C OMPOSITIO M ATHEMATICA
G ORDON M ASON
Injective and projective near-ring modules
Compositio Mathematica, tome 33, n
o1 (1976), p. 43-54
<http://www.numdam.org/item?id=CM_1976__33_1_43_0>
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43
INJECTIVE AND PROJECTIVE NEAR-RING MODULES
Gordon Mason Noordhoff International Publishing
Printed in the Netherlands
1. Introduction
Maxson
[10]
hasgiven
sufficienthomological
conditions for a unitalnear-ring
R to besemi-simple,
conditions which in the case ofrings
arealso necessary. One of these conditions is that every R-module M be
injective,
another is that every M bestrictly projective (see
definition2.3).
Now thecategory
of groups, viewed as thenear-ring
modules overthe
integers,
is anexample
where there exist no non-trivialinjective
orstrictly projective
modules. In this paper we define alternate forms of"injectivity"
whichgive
necessary and sufficient conditions for all modules over anear-ring
to besemi-simple.
We also prove that non-trivialinjectives
do not exist in thecategory
ofnear-ring
modulesover R for
any R and
discuss thepossibilities
incategories
of unitalnear-ring
modules.Finally
theinjectivity
conditions are related to essential extensions.With the
exception
of Thm.2.6,
all modules M will be unital left modules over the unitalright near-ring R
where R satisfies x · 0 = 0 forafi x
(see
e.g.[3], [6]
for basicterminology).
We use the term"R-subgroup
of M" to mean asubgroup
of(M, +)
closed under leftR-multiplication,
and "R-submodule" to mean a normalR-subgroup
Asatisfying r(m
+a) -
rm E A for all r ER, m
EM,
a E A. M issimple
if it has no proper non-zero R-submodules and irreducible if it has no
proper non-zero
R-subgroups.
The R-submodules of R are called left ideals. The wordsinjective
andprojective
are used in the usualcategorical
sense.By
thephrase
"in thed.g.
case" we mean(see [6])
that(R, +)
isgenerated by
amultiplicative semigroup
S of left distributive elements and thecategory
of(R, S )-modules
consists of all unital left modules M such that s(x
+y )
= sx + sy forall s E S,
x, y E M. Thus thecategory
of groups is an
example
of a"d.g.
case" with R = Z and S ={1}.
Notehowever,
that if S is taken to be R = Z itself the(R, S)-modules
arejust
the abelian groups. In this paper it may be assumed that S is aminimal distributive
generating semigroup
of R whereas in[2]
results have been obtainedassuming
S is the set of all distributive elements of R.Since references
[3]
and[10]
are noteasily available, proofs
havebeen sketched for results taken from these sources.
2.
Injectivity
We
begin
with thefollowing
definitions andresults,
some taken from[10].
DEFINITION 2.1
[10]: (a)
AnR-homomorphism f :
M ~ N is normal iff (M)
is an R-submodule of N.(b)
An exact sequenceM ~
N ~ 0splits
if there exists a normal g : N - M such that
fg
= 1N.(c)
The short exactsequence
(s.e.s.)
0 ~ L ~ M ~ N ~ 0splits
if the sequence M - N - 0splits. (d)
The exact sequence 0 -L 1
Msplits
if there exists g : M - L such thatgf
= 1L.If L and N are R-submodules of M such that M = L + N and
L ~ N = 0 we write M = L ~ N.
LEMMA 2.1
[10]:
Thefollowing
areequivalent.
(a )
The s.e.s.0 ~ L ~ M ~ N ~ 0 splits.
(b) M = h (L) ~ g (N) ~
L~
N where g is the normalsplitting
mapfor f.
(c )
The exact sequence0 ~L~ M splits.
PROOF:
(a)~(b)
If g : N ~ M is the normalsplitting
map thenfg
= 1Nand
g (N)
is an R-submodule of M. For all m EM, f(m - gf(m»
= 0 som
E ke f + g (N).
It iseasily
checked thatke f f1 g (N) = 0
so M =ke f ~g(N).
Since the sequence is exact at Mke f
=h (L )
so M =h(L)~g(N).
Moreover since both h and g aremonomorphisms, M~L~N.
(b) ~ (c)
For all mE M, m = h(x)+g(y), x E L,
y ~ N. Define k : M ~ Lby k(m) = h-1(g(y)).
Then kh = 1L.(c) ~ (a)
Let k : M ~ L be such that kh = IL. As in the firstpart
of theproof, M = h (L) + ke k
andh(L)nkek
= 0. Since L= h (L ) = ke f, L
is an R-submodule of M so M =
h(L)~ke k
= kefEt)ke
k. ThereforeN = f(m) = f(ke k).
Iff
i is the restriction off
toke k,
it is anisomorphism. Putting
g= (f1)-1
we haveg (N) = ke k
so g is normal andclearly fg
= 1N.DEFINITION 2.2: The s.e.s.
0 ~ L ~ M ~ N ~ 0
almostsplits
if there exists g : N - M(not necessarily normal)
such thatfg =
1N.DEFINITION 2.3
[10]:
The module P isstrictly projective
if everys.e.s. 0- L - M - P - 0
splits.
PROPOSITION 2.2
[10]: Every strictly projective
module isprojective
inthe
d.g.
case.PROOF: It is shown in
[6]
that in thed.g.
case every module P is afactor of a free module F. Thus F - P - 0 is exact, and if P is
strictly projective,
the sequencesplits.
Hence F ~ P@ Q
for someQ
andby
the usual
argument ([6])
a direct summand of a free module isprojective.
The
question
arises as to the existence ofstrictly projective
mod-ules. Maxson shows
[10,
p.53]
that a free group A is notstrictly projective
as follows: Let G be the free group sum of A with itself.Then there is a s.e.s.
0-K - G iA -0
where K is the normalsubgroup
of Ggenerated by
A. Since A is free it isprojective.
If thesplitting
g : A ~ G were normal we would have G = Im gEt)
kef.
But agroup cannot be both a free sum and a direct sum. Hence A is not
strictly projective.
Now since free groups are the
only projective
groups[5],
we concludeby Prop.
2.2 that there are nostrictly projective
groups.Dualizing
definition 2.3 we haveDEFINITION 2.4: A module I is
loosely injective
if every s.e.s.0 ~ 1 ~ M ~ N ~ 0
splits.
Finally,
it is useful todistinguish
two other forms of"injectivity."
DEFINITION
2.5(a)
I isn-injective (for "normal-injective")
if forevery
diagram
of R-modules and R-module
homomorphisms
in which g isnormal,
there exists h : B ~ I such thathg
=f.
(b)
I is called almostinjective
if every s.e.s. 0 ~ 1 ~ M ~ N ~ 0 almostsplits.
PROPOSITION 2.3:
(1)
1 isinjective ~ (2)
1 isn-injective ~ (3)
1 isloosely injective ~ (4)
1 is almostinjective.
PROOF:
(1) ~ (2)
and(3) ~ (4)
followdirectly
from the definitions.(2) ~ (3)
If 0 ~1 ~ M ~
N - 0 is a s.e.s., thenf
is normal sinceIm f
= ker g. Since I isn-injective
there exists h : M ~ I such thathf
= 1,.Examples :
In thecategory
of groups,(regarded
asZ-near-ring
mod-ules)
1. There are no non-trivial
injectives [5].
2. If G is
n-injective
then everydiagram
of thefollowing type
can be
completed.
Thus for every n F- Z and x E G there is anh : Z ~ G such that
h (n )
= x i.e.h (1)"
= x. Therefore G is a divisiblegroup.
3. G is
loosely injective
iff G is a direct summand of every group in which G is a normalsubgroup. Equivalently
G is acomplete
group[1,
Thm.1]
i.e. G has trivial center and everyautomorphism
is inner.Since
S3
iscomplete
but notdivisible,
this classproperly
contains the class ofn-injectives.
4. G is almost
injective
iff G has trivial center and 0 ~ Inn G ~ Aut G - K - 0 almostsplits [11,
Thm.2.7].
In
general
we say the module M is the semi-direct sum of itsR-subgroups
A andB,
and write M =A +
B if A is anR-submodule,
M = A + B and A n B =
(0).
In this case B is called a semi-direct summand of M. Then every m can beexpressed uniquely
as a + b forsome a E
A,
b E B. Moreover the canonicalprojection
p : M ~ Bgiven by p (a
+b )
= b is anR-homomorphism
sincep (a
+ b + a +bl)=p(a +b+a1-b
+b+bl)
Also
r (a
+b )
=r (b
+a1)
since A is a normalsubgroup
andr (b
+a1) -
rb E A since A is an R-submodule so
by uniqueness, r(b + a1) =
a’ + rb for some a’ and thereforep(r(a+b)) = p(a’+rb) = rb = rp(a+b).
PROPOSITION 2.4:
(a ) If M
= A+ B,
and M isn-injective
then so is B.(b) If M = A (3) B and
M isloosely injective,
then so are A and B.PROOF:
(a)
Adiagram
where g is normal
can be embedded in a
diagram
where p and i are the canonical
projection
andinjection
associatedwith M =
A + B,
sopi
= 1B. Since M isn-injective
there existsh : Y - M such that
hg
=if
andph
is therequired
map Y - B.(b)
Consider a s.e.s.f (a
+b ) = (f (a ), b )
and K is thequotient
module to make the sequence exact(the normality
off yields
thenormality
off ).
Since thissplits by hypothesis, P B = A ~ B ~ K. Identifying
P with P * = P x 0 inP x B every
p E
P can beexpressed uniquely
as p = a + b + k. Hence- a + p = b + k ~ P * ~ (K ~ B) = Y
say. Thus P * = A(DY
and soA is a direct summand of P as
required.
Theargument
issymmetric
f or B.COROLLARY: A direct summand
of
acomplete
group iscomplete.
LEMMA 2.5:
If
the s.e.s.0 ~ L ~ M ~ N ~ 0
almostsplits
thenM =L+N.
PROOF: This is almost identical to the
proof
that(a) ~ (b)
in lemma2.1. This time however the
splitting
mapg : N - M
is notnecessarily
normal so N
= g (N)
is ingeneral only
anR-subgroup
of M and thesum L + N is
only
semi-direct.We now
drop
therequirement
that modules be unital(though R
maybe)
and show there are noinjectives
in thiscategory.
To do this weneed to embed any group G in a
simple
group ofarbitrarily high cardinality.
Theprocedure
of Baer[5]
forembedding
a group in asimple
group does not sufHce but there is an alternate method due to Karrass and Solitar[8].
We may assume G is infinite for otherwise itcan be embedded
by Cayley’s
Theorem in a finitesymmetric
group and hence in an infinitesymmetric
group.If p is an infinite cardinal whose s,,.iccessor is denoted
p *,
letS (p )
denote the full
symmetric
group on a set X ofcardinality
p, andSp = {03C3 ~ S(p)|#{x|03C3(x)~x}p}.
Then([8]) S*p/Sp
is asimple
group of order 2p which contains
copies
of all groups of order p.Iterating
thisprocedure gives
a distinct chain ofsimple
groups(distinct
because the orders are
strictly increasing) containing
a group of order P.Now let R be a
near-ring
withunderlying
group(R, +).
If G is agroup
containing R,
G has a natural R-module structuregiven by
THEOREM 2.6: There are no
injective
R-modulesfor
anynear-ring
R.PROOF:
Suppose
I isinjective.
We reach a contradictionby showing
I has to contain
simple
R-modules ofarbitrarily high cardinality.
Let Gbe any
simple
groupcontaining (R, +)
and endow it with its R-module structure. It remainssimple
as anR-module,
for any R-submodule is anormal
subgroup.
If x ~ 0 isarbitrarily
chosen in I the mapf :
R ~ Igiven by f (r)
= rx is anR-homomorphism
well-defined on thecyclic R-subgroup
R of G. Since I isinjective,
this can be extended to all of G. Since G issimple
the extension map has no kernel so G is anR-subgroup
of I. Since we can find suchsimple
G ofarbitrarily high cardinality,
theproof
iscomplete.
The theorem is false in
general
forcategories
of unital R-modules asshown in
[2].
However thatexample
is acategory
of(R, S)-modules
with S = all distributive elements of
R,
and the existencequestion
remains open in the
general (unital) d.g.
case as well as the(unital)
nond.g.
case.3.
Semi-simplicity
There are several
equivalent
conditions for asemi-simple
module.PROPOSITION 3.1
([3]):
Thefollowing
areequivalent (M
is calledsemi-simple).
(a) Every
R-submoduleof
M is a direct summand.(b)
M is a sumof simple
submodules.(c)
M is a direct sumof simple
submodules.DEFINITION: The
near-ring
R issemi-simple
if RR is asemi-simple
module.
PROOF: The
proof parallels
the usualring
theoreticproof
as founde.g. in
[9,
p.59-61].
Similarly
thefollowing proposition
iseasily
verifiedfollowing ring
theoretic
proofs.
PROPOSITION 3.2:
(a) [3,
Cor.4.14]
Afactor
moduleof
asemi-simple
module is
semi-simple.
(b) [3,
p.63]
An external direct sumof simple
modules issemi-simple.
THEOREM 3.3
(cf. [ 10,
Thm.II 2.9]):
Thefollowing
areequivalent for a near-ring
R.(a) Every
M isn-injective.
(b) Every
M isloosely injective.
(c) Every
M isstrictly projective.
(d) Every
s.e.s.of
R-modulessplits.
(e) Every
M issemi-simple.
Each
of
theseimplies (f)
R issemi-simple.
PROOF:
(a) ~ (b) by Prop.
2.3(c)~(d)~(e) ~(f) by [10,
Thm.II, 2.9].
(b) ~ (c)
IfM À
P ~ 0 is exact, then so is the s.e.s. 0 ~ kerf ~
M -P -+0
which, by (b), splits.
(e) ~ (a)
Givenwith g
normalthen
by (e)
B= A~C
for some C so we can define h : B - Mby
h(a + c) = f (a). ’
It remains an open
question
whether Rsemi-simple ~
every R- module issemi-simple.
We do have thefollowing partial
results.PROPOSITION 3.4:
(a)
R issemi-simple iff
every direct sumof cyclic
R-modules issemi-simple.
(b) If
everyR-subgroup of
M is an R-submodule then Rsemai-simplet
M
semi-simple.
PROOF:
(a) (~)
Since 1 ER,
R iscyclic.
(~)
Let Ann m ={r
ER /rm
=01.
If M = Rm then M =R /Ann
m soby Prop. 3.2(a),
Rsemi-simple ~
Msemi-simple. By Prop. 3.2(b)
every direct sum ofsemi-simple
modules issemi-simple.
n
(b) (Cf. [3,
p.74])
Since R issemi-simple,
R =~Ji
where theJi
aresimple
R-submodules of R. If m E M then Jimis an R-subgroup
of Mwhich
by hypothesis
is an R-submodule. Themapping 03C8:Ji ~ Jim given hy tf¡(j)
=jm
is an R-modulehomomorphism
and sinceJi
issimple,
eitherJim
= Ji orJim
= 0. Thus M is the direct sum of thenonzero
Jim
which aresimple
R-submodules of M.Recently Choudhary
and Tewari[4]
havegiven
severalequivalent
conditions for a
near-ring
to bestrictly semi-simple.
The above resultsyield
acomparable
theorem in thesemi-simple
case.(Thm. 3.6)
Forcomparison
we firstrephrase
their theorem in ourterminology.
THEOREM 3.5
[4]: If
R is anear-ring
withright identity
thefollowing
are
equivalent (R
is calledstrictly semi-simple).
(a)
R is a direct sumof
irreducibleleft
ideals.(b) Every R-subgroup of
acyclic
R-module is a semi-direct sum -mand.
(c) Every R-subgroup of
R is a semi-direct summand.(d) If L1, L2
areR-subgroups of
R and 0 ~ L1 ~L2
is exact, then itsplits.
(e) If L1, L2
areR-subgroups of
R with Li~L2
then every R-homomorphism of Li
into any M can belifted
to L2.(f)
R has the d.c.c. onleft
ideals andJ(R)
= 0.(g)
R has the d.c.c. onR-subgroups
and nonilpotent
nonzeroR-subgroup.
Here
J(R)
= flfleft
ideals which are maximalR-subgroups}.
THEOREM 3.6:
If
R is aunitary near-ring,
thefollowing
areequivalent :
(a)
R is a direct sumof simple left
ideals.(b ) Every
R-submoduleof
acyclic
R-module is a direct summand.(c) Every left
idealof R is a
direct summand.(d)
For everyleft
ideal L andR-subgroup
Lof R
the exact sequence 0 - L - L 1splits.
(e) If L is a left
ideal and L anR-subgroup of
R such that LCL,,
then everyR-homomorphism of L
into any M can belifted
to L1.( f ) R
has the d.c.c.on left
ideals andN(R) = 0.
HereN(R)
= n{maximal left ideals.
PROOF:
(a) ~ (b) By Prop. 3.4(a)
everycyclic
R-module is semi-simple
soby Prop.
3.1 every R-submodule of acyclic
R-module is a direct summand.(b) ~ (c)
R is acyclic
R-module in the unital case soby (b)
every R-submodule(i.e.
leftideal)
is a direct summand.(c) ~ (a)
This ispart
ofProp.
3.1.(c) ~ (d)
Since L is a direct summand ofR, R
=L~K
with canoni-cal
projection p:R ~ L.
The restriction of p toL
1 is therequired splitting.
(d) ~ (e)
Iff : L ~ M
is anR-homomorphism
andg : L1 ~ L
is thesplitting
for0 - L 4 Li where
i is the inclusion map,then fg
is therequired lifting
since for all x E Lfg (x ) = fgi (x ) = f (x ).
(e) ~ (c)
If L is any leftideal,
theidentity
map on L can be lifted(by (e))
to all of R. Thus0 - L 1 R splits
and soby
lemma 2.1 L is a direct summand of R.n
(a) ~ (f)
R =~Ji, 1ï simple ~
R has d.c.c. on left ideals.Put Si
=~Ji.
i iÉj
Then
N(R ) C ~ Sj
= 0.(f) ~ (a) (Cf. [3,
p.87])
Let S be thefamily
of all finite intersections of maximal left ideals.By
d.c.c. let C be a minimalelement,
so for every maximal left idealB,
B ~ C = C i.e. C C B. Therefore C CN(R )
= 0.n
Write 0 = C = Si where the Si are maximal left ideals such that Ci = ~
Sj ~
0 for all i.By maximality, R
= Ci~
Si soby [3,
p.86]
R
Ci.
Since Ci ~ RIBi,
the Ci aresimple.
Example
Let G be a finite nonabelian group and letE(G)
be theunital
near-ring d.g. by
theendomorphisms
of G. ThenE (G )
is finite sohas d.c.c. on left ideals. Johnson
[7]
has shown that if G is the sum of its minimalfully
invariantsubgroups
thenN(E(G))
= 0. Thus for suchG, E(G)
issemi-simple.
The
question
arises as to when asemi-simple near-ring
isstrictly semi-simple (the
converseclearly
isalways true).
It is known(see [3])
that asemi-simple
module isstrictly semi-simple
if every maximalR-submodule is a maximal
R-subgroup.
PROPOSITION 3.7: A
semi-simple near-ring
R isstrictly semi-simple if
every non-zero
R-subgroup
contains a non-zero R-submodule(left ideal ).
PROOF: Let
L 1 C L2 be R-subgroups
andf : L1 ~ M. By hypothesis
there is a left ideal L
c Li. f
can be restricted to L andby
Theorem3.6(e)
extended to L2. Thusby
Theorem3.5(e)
R isstrictly
semi-simple.
REMARK: For
rings,
a module M isinjective
iff it satisfies Baer’scondition, namely
everyhomomorphism f :
L - M for L a left ideal ofR,
can be extended to R. Fornear-rings,
this is notapparently
true forn-injectives,
but Theorem 3.6 shows that R issemi-simple
iff every M satisfies Baer’s condition.In this connection it can be noted that the Theorem of
[13]
to the effect that in thed.g.
case a module M isinjective
iff it satisfies Baer’scondition,
is false.(This
is also noted in[2])
Acounter-example
isprovided by
thecategory
of groups which has noinjectives
but inwhich the divisible groups
satisfy
Baer’s condition.4. Essential extensions
DEFINITION 4.1
[10]:
The R-module M is an essential(resp. strictly essential)
extension of its R-submodule A if K ~ A 0(0)
for allnon-zero R-submodules
(resp. R-subgroups)
K of M.In standard module
theory
theinjective
modules areprecisely
thosewith no proper essential extensions. The
analogous
results for near-rings
are:PROPOSITION 4.1:
If
A isloosely injective (resp.
almostinjective)
then A has no proper essential
(resp. strictly essential)
extensions.PROOF:
Suppose
M is an essential extension ofA,
so we have a s.e.s.0 ~ A ~ M ~ B ~ 0. This
splits (resp.
almostsplits) by hypothesis
soby
lemma 2.1 M is a direct sum(resp. by
lemma 2.5 M is a semi-directsum)
of A and B. Thus A n B =(0)
where B is an R-submodule(resp.
R-subgroup)
of M.COROLLARY: If
A isn-injective,
it has no essential extensions.As a
pseudo
converse to thiscorollary
we havePROPOSITION 4.2:
If A is an
R-submoduleof
ann-injective
module Iand A has no essential
extension,
then A isn-injective.
PROOF:
(Cf. [9, Prop. 8,
p.91])
Let A’ be an R-submodule of I maximal withrespect
to theproperty
A ~ A’ =(0).
Then A +A’/A’ ~
A is an R-submodule of
I /A’
and we show this is an essential extension. It will follow thatI/A’ ’=A+A’/A’soJ=A@AB
Since Iis
n-injective,
so is Aby Prop. 2.4(a).
Now if there is an R-submodule K with A’ C K C I such that
K/A’nA+A’/A’=(0),
thenK n (A + A ’ ) c A ’
so K ~ A ~ A ’ ~ A= (0)
soby
themaximality
ofA’,
K = A’ . ThusI /A’
is essentialover A as claimed.
In view of this
proposition,
it would be nice to know if every A is anR-submodule of an
n-injective
module(for
somenear-rings R ).
In thecase of groups, this means
knowing
whether every group can be embedded as a normalsubgroup
of ann-injective
group.Now,
the class ofn-injective
groups is notknown,
but ann-injective
group is both divisible andcomplete.
It is known that every group is asubgroup
of a divisible group
[12,
p.99]
and of acomplete
group,namely
a fullsymmetric
group. Thissuggests trying
todrop
thenormality require-
ment in the
proposition.
This can bedone,
but at the expense ofchanging
theessentiality
condition. The revised result isPROPOSITION 4.3:
If A is an R-subgroup of
ann-injective
moduleI,
and A is not essential in any R-modulecontaining
it as a properR-subgroup,
then A isn-injective.
PROOF: Similar to
Prop.
4.2. This time A+ A’/A’
is anR-subgroup
instead of an R-submodule of
I /A’ .
Then I = A’+ A
soProposition 2.4(a)
can still be used.Because
Proposition 2.4(b)
was not proven for semi-direct sums, butonly
direct sums, it is notpossible
togive
such arewording
of the nextproposition.
PROPOSITION 4.4:
If
A is an R-submoduleof
aloosely injective
module I and A has no proper essential extension then A is
loosely injective.
PROOF: As for
Proposition 4.2,
but invokeProposition 2.4(b)
inplace
of
Proposition 2.4(a).
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