the set T of matrices commuting with (1) y g E G. If T = ct~.v') 1 1 Then J t (~1 ;v) ( cj, y g G- G and \r' ~~ Ir-' , t.:~ (]\,. g -1 = t~!""' -1 a ) let ~ = 1 and a---g ""' 1 also. Or t = t .. 'r/ i ( For doubly transitivity 11 J.J. of the group ~ transitivity also.) -1 ~ a' ) let ~ = 1 But ~g r 1 say 2. As the group is doubly transitive. 47 So the rank of the cornmuting algebra of (1) is 2. So there are two irreducible representations in the above representation and the sum of their degrees = n. But one of these is identity which can be seen by taking a new basis for l'rt.. n Let y = i_x. 1 ' l. 1:=1 • or t!,.l-is identity, where ~l is the top constituent of tl,. Hence a representation ~ similar to the natural representation contains the identity representation and ~s proved an irreducible repres- entation of degree n-1. Corollary - ·: Every symmetric group S is doubly transitive, its n natural representation determines the irreducible representation of degree n-1. Theorem 4-11: (Fundamental theorem for Abelian groups) Every finite abelian group is the direct product of cyclic sub-groups. Proof: Let n the order of the group be g'pa, (g•, p) = 1 Pick an element H é- G such that the order of H is a p:>wer of p s~ j=l~ and G does not contain elements of order p8 s > r. It is always p:>ssible to find such H, for either any g .~ G can be l. tts tt of order g p , g (g' or directly a power of p. 48 ·: G is an abelian group ail its irreducible characters are linear. r p :a J.i of G ~ such that [f(H) ] ""' L . s but [ fJ. (H)] P t l ifS< r For if not . s ( )'J.(H)p ) = 1 s < r vi S. > F. (HP ) = 1 vi ' -J. s or !:i((H)p - 1.1 "" 0 'Vi s or Hp - 1. <:::: N" s or Hp = 'L ( Contradiction for H is of order p:) Consider the map. gi -?( )' "(~) )gl 1 ·: (g1 p) = 1 ( )"i(H)gt is a primitive p-r th root of unity and when "'ti is restricted to l H} it gives rise to a cyclic group of orcier r p • a,primiÏve If sorœ element g1 is mapped-onip î"h lroot of unity 1 > r. Then we shall have t Hence G -->cyclic group generated by prth roots of unity. 49 Sub-group ~ H' ~ 3 Let N be the kernel of the mapping Then G == N. f Hf but N 1"1 ~ HS identity element of G. : . G== ~H~XN. a a-r N is of order g' p == g 1 p r p By applying the same argument to N, we get the theorem. Theorem 4-12: If G JH and an irreducible representation 'IJ of H is known then it is possible to construct ~~of G, called the induced character of G\which however may not be irreducible. The formulais 11.:'~ ( g) == m ( ~ 'f(g')) where m == (G·.H) h g gt...,.. g hg,= class number of g gt E: H Proof: Form the group algebra of H (a sub-algebra of the group algebra of G) The rrodule of the irreducible representation giving 't'Of H is some minimal right ideal of this algebra. Let x , ~' • • • , xs be a basis of this ideal B. 1 • • HT be the coset decomposition of G. Let G = HT l + • • m Consider the set xiT J , 1 ~ i ~ s, 1 ~ J ~ m. I+ ClJ E- K. m s ~ c.Jx.TJ == o me ans ~ ( ~ C. J x. )T J == 0 . J 1 1 1 1 1, J=l i=l 50 Each summand of the outer sum consists of linear combinations of group elements of the form hT J' h ~ H and are different for distinct J. ~. ( ~ C. J x.) TJ = 0 . ~ ~ V'J ~ or iC . .x. = 0 ~J ~J ~ But ~ are linearly independent. ,. a linear combination of xiTJ forms a .roodule in the group algebra of G1 with a linearly independent basis. Moreover it is a right ideal also: s ~T J.g = xilil'a = ( J:l 1J' J) Ta s e.,....; ence wr..;te ~ T ~ = -y h T - For Conv u...... i J-a .._ i ŒlJ) J(.a.) - ~Ci~ a(J)) xk T J(a.) k By ordering the basis by i for each fixed J, it is easy to 3ee that the representation can be partitioned into ~ blocks of sxs matrices, so that the matrix in the Jth row block is in the J(a)th column block. Or -1 -1 the ( J ,k ) th block is a zero block unless TJgaTk ~ H and if T?aTk c:- H, then in ( J ,k)th block we have a matrix. which corresp:mds to the given representation of that element of H• .. 51 m ( -1 -1 = ~ - T~a.T J ) = ... 't if TJga.T J E: H Jml '+' 't -1 =0 if T~a.T J f H m 'f- s:: 1 1 -\ -1) (hT?a.TJ-\- ) - ~ m~i:- T?a.T J i ) h,.:.. H : 1 = ~ 't - h J:~l i J=l hi-EH 1 ~ ... ~ (g ga -g-1) h g'€ G N (gq,) h - 1 where N ( g ) is order of = h ~ g ~ ( S~a S J- ) , J""'l a norma.lizer of g • h a tvCga.) •Dt ~ g = hg J'l(~) J""l m .. - ~ 'i-'(gt) hg gl = 0 if no such gl exists. Theorem 4-1): (Frobenius Reciprocity Theorem). If )H contains 't' with multiplicity a then 't'* contains )' with the same multiplicity. where J is an irreducible character of G. o/ is an irreducible character of H. '/. H is the character J restricted to H. 'f* is the character induced on G by 't o·f H. m 1 Proof: As we saw '+'*(g) = ~ 't- (TJgrJ- ), where the TJ are Jml representatives of the cosests of G relative to H and ~(x) ... 't(x) 1 x fr H = 0 xtH• m -1 ~(g) = J-~ ~(hTJ)g(hTJ) 52 If a = multiplicity of l in ~*· Then a = ~ ~ t(glJ (g) = 1 z n(H:l) x~ G = ! ~b n x<:G where b = H;l~ 't(y) 'J (y) is the multiplicity of 't' in f H. = ! .n.b = b n Le.lllilB 4-C: If F is an irreducible representation of degree > 1 of a p-group G then "3 a normal sub-group H of index p in G such that f on H is reducible. ~f: Let fbe the irreducible character of G corresponding to !:. Let f T = fl + • • • • + ai J i + • • • • (1) "3 i ; fi f fl which is linear ( -. · degree of an irreducible character divides the group order). New r r contains fi as often as rTi contains 1 . -.·the degree of J J i is f as is that of J (2) or From (2) (2)1 53 'Y (G) s t f · b s-l Now 3 s such t h at 11 i are p roo s o un~ty ut not p roots of unity. s-1 Then t~ P_ - (G) are pth roots of unity not ail = 1 .l. s-1 Gall (J J As discussed in case of contained in (1) just once. Or 1 l = )'1 + • • • • + Xs + • • • (3) Consider the mapping of G onto a cyclic group of pth roots of unity namely g -) fs(g) Let H be the kerne1 of this ma.pping. Then H Restrict )' to H and denote this new character by jH). Then from (3) ~ (H) J (H) = /l(H) + •••• + /s(H) + • • • But /s(H) = /1(H) ... H is the keme1 • -·- ,! J restricted to H contains the identity representation at 1east twice. So )'(H) is not an irreducib1e character of H. Theorem 4-14: Every representation of a p-group is similar to a mnomial representation. ~: It suffices to prove for irreducib1e representations, for every representation is a 1inear combination of the irreducib1e ones. 54 If G is abelian it is obvious, for then avery irreducible character is linear. Let J be non-linear (i.e. not linear) irreducible character of G, 5 hence degree of Y= p , for some s. Now choose H .d G as in the above lenma. Then f(H) is reducible. Let 't be an irreducible constituent of f(H) • :. Degree "r = 1 or p-t • Let 'r * be character induced upon G by 't of H. Then 't * contains y by Theorem 4-13. ·: (G:H) = p : . p deg. 't' = deg. '+- * But deg. 't- * ~ deg.l > deg. 't , and p deg. 't' ~ deg.f, deg. 1.- and deg. 't' are powers of p • .· . deg. 't * = deg. 'f . But J is contained in 't'* ~ 't"'} ~ ~ By using as inductional hypothesis the truth of the theorem for H that is assuming '+' to be m:>nomi.al, we get J = 't * also monomial by the very method of its construction. Hence the theorem. Theorem 4-15: Every group of order paqb is solvable, if p and q are primes. Proof: ·: Any sub-group will have an order of the same form it suffices to show that G cannot be simple. By the Sylow theorem 3 H Let ?"t(h) be the normalizer of h in G, then 55 rt(h) JH, ~- rt(h):l ~ H:l .. pa. b c :. G: -re. ( h) ~ q = q c ~ b But G: l1. (h) == number 'of elements of G in conjugate class of h. The theorem is now established by the following Lemma of Burnside: If the number of elements of a conjugate class is a prime qc then the group is not simple. Proof: Recall that hi).,. is an algebraic integer where hi "" number of elements f J in a conjugate class Ci and )' i is an irreducible character of degree f .• l. Now ~ f J • J /x) = 0 or 1 + ~ r J lJ (x) == o J:f:l The summands on L.H.S. are algebraic integers. ~· . 311 f 1, (q, f Jl "" 1 and so also (~', f J) = 1 and fJ + 0 But qc t /gi) is an algebraic integer. fJ ·; (fJ" qc) = 1 ~ ]. igi)/fJ to be an algebraic integer. · : ~(g) is a sum of r J roots of unity~ rxigi) 1 ~ fJ If 1 XJ(gi) f = f J then by Lemma 4-B FJ(g ) is a scalar matrix. is not faithful then its 1 ~J kernel is normal in G, hence G is not simple. 56 If. FJ is :taitb.f'ul.., then is a central ele.rœnt of a . '. G has a non-trivial centre and is again not simple • On the other hand l it J~;) 1 { 1 as the following shows: XJ - (g.i) is an algebraic integer fJ 3 an irreducible polynomial m m-1 g(x) = x + a.,x • • • • + am' ai rational integers of which w is a root. • w be the remaining roots of this 1 • • m k primitive é 1 + •• • • • + , ~ some jsth root of unity say 0 ~ kl ~ s-1 LI) Ls-1) . Form the set wl" w •••• , w by replacl.l'lg lé:. by its 1 1 1 conjugates E. JH., J = O., 1, • • • s-1. for each of these 1 fw1T/ ~ 1 s-1 Consider the polynomial f(x) (x - w ) (x - w ) • • • (x - w ) = 1 1 1 who se coefficients are synmetric ftmctions of f:. J, J = 0 1 1., • • • 1 s-1, and hence rational. ·: w is a root of this so g(x) f(x) 1 J . . roots of g(x) are among roots of f(x). 1ws1 ~ 1 , s = 2, • • • • , m. 57 But L.H.S 1 But a is an integer. < m -.· a m = 0 Or w ~ 0 impossible. 1 Section 5 Concluding Remarks 1) Theorem 4-15 has apparently never been proved except in this way1 using representation theory. Another classical application of the theory is the following result of Frobenius: If G is a transitive per- mutation group of degree n whose permutations other than the identity leave at most one of the syrnbols invariant1 then those permutations of G which displace ali the symbols form together with the identity a normal sub-group of order n. [ 'l) 2) The basis theorem for abelian groups and the solvability of p-groups given above can of course be proved directly although it may be observed that the usual proof of the latter makes use of the class equation of the group which is actually a special case of the first character relation. 3) Theorem 4-6<5hows that the character table of a group reveals the existence or non-existence of normal sub-groups. In fact since f g <= G \Ji(g) • fi j is the kernel of !:! an actual normal sub-group can be written down. However, as not ail normal sub-groups need occur as the kernels of irreducible representations it may be necessary to work with 58 linear combinations of the irreducible characters if the determination of k ali normal sub-groups is sought; take Y "" < a. /( ., a. integers ~ 0 t l,="ll. l. l. and then l g f: G jl (g) = f degree of/} is a normal sub-group of G. Since k f may be restricted to divisors of G:l and since f = 5 a.f., the i~ l. l. possible a. are finite in number and the zœthod involves a finite number l. of trials. 4) If Hi' i = 1, 2, • • • are sub-groups of G and ~iJ' J = 1, 2, irreducible characters of H. we may use Theorem 4-l~ to find ••• are . l. induced characters 'f* of G which bowever may not be irreducible. If iJ now v= ~a.J 't'"* , a.J integers, is such that .l) c: 1, ')1 (I) > 0 /-' l. iJ l. cr ' 1'- we have seen that Jr will be an irreducible character of G. The question as to what sub-groups Hi and what characters 't'iJ are sufficient to deter 6 mine !f!:. the characters of G has be en answered by R. Brauer ( ) : it is sufficient to take for Hi ali elezœntary sub-groups of G and for '1-' iJ all linear characters of H. • - l. 5) There are a number of special zœthods of finding the characters 13 of particular groups. See for example references in ( ). .For the symmetric groups S there is the classical method of A. Young an account of which may n 14 15 be foum, together with references in ( ' ;1.6). What are to soma extent generalizations of the method and applications to certain special groups can be found in ( 10), (11) and ( 12) • 59 6. BIBLIOGRAPHY 1. G. Frobenius: Uber Gruppencharaktere. S.-B. preusz Akad. Wiss 1g96 9g5-1021 - Uber die Prirnfaktoren der Gruppendeterminante. Ebenda 1g96 1343-13g2; 1903 401-409 - Uber die Darstellung der end.lichen Gruppen durch Lineare Substitutionen. Ebenda 1g97 994-1015; 1g99 482-500 - Uber die Komposition der Charaktere einer Gruppe. Ebenda 1g99 330-339 G. Frobenius u. I Schur; Uber die Aquivalenz der Gruppen Linearer Substitutionen. Ebenda 1906 209-217 - Zur Entstechung der Darstellungstheorie. Vgl. auch den Briefwechsel Zwischen Dedekind und Frobenius in Dedekinds -~Jerken 2. 2. H. Burnside: On the continuous 8roup that is defined by any group of finite order. Froc. Lon:lon Nath. Soc. 29 (1g9g) 207-224 arrl (546-565) - On group characteristics. Ebenda 33 (1901) 146-162 - On the composition of group characteristics. Ebenda 34 (1901) 4l-4g - On the representations of a group of finite order as an irreducible group of linear substitutions and the direct estab lisrment of the relation bebreen group-characteristics. .Ebenda (2) I (1903) 117-123 - Theory of Groups. 2nd edition. Cambridge 19ll. 3. I. Schur: Heue Begrundung der Theorie der Gruppencharaktere. S.-B. preusz Akad. Wiss. 1905 406-432. Arithmetische untersuch ungen uber end-liche Gruppen Linearer Substitutionen. .Ebenda 1906 164-184 - Uber die Darstellung der endlichen Gruppen durch gebrochene Lineare Substitutionen. I reine ange>v. Hath 127 (1904) 20-50; 132 (1907) g5-137. 4. T. Lolien, I' ~ath. Ann. 41 (lg92), 83-156. 5. z. Noether; Hyperkomplexe Grosz en und Darstellungstheorie. Ha th. 2.30 (1929) 61+1-692. 6. R. Brauer; Annals of Naths. 4J3 (1947) 502-514. 7. Nagao, Hirosi. A remark on the orthogonality relations in the representation theory of finite groups. Canadian Journal of I-1atha'-!e.tics il (1959) 59-6o. g. Van Der Vlaerden. 1-lod.ern Algebra Vol. II, 151-154. 9. N. Hall.J'C"'.The Theory of Groups. Macmillan, N.Y., 1959. pp. 247-298. 60 10. M.D. Burrow, A generalization of the Young diagram. Canadian J. Math. 6 (1954) 498-508. ll. J .s. Frame, An irreducible representation extracted from two permutation groups. Ann~ath., 55 (1952), 85-lOO. 12. B. French 11 The Representations of Particular Groups". Li.near Groups in One Variable in GF(pn) and Linear Fractional Groups LF(2,pP). McGill Thesis 1956. 13. B.L. Van der Waerden, Gruppen von linearen transforrœ.tionen (Chelsea, N.Y., 1948). 1.4. A. Speiser, Theorie der Gruppen von end1icher Ordnung (Berlin, 1927). 15. D.E. Rutherford, Substitutional Analyses (Edinburgh U.P. 1948). 16. D.E. Littlewood, Group Representations (Oxford, 1940). NOTE: a) )'l'{ , ")"1"(. (K, -&) mdules are operator isomorphic if the 1 2 operator homomorphism ~ 0'-' U~) V a. u '=----) a v and a1so ua ~ -7 v. a u E~, v e:- '2" a~ K, a ~-cr. is one-one and onto.