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### Indecomposable S6(2, 3, v)'s

Salvatore Milici
1991 Journal of combinatorial theory. Series A
In this paper, we determine the spectrum for indecomposable S,(2, 3, u))s, and we prove the existence of MPT-free Ss(2, 3, v) for all v > 9 and u = 7. 0 1991 Academic Press, Inc.  ...  Let now u 2 17. If (u -1)/2 = 0 (mod 2), Lemma 4.4 gives a simple and MPT-free S, (2, 3, u).  ...  If (u -1)/2 E 1 (mod 2), since there exists a simple and MPT-free S6 (2, 3, u) for u = 23, 5 + 12k, k > 1, Lemma 4.5, by induction, gives a simple and MPT-free F& (2, 3, u).  ...

### The new stems (πS N, 46≤N≤64) [chapter]

Stanley O. Kochman
1990 Lecture notes in mathematics
(a) o'2A[32,1] ~ A[32,1] = u c v'~ S = 0. 43 (b) Since A[39,3] ~ , ~A[39,3] e ~ = <~,u,B[34]>n = <~,v,>n D <~,u,n,2>nA[32,1] = 0 because S <¢,v,w,2> e ~13 = O.  ...  S while the other elements of ~ are divisible by n and thus have order two.) 46 Now "~<2, A[40,1],v> = <'~,2,A[40,1]>v = O. Thus, <2,A[40,1],v> = 2kC[44].  ...  [7.41] Now 2A[62,2] • 2<A[S6],2,n,v> c <<2,A[S6],2>,n,v> c <nA[S6],n,u> = <vA[S4,1],n,v> D A[S4,1]<v,n,v> = A[8]A[54,1] = <n,v,2v>A[54,1] = n<v,2u,A[54,1]>.  ...

### Certain representation algebras

S. B. Conlon
1965 Journal of the Australian Mathematical Society
We have the following P r o p o s i t i o n 3. The projective ideal 2i is semi-simple and finite dimen sional. R e m a r k .  ...  In these cases there are only a finite number of different indecomposable classes and so sé is a direct sum of copies of .  ...  * 1 splits up into 3 indecomposable, non-isomorphic 1F(S4 i)-modules £C A (all superscripts will be considered to be in tegers modulo 3), such that (^a )^4 ^* S6', as in proposition 3 of [2] .  ...

### Vertices and sources

John G Thompson
1967 Journal of Algebra
For each ring S and group 6, S6 is the group ring of 6 over S. If S is a subring of S' and g is a subgroup of @, we view Sg as a subring of S'6.  ...  Since L is indecomposable it follows from Lemma 1 that no component of L -.is projective. Since (Ms,) 'v A& , it follows from the indecomposability of S that 1 !x S. LEMMA 4.  ...

### On the indecomposability of induced modules

Reinhard Knörr
1986 Journal of Algebra
Condition (i) is then just that S is indecomposable, while (ii) asserts that HZ TJS). If R is a field, the theorem is therefore precisely Theorem 9.6(a), (c) in [3, Chap. VII].  ...  Then v is indecomposable for every indecomposable direct summand W of SH. Moreover, if W, and W2 are two surh .summands, then WY z WF if and only 17 W, E W,. Remarks.  ...  Why some condition is needed here will become clear in the proof. (3) In case R = 0 or R = F, Green's Indecomposability Theorem asserts that SN is indecomposable provided S is and O"(N) s U.  ...

### Concealed-Canonical Algebras and Separating Tubular Families

Helmut Lenzing, José Antonio De La Peña
1999 Proceedings of the London Mathematical Society
(v) For 1 < i < t there are short exact sequences 0 3 L « f i 3 L i 1 3 t À1 S i 3 0; 0 3 L i 1 3 L i 2 3 t À2 S i 3 0; . . . 0 3 L i p i À 2 3 L i p i À 1 3 t À p i À 1 S i 3 0; 0 3 L « 3 L 3 S 3 0; 0  ...  By the dual of (S16) we get a connecting sequence h: 0 3 N 3 V x 3 N 3 0 with V x P U x and N P mod S.  ...  translation for H; (v) H admits S and the corresponding canonical algebra L as torsion-free  ...

### Classification of stable homotopy types with torsion-free homology

Hans-Joachim Baues, Yuri Drozd
2001 Topology
We compute the number of indecomposable stable homotopy types with "nitely generated torsion free homology of stable dimension k*0.  ...  We de"ne the spaces > G and >T H with i3+1, 2 ,5,, j3+1,2,3,, v3+1, 2 ,6, by > "S"X(S), > "S6 E e"X( ) , > "S6 E e"X( ) , > "S"X(S), > "S"X(S), >T "X( v ) , >T "X( v ) , >T "X(v ) .  ...  Indeed, the ring consists of the matrices of the same form as in for which rows and columns only correspond, however, to the indexes i"1, 2, 3, 4, 5 and the pairs (v, j) with v"3, 6 and j"1, 2, 3.  ...

### Algebras stably equivalent to Nakayama algebras of Loewy length at most 4

Idun Reiten
1979 Journal of the Australian Mathematical Society
Idun Reiten [2] of use, available at https://www.cambridge.org/core/terms. https://doi.  ...  ( T \ I ' ) V-'i+i/ IT \ i i + 1 \ so that t + l W S i + 2 ).  ...  Hence (T \ i*\ (T \ we must have ( J^W S j l and ( 4 )<-+S 5 . It further follows that T 6^S6 and l Ss \ W (e) Assume that P 3 has length 3. We must then have If T A^TU P^P X is not uniserial.  ...

### On saturated fusion systems and Brauer indecomposability of Scott modules [article]

Radha Kessar, Naoko Kunugi, Naofumi Mitsuhashi
2010 arXiv   pre-print
So,M = Ind S6 S5 (O). Now, if u = (1, 3), then χ(u) = 4 and if u = (1, 2)(3, 4) or u = (1, 2, 3, 4) then χ(u) = 2.  ...  Case: P = (1, 2, 3, 4), (1, 3)(5, 6) . The image of P under the exceptional noninner automorphism of S 6 is S 6 -conjugate to (1, 2, 3, 4) , (1, 3) .  ...

### A complete classification of homogeneous plane continua [article]

L. C. Hoehn, L. G. Oversteegen
2016 arXiv   pre-print
The main technical result in this paper is a new characterization of the pseudo-arc: a non-degenerate continuum is homeomorphic to the pseudo-arc if and only if it is hereditarily indecomposable and has  ...  Moreover, by (S6 ) for S and since ϕ(F 1 ∪ F 2 ) = π(P 1 ∪ P 2 ), we have that ϕ −1 (α i0 ) ⊂ (F 3 ) , so that ϕ −1 (α i0 ) ∩ (F 3 ) = ϕ −1 (α i0 ) = α i0 . Thus E(S i0 ) = α i0 ∪ β i0 .  ...  ⊂ V , X 2 ∩ X 3 ⊂ U , and X 1 ∩ X 3 = ∅.  ...

### Subcontinua with degenerate tranches in hereditarily decomposable continua

Lex G. Oversteegen, E. D. Tymchatyn
1983 Transactions of the American Mathematical Society
S. Thomas, Jr. Theorem. Every hereditarily decomposable continuum contains a subcontinuum with a degenerate tranche. Corollary.  ...  Let (U, V) be a subcollection of A such that (l)U,VC(U,V), (2) (U,V) is a cover of some subcontinuum K of M such that U n K =£ 0 ¥= V n K, and (3) if W C (U, V) is a cover of a subcontinuum L of M such  ...  Then Cl(U/T) C U/S n UR.To see this, let W E fT. There exists Z E C\F such that ZniF^0. Let FE/1 such that S\Z,C) C V. Since C1(Z) C F, F <2 S. Let F"...,F" E S such that Cl(FF) C F, U • • • U V".  ...

### Subcontinua with Degenerate Tranches in Hereditarily Decomposable Continua

Lex G. Oversteegen, E. D. Tymchatyn
1983 Transactions of the American Mathematical Society
S. Thomas, Jr. Theorem. Every hereditarily decomposable continuum contains a subcontinuum with a degenerate tranche. Corollary.  ...  Let (U, V) be a subcollection of A such that (l)U,VC(U,V), (2) (U,V) is a cover of some subcontinuum K of M such that U n K =£ 0 ¥= V n K, and (3) if W C (U, V) is a cover of a subcontinuum L of M such  ...  Then Cl(U/T) C U/S n UR.To see this, let W E fT. There exists Z E C\F such that ZniF^0. Let FE/1 such that S\Z,C) C V. Since C1(Z) C F, F <2 S. Let F"...,F" E S such that Cl(FF) C F, U • • • U V".  ...

### Generating large indecomposable continua

Michel Smith
1976 Pacific Journal of Mathematics
Bellamy that every metric continuum is homeomorphic to a retract of some metric indecomposable continuum. This result was later extended by G. R.  ...  Suppose S 2 and S 2 are two distinct subsets of a and a is an element of Si not in S 2 .  ...  From condition (2) it follows that I a C V, so I a C V= V-/ α Π VCM-/ α . Now suppose M is the union of two proper subcontinua H and K.  ...

### Morita equivalence for semigroups

S. Talwar
1995 Journal of the Australian Mathematical Society
The conditions (2), (3) and (4) follow from the fact that for a monoid R and aleft/?-actM, R® R M = M. Since S is a monoid, it is an indecomposable projective left S-act.  ...  T h e n either (s, t • <i>) = (M, v • r)), in which case s ® t ® = u<g>v-r)inS<8) 5Hom(f/, N), or for some n > 2 there exists a sequence, (5, *•<!  ...  Formally, let d 2 ), (a,ba), (b,ab), (b,bc), (c,cb), (c,dc) , (d, cd) , (d, da)}, let A be the congruence generated by A o on the free semigroup F x and let S P4 = F x /A.  ...

### Blocks with a cyclic defect group

R.M Peacock
1975 Journal of Algebra
Now for the Green correspondence (m, , 2,) + (m, B) notice that the set 3 = {D n D: x E Nt\rn} = {l}, and hence 41 = {S: 1 < S < D).  ...  Moreover there exists a permutation 8 of I so that the "socle" E(f Vi) s S6-lti) for all i E I.  ...
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