Subgroups of direct products of groups, ideals and subrings of direct products of rings, and Goursat’s lemma

12 Pages·2009·0.217 MB·English

by  D. D. Anderson, V. Camillo

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Volume480,2009
Subgroups of Direct Products of Groups, Ideals and
Subrings of Direct Products of Rings, and Goursat’s Lemma
D. D. Anderson and V. Camillo
Abstract. We give an exposition of Goursat’s Lemma which describes the
subgroupsofadirectproductoftwogroups. Aringversiongivingthesubrings
andidealsofadirectproductoftworingsisalsogiven.
In group theory there are three important constructions of new groups from
oldgroups: (1)thesubgroupH ofagroupG,(2)thequotientorfactorgroupG/H
where H is a normal subgroup of G (denoted H (cid:1)G), and (3) the direct product
G ×G of two groups G and G . For each of these three constructions, we can
1 2 1 2
ask what are the subgroups? The answer in the first two cases is easy. A subgroup
L of H is just a subgroup L of G contained in H (a subgroup of a subgroup is a
subgroup!) and by the Correspondence Theorem a subgroup of G/H has the form
J/H where J is a subgroup of G with H ⊆ J ⊆ G (moreover, J/H (cid:1)G/H if and
onlyifJ(cid:1)G). Thethirdcaseismoredifficultandisthefocusofthisarticle: Given
groups G and G , find all the (normal) subgroups of G ×G .
1 2 1 2
GiventwogroupsG andG ,thedirectproductG ×G ofG andG istheset
1 2 1 2 1 2
of ordered pairs {(g ,g )|g ∈ G } with coordinate-wise product (g ,g )(h ,h ) =
1 2 i i 1 2 1 2
(g h ,g h ). Here (1,1) is the identity element and (g ,g )−1 = (g−1,g−1). If H
1 1 2 2 1 2 1 2 i
is a subgroup of G , then H ×H is easily checked to be a subgroup of G ×G .
i 1 2 1 2
Moreover, H ×H is a normal subgroup of G ×G if and only if each H (cid:1) G .
1 2 1 2 i i
Let us call a subgroup of G ×G of the form H ×H a subproduct of G ×G .
1 2 1 2 1 2
A beginning abstract algebra student may be tempted to conjecture that every
(normal)subgroupofadirectproductoftwogroupsisasubproduct. Thestandard
counterexample is Z ×Z with normal subgroup {(¯0,¯0),(¯1,¯1)} where (Z ,+) is
2 2 2
the integers mod2 under addition.
There is a way to describe the subgroups of G ×G going back to E´. Goursat
1 2
[4] in 1889 which involves isomorphisms between factor groups of subgroups of G
1
andG . Briefly,givensubgroupsH (cid:1)H ⊆G andanisomorphismf:H /H →
2 i1 i2 i 12 11
H /H , H ={(a,b)∈H ×H |f(aH )=bH } is a subgroup of G ×G and
22 21 12 22 11 21 1 2
each subgroup of G ×G is of this form; moreover, a criterion for H to be normal
1 2
is given. This is presented in Theorem 4.
Math SubjectClassification (2000): primary20-02,secondary20D99,16-02
Keywords: Directproduct,Goursat’sLemma
11
22 D.D.ANDERSONANDV.CAMILLO
In contrast, if R and R are rings with identity, then every ideal of R ×R
1 2 1 2
hasthe formI ×I where I isanidealofR . Ofcourse, ifR andR donothave
1 2 i i 1 2
an identity, this result is no longer true. Indeed, if we endow (Z ,+) with the zero
2
product ¯0·¯0 = ¯0·¯1 = ¯1·¯0 = ¯1·¯1 = ¯0, then {(¯0,¯0),(¯1,¯1)} is an ideal of Z ×Z .
2 2
InTheorem11we give aring versionof Goursat’s Lemma thatdescribesthe ideals
andsubrings ofthedirectproductR ×R oftworingsandinTheorem14we give
1 2
a module version.
We also consider the following two questions:
(1) What pairs of groups G ,G have the property that every (normal) subgroup
1 2
of G ×G is a subproduct of G ×G (recall that a subgroup H ×H is called a
1 2 1 2 1 2
subproduct of G ×G )?
1 2
(2) What pairs of rings R ,R have the property that every subring (ideal) of
1 2
R ×R is a subproduct of R ×R ?
1 2 1 2
At the end of the paper we give a brief biographical sketch of E´. Goursat and
a history of his lemma.
It is our contention that Goursat’s Lemma is a useful result that deserves to
be more widely known. Indeed, we show that the Zassenhaus Lemma (Theorem 8)
is a corollary of Goursat’s Lemma. This article could be used in a first abstract
courseasitmakesgooduseofthethreeimportantconstructions,subgroups, factor
groups, and direct products, and of isomorphisms. The only prerequisites from
group theory are a good understanding of subgroups, normal subgroups, factor
groups, direct products, and isomorphism. In one place (the proof of Theorem
4(1)) the First Isomorphism Theorem for groups is used, but this can easily be
avoided. For the ring portion one only needs to be familiar with rings, subrings,
ideals, ring isomorphisms, and direct products of rings.
Goursat’s Lemma
As a warm up, we first determine the pairs of groups G ,G for which every
1 2
subgroup of G ×G is a subproduct. A group G is nontrivial if |G| > 1. The
1 2
proof is based on the well known result given below that a direct product G ×G
1 2
of two nontrivial groups is cyclic if and only if G and G are finite cyclic groups
1 2
with |G | and |G | relatively prime.
1 2
Lemma 1. Let G and G be nontrivial groups. Then G ×G is cyclic if and
1 2 1 2
only if G and G are finite cyclic groups with gcd(|G |,|G |)=1.
1 2 1 2
Proof. (⇒) Suppose that G ×G is cyclic, say G ×G = (cid:6)(g ,g )(cid:7). Let
1 2 1 2 1 2
g ∈ G , so (g,1)=(g ,g )n for some integer n. So g =gn and gn =1. This gives
1 1 2 1 2
that G =(cid:6)g (cid:7) and g has finite order. Likewise, g has finite order andG =(cid:6)g (cid:7).
1 1 2 1 2 2
Let o(g )=n where o(g ) denotes the order of g . Then o((g ,g ))=|G ×G |=
i i i i 1 2 1 2
n n . However, if (cid:2) = lcm(n ,n ), then (g ,g )(cid:1) = (1,1). So (cid:2) = n n , that is,
1 2 1 2 1 2 1 2
gcd(n ,n )=1. (⇐)SupposethatG =(cid:6)g (cid:7)where|G |=n withgcd(n ,n )=1.
1 2 i i i i 1 2
Certainly (g ,g )n1n2 = ((gn1)n2,(gn2)n1) = (1,1). But (1,1) = (g ,g )(cid:1) implies
1 2 1 2 1 2
g(cid:1) =1; son |(cid:2). Since gcd(n ,n )=1, n n |(cid:2). So (g ,g ) has ordern n and hence
i i 1 2 1 2 1 2 1 2
(cid:6)(g ,g )(cid:7)=G ×G . (cid:2)
1 2 1 2
Theorem 2. Let G and G be nontrivial groups. Then every subgroup of
1 2
G ×G is a subproduct if and only if for g ∈ G , g has finite order o(g ) and
1 2 i i i i
gcd(o(g ),o(g ))=1.
1 2
SUBGROUPS OF DIRECT PRODUCTS OF GROUPS 33
Proof. (⇒) Letg ∈G −{1}. Then (cid:6)(g ,g )(cid:7) is a subproduct of G ×G , so
i i 1 2 1 2
(cid:6)(g ,g )(cid:7)=(cid:6)g (cid:7)×(cid:6)g (cid:7). By Lemma 1, g has finite order and gcd(o(g ),o(g ))=1.
1 2 1 2 i 1 2
(⇐) LetH be a subgroup of G × G and let (g ,g ) ∈ H. Since gcd(o(g ),
1 2 1 2 1
o(g )) = 1, the proof of Lemma 1 gives that (cid:6)(g ,g )(cid:7) = (cid:6)g (cid:7)×(cid:6)g (cid:7). So (g ,1),
2 1 2 1 2 1
(1,g ) ∈ H. Thus H = H ×H where H = {g ∈ G |(g,1) ∈ H} and H = {g ∈
2 1 2 1 1 2
G |(1,g)∈H}. (cid:2)
2
We shift gears for a moment and look at subrings of R ×R where R and
1 2 1
R are rings with identity. Since {(¯0,¯0),(¯1,¯1)} is a subring of Z ×Z , a subring
2 2 2
of R ×R having the same identity as R ×R need not be a subproduct. With
1 2 1 2
Theorem2(orLemma1)inmind,itiseasytocharacterizethepairsofringsR ,R
1 2
withidentitysuchthateverysubringofR ×R containingtheidentityofR ×R
1 2 1 2
is a subproduct. Recall that the characteristic of a ring R, denoted by charR, is
the least positive integer n with na=0 for all a∈R (or just with n1=0 if R has
an identity) or 0 if no such n exists.
Theorem 3. Let R and R be rings with identity. Then every subring of
1 2
R ×R with identity (1,1) is a subproduct of R ×R if and only if each R has
1 2 1 2 i
nonzero characteristic charR and gcd(charR , charR )=1.
i 1 2
Proof. (⇒) The prime subring Z(1,1) = {n(1,1)|n ∈ Z} of R ×R is a
1 2
subring of R ×R containing (1,1). So Z(1,1) = S ×S where S is necessarily
1 2 1 2 i
theprimesubringofR . Hencethedirectproductofthetwoadditivecyclicgroups
i
S and S is cyclic. By Lemma 1, S and S are finite with gcd(|S |, |S |) = 1;
1 2 1 2 1 2
that is, charR (cid:9)=0 and gcd(charR , charR )=1.
i 1 2
(⇐) Conversely, by Lemma 1 the conditions on charR give that Z(1,1) =
i
Z1 ×Z1 . Let S be a subring of R ×R with (1,1) ∈ S. Then (1,0),(0,1)∈
R1 R2 1 2
Z(1,1) ⊆ S. So if (s ,s ) ∈ S, then (s ,0), (0,s ) ∈ S. Thus S = S ×S where
1 2 1 2 1 2
S = {s ∈ R |(s,t) ∈ S for some t ∈ R } (resp., S = {t ∈ R |(s,t) ∈ S for some
1 1 2 2 2
s∈S })isasubringofR (resp.,R )containingtheidentityofR (resp.,R ). (cid:2)
1 1 2 1 2
We next give a version of Goursat’s Lemma for groups.
Theorem 4. (Goursat’s Lemma for Groups) Let G and G be groups.
1 2
(1) Let H be a subgroup of G ×G . Let H = {a ∈ G |(a,1) ∈ H}, H =
1 2 11 1 21
{a ∈ G |(1,a) ∈ H}, H = {a ∈ G |(a,b) ∈ H for some b ∈ G }, and H =
2 12 1 2 22
{b ∈ G |(a,b) ∈ H for some a ∈ G }. Then H ⊆ H are subgroups of G with
2 1 i1 i2 i
H (cid:1)H and the map f :H /H →H /H given by f (aH )=bH where
i1 i2 H 12 11 22 21 H 11 21
(a,b)∈H is an isomorphism. Moreover, if H (cid:1)G ×G , then H ,H (cid:1)G and
1 2 i1 i2 i
H /H ⊆Z(G /H ), the center of G /H .
i2 i1 i i1 i i1
(2) Let H (cid:1) H be subgroups of G and let f:H /H → H /H be an iso-
i1 i2 i 12 11 22 21
morphism. Then H = {(a,b) ∈ H × H |f(aH ) = bH } is a subgroup of
12 22 11 21
G ×G . Further suppose that H ,H (cid:1) G and H /H ⊆ Z(G /H ). Then
1 2 i1 i2 i i2 i1 i i1
H (cid:1)G ×G .
1 2
(3) The constructions given in (1) and (2) are inverses to each other.
Proof. (1)ItiseasilycheckedthatH ⊆H aresubgroupsofG ;orobservethat
i1 i2 i
we may identify H with H ∩G and H =π (H) where π :G ×G →G is the
i1 i i2 i i 1 2 i
projectionmap. Also,H (cid:1)H . Wedothecasei=1. Nowa∈H ⇒(a,1)∈H.
i1 i2 11
44 D.D.ANDERSONANDV.CAMILLO
Letc∈H ;sothereexistsbwith(c,b)∈H. Now(c−1ac,1)=(c,b)−1(a,1)(c,b)∈
12
H so c−1ac ∈ H . Define fˆ:H → H /H by fˆ(a) = bH where (a,b) ∈ H.
11 12 22 21 21
If (a,b),(a,c) ∈ H, then (1,b−1c) = (a,b)−1(a,c) ∈ H implies b−1c ∈ H and
21
hence bH =cH . So fˆis well-defined. It is easily checked that fˆis a surjective
21 21
homomorphismwithkerfˆ=H . Indeed,ifa∈kerfˆ,thenfor(a,b)∈H,b∈H .
11 21
But then (1,b) ∈ H, so (a,1) = (a,b)(1,b)−1 ∈ H which gives a ∈ H . So by
11
the First Isomorphism Theorem, f :H /H →H /H is an isomorphism. It is
H 12 11 22 21
easilycheckedthatH (cid:1)G ×G impliesH (cid:1)G ;orusethepreviousidentification
1 2 ij i
withtheintersectionwithG andprojectionontoG . Also,forg ∈G anda∈H
i i 1 12
with (a,b) ∈ H, (g−1ag,b), (a−1,b−1) ∈ H give (a−1g−1ag,1) ∈ H and hence
a−1g−1ag ∈H . So H /H ⊆Z(G /H ). Likewise, H /H ⊆Z(G /H ).
11 12 11 1 11 22 21 2 21
(2) It is easily checked that H is a subgroup of H ×H and hence of G ×G .
12 22 1 2
Indeed, if (a,b),(c,d) ∈ H, then f(aH ) = bH and f(cH ) = dH . So
11 21 11 21
f(acH )=f(aH cH )=f(aH )f(cH )=bH dH =bdH andf(a−1H )=
11 11 11 11 11 21 21 21 11
f((aH )−1) = (f(aH ))−1 = (bH )−1 = b−1H . So (ac,bd), (a−1,b−1) ∈ H.
11 11 21 21
Further, suppose that H (cid:1) G and H /H ⊆ Z(G /H ). We show that
ij i i2 i1 i i1
H (cid:1) G × G . Let (a ,a ) ∈ H and (g ,g ) ∈ G × G . So H (cid:1) G gives
1 2 1 2 1 2 1 2 i2 i
(g−1a g ,g−1a g ) ∈ H ×H . And H /H ⊆ Z(G /H ) givesg −1a g H =
1 1 1 2 2 2 12 22 i2 i1 i i1 i i i i1
a H , so f(g−1a g H )=g−1a g H . Thus (g−1a g , g−1a g )∈H.
i i1 1 1 1 11 2 2 2 21 1 1 1 2 2 2
(3) Clear. (cid:2)
The reader may have noticed that the proof in Theorem 4(2) that H is a
subgroup did not use the fact thatf is a bijection. Indeed, for any homomorphism
f:H /H → H /H , H is a subgroup. However, if we set imf = H(cid:2) /H
12 11 22 21 22 21
where H(cid:2) is a subgroup of H with H (cid:1)H(cid:2) and kerf =H(cid:2) /H where H ⊆
22 22 21 22 11 11 11
H(cid:2) (cid:1)H , then f induces the isomorphism f(cid:2):H /H(cid:2) →H(cid:2) /H and the same
11 12 12 11 22 21
subgroup H.
Note that the subproduct H = H × H corresponds to H = H = H
1 2 11 12 1
and H = H = H ; so f :H /H → H /H is the trivial map. We illustrate
21 22 2 H 1 1 2 2
Goursat’s Lemma by finding all (normal) subgroups of S ×S .
3 3
Example 5. The subgroups of S ×S . First, the subgroups of S are (cid:6)(1)(cid:7),
3 3 3
(cid:6)(12)(cid:7),(cid:6)(13)(cid:7),(cid:6)(23)(cid:7),(cid:6)(123)(cid:7) =A ,andS . Ofthese,(cid:6)(1)(cid:7),A ,andS arenormal.
3 3 3 3
We have the following subnormal quotient groups H/K where K(cid:1)H ⊆S grouped
3
by order: (a)|H/K| = 1; (cid:6)(1)(cid:7)/(cid:6)(1)(cid:7), (cid:6)(12)(cid:7)/(cid:6)(12)(cid:7), (cid:6)(13)(cid:7)/(cid:6)(13)(cid:7), (cid:6)(23)(cid:7)/(cid:6)(23)(cid:7),
A /A , S /S ; (b)|H/K| = 2: (cid:6)(12)(cid:7)/(cid:6)(1)(cid:7), (cid:6)(13)(cid:7)/(cid:6)(1)(cid:7),
3 3 3 3
(cid:6)(23)(cid:7)/(cid:6)(1)(cid:7), S /A ; (c)|H/K| = 3: A /(cid:6)(1)(cid:7); (d)|H/K| = 6: S /(cid:6)(1)(cid:7). Note
3 3 3 3
that within each of the four classes, the quotient groups are all isomorphic. Class
(a)has only the identity maps between the 6 different quotients; so there are 36
different isomorphisms f:H /H → H /H yielding the 36 different subprod-
12 11 22 21
ucts (cid:6)(1)(cid:7) × (cid:6)(1)(cid:7),··· ,S × S . Of these 9 are normal. Class (b)has 4 groups
3 3
of order 2. Since there is a unique isomorphism between two groups of order 2,
there are 16 different isomorphisms f:H /H → H /H yielding 16 distinct
12 11 22 21
subgroups. For example, the isomorphism (cid:6)(12)(cid:7)/(cid:6)(1)(cid:7) → (cid:6)(13)(cid:7)/(cid:6)(1)(cid:7) gives the
subgroup {((1),(1)), ((12),(13))}. There are 9 such subgroups each of order 2.
The isomorphism (cid:6)(12)(cid:7)/(cid:6)(1)(cid:7)→S /A gives the subgroup {((1),(1)), ((1),(123)),
3 3
((1),(132)), ((12),(12)), ((12),(13)), ((12),(23))} and there are 6 such subgroups
SUBGROUPS OF DIRECT PRODUCTS OF GROUPS 55
each isomorphic to S . The isomorphism S /A → S /A gives rise to the sub-
3 3 3 3 3
group E = {(a,b) ∈ S ×S |(a,b) ∈ A ×A or o(a) = o(b) = 2}. Note that E
3 3 3 3
has 9 elements of order 2 and hence is not isomorphic to a subproduct of S ×S .
3 3
So this gives a total of 16 subgroups in this class and as only the last one satisfies
H /H ⊆ Z(G/H ), onlyE is a normal subgroup. Class (c)has only one group
i2 i1 i1
of order 3 but two isomorphisms (cid:6)(123)(cid:7)/(cid:6)(1)(cid:7) → (cid:6)(123)(cid:7)/(cid:6)(1)(cid:7). The identity map
gives the subgroup {(x,x)|x∈A } which is isomorphic to A and the isomorphism
3 3
given by (123) →(132)gives thesubgroup{((1),(1)),((123),(132)),((132),(123))}
which is isomorphic to A . Neither subgroup is normal. Finally, for class (d)there
3
are 6 isomorphisms S → S , each given by conjugation. (It is well known that
3 3
for n (cid:9)= 2, 6 each automorphism of S is inner, that is, a conjugation on S , and
n n
Aut(S )≈S ; see for example, [9, Theorem 7.4].)Thus this class has 6 subgroups:
n n
G ={(a,σ−1aσ)|a∈S } forσ ∈S . Note that each is isomorphicto S , but none
σ 3 3 3
are normal.
In summary S ×S has 60 distinct subgroups, 10 of which are normal. More-
3 3
over, each of these subgroups except E is isomorphic to a subproduct.
This example raises the following question.
Question 6. What pairs of groups G and G have the property that every
1 2
(normal)subgroup of G ×G is isomorphic to a subproduct of G ×G ?
1 2 1 2
It is not hard to prove that a pair of finitely generated abelian groups G and
1
G satisfytheconditionofQuestion6. ButwehaveshownthatthepairS ,S does
2 3 3
not. Also, if G is a rank two indecomposable abelian group, then G is isomorphic
to a subgroup of Q×Q, but not to a subproduct of Q×Q.
We next show how you can use Goursat’s Lemma to prove Theorem 2. By
Goursat’sLemmaeverysubgroupofG ×G isasubproductifandonlyiftheonly
1 2
pairs of normalsubgroups H (cid:1)H ⊆G withH /H andH /H isomorphic
i1 i2 i 12 11 22 21
are the trivial ones H = H . So suppose that every subgroup of G ×G is
i1 i2 1 2
a subproduct. Then for g ∈ G and m > 1 with (cid:6)gmi(cid:7) (cid:9)= (cid:6)g (cid:7), (cid:6)g (cid:7)/(cid:6)gm1(cid:7) and
i i i i i 1 1
(cid:6)g (cid:7)/(cid:6)gm2(cid:7) can not be isomorphic. From this it follows that each order o(g ) is
2 2 i
finite and gcd(o(g ), o(g )) = 1. Conversely, suppose that each g ∈ G has finite
1 2 i i
order and gcd(o(g ),(o(g ))=1. Then H /H are groups in which every element
1 2 i2 i1
has finite order and the order of each element of H /H is relatively prime to
12 11
the order of each element of H /H . Thus we can have H /H and H /H
22 12 12 11 22 12
isomorphic only in the trivial case that H = H . But then every subgroup of
i1 i2
G ×G is a subproduct.
1 2
We next use Goursat’s Lemma to determine the pairs of nontrivial groups G
1
and G with every normal subgroup of G ×G a subproduct. Certainly if every
2 1 2
subgroup of G × G is a subproduct, every normal subgroup is a subproduct.
1 2
However,notethatG ×G mayhaveeverynormalsubgroupasubproductwithout
1 2
having every subgroup a subproduct. For let G be any nonabelian simple group.
Then certainly G×G has subgroups that are not subproducts. However, the only
proper, nontrivialnormalsubgroups ofG×GareG×{1}and{1}×G. Thiseasily
follows from Theorem 4.
Theorem 7. Let G and G be nontrivial groups. Then the following are
1 2
equivalent.
(1) Every normal subgroup of G ×G is a subproduct.
1 2
66 D.D.ANDERSONANDV.CAMILLO
(2)TheredonotexistnormalsubgroupsH (cid:1)H ⊆G withH /H ⊆Z(G /H )
i1 i2 i i2 i1 i i1
and H /H isomorphic to H /H .
12 11 22 21
(3) One of the following two conditions holds.
(a) G or G is residually centerless. (A group G is residually centerless if for
1 2
each homomorphic image G¯, Z(G¯)={1}.)
(b) For each normal subgroup H (cid:1) G , Z(G /H ) is torsion and gcd(o(g ),
i i i i 1
o(g ))=1 for g ∈G /H .
2 i i i
Proof. (1)⇔(2) This follows from Theorem 4. (2)⇒(3) Suppose that (a) does
not hold. Suppose that there is a normal subgroup H (cid:1)G with Z(G /H ) hav-
1 1 1 1
ing an element g¯ (g ∈ G) of infinite order. Let H (cid:1) G with Z(G /H ) (cid:9)= {¯1}.
2 2 2 2
Let ¯1 (cid:9)= h¯ ∈ Z(G /H )(h ∈ G ). Since g¯ ∈ Z(G /H ) (resp., h¯ ∈ Z(G /H )),
2 2 2 1 1 2 2
H (cid:1) (cid:6)g(cid:7)H (cid:1) G with (cid:6)g¯(cid:7) = (cid:6)g(cid:7)H /H ⊆ Z(G /H ) and H (cid:1) (cid:6)h(cid:7)H (cid:1) G
1 1 1 1 1 1 1 2 2 2
with (cid:6)h¯(cid:7) = (cid:6)h(cid:7)H /H ⊆ Z(G /H ). If h¯ has infinite order, then (cid:6)g(cid:7)H /H and
2 2 2 2 1 1
(cid:6)h(cid:7)H /H are isomorphic, a contradiction. So o(h¯) = n > 1. But then (cid:6)gn(cid:7)H
2 2 1
and (cid:6)g(cid:7)H are normal subgroups of G with (cid:6)g(cid:7)H /(cid:6)gn(cid:7)H ⊆ Z(G /(cid:6)gn(cid:7)H ) and
1 1 1 1 1 1
(cid:6)g(cid:7)H /(cid:6)gn(cid:7)H is cyclic of order n. Thus (cid:6)g(cid:7)H /(cid:6)gn(cid:7)H and (cid:6)h(cid:7)H /H are isomor-
1 1 1 1 2 2
phic; acontradiction. Thus foreachnormalsubgroupH (cid:1)G , Z(G /H ) must be
i i i i
torsion. Suppose that there are normal subgroups H (cid:1) G with elements g ∈ G
i i i i
such that g H ∈ Z(G /H ) and gcd(o(g H ), o(g H )) (cid:9)= 1. Then Z(G /H )
i i i i 1 1 2 2 1 1
and Z(G /H ) have elements g(cid:1)H (g(cid:2) ∈ G ) with o(g(cid:1)H ) = o(g(cid:1)H ) > 1. But
2 2 i i i i 1 1 2 2
then (cid:6)g(cid:1)(cid:7)H /H and (cid:6)g(cid:1)(cid:7)H /H are isomorphic; a contradiction. (3)⇒(2) Sup-
1 1 1 2 2 2
pose there exist normal subgroups H (cid:1) H ⊆ G with H /H ⊆ Z(G /H )
i1 i2 i i2 i1 i i1
and H /H isomorphic to H /H . Then there are elements h¯ ∈H /H with
12 11 22 21 i i2 i1
1<o(h¯ )=o(h¯ )<∞, a contradiction. (cid:2)
1 2
We next show how Goursat’s Lemma can be used to prove the Zassenhaus
Lemma [9, Lemma 5.8]: if A(cid:1)A∗,B (cid:1)B∗ are subgroups of a group G, then the
groups A(A∗ ∩B∗)/A(A∗∩B) and B(B∗ ∩A∗)/B(B∗ ∩A) are isomorphic. The
ZassenhausLemmaplaysakeyroleintheproofoftheSchreierRefinementTheorem
[9,Theorem5.9]whichstatesthattwosubnormalseriesforagrouphaveequivalent
refinementswhichinturnisusedtoprovetheJordan-Ho¨lderTheorem[9,Theorem
5.10]: any two composition series of a group G are equivalent.
Theorem 8. (Zassenhaus Lemma)Let G be a group and A(cid:1)A∗ and B (cid:1)B∗
subgroups of G. Then A(A∗∩B)(cid:1)A(A∗∩B∗), B(B∗∩A)(cid:1)B(B∗∩A∗) and the
quotient groups A(A∗∩B∗)/A(A∗∩B) and B(B∗∩A∗)/B(B∗∩A) are isomorphic.
Proof. Let H = {(ac,bc) ∈ G×G|a ∈ A, b ∈ B, c ∈ A∗ ∩B∗}. We first
show that H is a subgroup of G×G. Let (ac,bc), (a(cid:2)c(cid:2),b(cid:2)c(cid:2)) ∈ H where a,a(cid:2) ∈
A, b,b(cid:2) ∈ B, and c,c(cid:2) ∈ A∗ ∩ B∗. Now A (cid:1) A∗ gives ca(cid:2) = a¯c and c−1a−1 =
a∗c−1 for some a¯,a∗ ∈ A and B (cid:1) B∗ gives cb(cid:2) = ¯bc and c−1b−1 = b∗c−1 for
some ¯b,b∗ ∈ B. So (ac,bc)(a(cid:2)c(cid:2),b(cid:2)c(cid:2)) = (aca(cid:2)c(cid:2),bcb(cid:2)c(cid:2)) = (aa¯cc(cid:2),b¯bcc(cid:2)) ∈ H and
(ac,bc)−1 = (c−1a−1,c−1b−1) = (a∗c−1,b∗c−1) ∈ H. Now using the notation of
Goursat’s Lemma (Theorem 4), we determine the H ’s. Certainly H = A(A∗∩
ij 12
B∗) andH = B(B∗ ∩A∗) (which shows that they are subgroups of G). Now
22
H ={ac|a∈A, c∈A∗∩B∗, (ac,1)∈H}= {ac|a∈A, c∈A∗∩B∗, c=b−1 for
11
some b∈B}= {ac|a∈A, c∈A∗∩B}= A(A∗∩B). Likewise H =B(B∗∩A).
21
Thus from H (cid:1)H we get that A(A∗∩B) is a normal subgroup of A(A∗∩B∗),
i1 i2
SUBGROUPS OF DIRECT PRODUCTS OF GROUPS 77
B(B∗∩A) is a normal subgroup of B(B∗∩A∗) and since H /H and H /H
12 11 22 21
are isomorphic, the proof is complete. (cid:2)
We next turn to direct products of rings and their ideals and subrings. If R
1
and R are rings, and I and I are ideals of R and R , respectively, then I ×I
2 1 2 1 2 1 2
is an ideal of R ×R . Similar statements hold for right and left ideals. If R and
1 2 1
R haveanidentity,thenitiswellknownthateveryideal(right,left,ortwo-sided)
2
has this form. This is our next proposition.
Proposition 9. Let R and R be rings with identity. Then every (right, left,
1 2
two-sided) idealof R ×R has the form I ×I where I is a (right, left, two-sided)
1 2 1 2 i
ideal of R .
i
Proof. Let I be a left ideal of R × R . Let I = {a ∈ R |(a,0) ∈ I}
1 2 1 1
and I = {a ∈ R |(0,a) ∈ I}. It is easily checked that I is a left ideal of R .
2 2 i i
Let (a,b) ∈ I ×I , then (a,0),(0,b) ∈ I, so (a,b) = (a,0)+(0,b) ∈ I. Hence
1 2
I ×I ⊆ I. Conversely, suppose that (a,b) ∈ I. Then (a,0) = (1,0)(a,b) ∈ I;
1 2
so a ∈ I . Likewise (0,b) = (0,1)(a,b) ∈ I; so b ∈ I . Hence (a,b) ∈ I ×I and
1 2 1 2
I ⊆I ×I . (cid:2)
1 2
Of course Proposition 9 may fail if R and R do not have an identity. For
1 2
example, if we take R = R = (Z ,+) where Z has the zero product, then
1 2 2 2
{(¯0,¯0),(¯1,¯1)} is an ideal of Z × Z that is not a subproduct. We next give a
2 2
partialconversetoProposition9. LetuscallaringRaleft e-ring ifforeachr ∈R
there exists an element e ∈R, depending on r, with e r =r. Note that R is a left
r r
e-ring if and only if RI =I for each left ideal I of R. The next result comes from
[1].
Theorem 10. For a ring R the following statements are equivalent.
(1) R is a left e-ring.
(2) For each ring S, every left ideal of R×S is a subproduct of left ideals.
(3) Every left ideal of R×R is a subproduct of left ideals.
Proof. (1)⇒(2) Let I be a left ideal of R. For (a,b)∈ I choose e ∈ R with
a
e a=a. Then (a,0)=(e ,0)(a,b)∈I. Hence (0,b)=(a,b)−(a,0)∈I. Then as
a a
intheproofofProposition9wehaveI =I ×I whereI ={a∈R|(a,0)∈I}and
1 2 1
I ={b∈S|(0,b)∈I}. (2)⇒(3)Clear. (3)⇒(1)Leta∈R. Thenthe principal left
2
ideal generated by (a,a), ((a,a)) ={(r,s)(a,a)+n(a,a)|r,s∈R,n∈Z}, is a left
(cid:1)
idealof R×R. So ((a,a)) =I ×I where I andI are leftidealsofR. Certainly
(cid:1) 1 2 1 2
a∈I , so (a) ⊆I . And as ((a,a)) ⊆(a) ×(a) , we have ((a,a)) =(a) ×(a) .
i (cid:1) i (cid:1) (cid:1) (cid:1) (cid:1) (cid:1) (cid:1)
Thus(a,0)∈((a,a)) ; so(a,0)=(e ,e )(a,a)+n(a,a)forsomee ∈Randn∈Z.
(cid:1) 1 2 i
Hencea=e a+naand0=e a+na;soa=e a+na=e a−e a=(e −e )a. (cid:2)
1 2 1 1 2 1 2
Weleaveittothereadertodefinearighte-ringandtostateversionsofTheorem
10 for right ideals and two-sided ideals. Note that if {R } is any nonempty family
α
of left e-rings, then their direct sum ⊕R with coordinate-wise operations is again
α
a left e-ring. In particular, an infinite direct sum of rings each having an identity
is both a left and right e-ring, but does not have an identity.
We next give versions of Goursat’s Lemma for ideals and subrings of a direct
product of rings.
88 D.D.ANDERSONANDV.CAMILLO
Theorem 11. (Goursat’s Lemma for Ideals and Subrings) Let S and S be
1 2
rings.
(1) Let T be an additive subgroup of S ×S . Let T = {s ∈ S |(s,0)∈ T},
1 2 11 1
T ={s∈ S |(s,t)∈T for some t∈ S }, T ={t∈S |(0,t)∈T}, and
12 1 2 21 2
T = {t ∈ S |(s,t) ∈ T for some s ∈ S }. Then T ⊆ T are subgroups
22 2 1 i1 i2
of(S ,+)andthemapf :T /T →T /T givenbyf (s+T )=t+T
i T 12 11 22 21 T 11 21
for (s,t)∈T is an abelian group isomorphism.
(a) Suppose that T is a left (resp., right, two-sided) ideal of S ×S .
1 2
Then T ⊆ T are left (resp., right, two-sided) ideals of S with
i1 i2 i
S T ⊆T (resp., T S ⊆T , S T ⊆T and T S ⊆T ).
i i2 i1 i2 i i1 i i2 i1 i2 i i1
(b) Suppose that T is a subring of S ×S . Then T is a subring of
1 2 i2
S and T is an ideal of T . Moreover, f :T /T → T /T is
i i1 i2 T 12 11 22 21
a ring isomorphism. Further, suppose that each S has an identity
i
1 . Then T contains (1 ,1 ) if and only if 1 ∈T and f (1 +
Si S1 S2 Si i2 T S1
T )=1 +T .
11 S2 21
(2) Conversely, suppose that T ⊆ T ⊆ S are additive subgroups with
i1 i2 i
abelian group isomorphism f:T /T →T /T .
12 11 22 21
(a) If S T ⊆T (resp., T S ⊆T , S T ⊆T and T S ⊆T ), then
i i2 i1 i1 i i1 i i2 i1 i2 i i1
T = {(a,b) ∈ T ×T |f(a+T ) = b+T } is a left (resp., right,
12 22 11 21
two-sided) ideal of S ×S .
1 2
(b) Suppose that T is a subring of S , that T is an ideal of T , and
i2 i i1 i2
f:T /T → T /T is a ring isomorphism. Then T = {(a,b) ∈
12 11 22 21
T ×T |f(a+T ) = b+T } is a subring of S ×S . Moreover,
12 22 11 21 1 2
if each S has an identity 1 , then (1 ,1 ) ∈ T if and only if
i Si S1 S2
1 ∈T and f(1+T )=1+T .
Si i2 11 21
(3) The constructions given in (1) and (2) are inverses to each other.
Proof. (1) The facts that T ⊆ T are subgroups of (S ,+) and that f is an
i1 i2 i T
abelian group isomorphism follow from Theorem 4.
(a) Suppose that T is a left ideal of S × S . Let s ∈ S and t ∈ T where
1 2 1 12
(t,t(cid:2))∈ T for some t(cid:2) ∈S . Then (st,0)=(s,0)(t,t(cid:2))∈ T implies st∈T . Hence
2 11
S T ⊆ T . This also shows that T ⊆ T are left ideals of S . The proofs of
1 12 11 11 12 1
the other cases are similar.
(b)SupposethatT isasubringofS ×S . Lett ,t ∈T where(t ,t(cid:2)),(t ,t(cid:2))∈T
1 2 1 2 12 1 1 2 2
for some t(cid:2),t(cid:2) ∈ S . Then (t t ,t(cid:2)t(cid:2)) = (t ,t(cid:2))(t ,t(cid:2)) ∈ T gives t t ∈ T . Hence
1 2 2 1 2 1 2 1 1 2 2 1 2 12
T is asubring ofS . Ifactuallyt ∈T , thenwe cantaket(cid:2) =0, so (t t ,0)∈T
12 1 2 11 2 1 2
gives t t ∈ T and hence T is a left ideal. Similarly T is a right ideal and
1 2 11 11 11
hence is a two-sided ideal. Likewise, T is a subring of S and T is an ideal of
22 2 21
T . Now f (t +T ) = t(cid:2) +T ; so f ((t +T )(t +T )) = f (t t +T ) =
22 T i 11 i 21 T 1 11 2 11 T 1 2 11
t(cid:2)t(cid:2) +T = (t(cid:2) +T )(t(cid:2) +T ) = f (t +T )f (t +T ); hence f is a ring
1 2 21 1 21 2 21 T 1 11 T 2 11 T
isomorphism. Suppose that each S has an identity 1 . Certainly (1 ,1 ) ∈ T
i Si S1 S2
gives 1 ∈ T and that f preserves identities. Conversely, if 1 ∈ T , then
Si i2 T Si i2
f (1 +T )=1 +T gives (1 ,1 )∈T.
T S1 11 S2 21 S1 S2
(2) Suppose that T ⊆T are additive subgroups of (S ,+) and that f:T /T →
i1 i2 i 12 11
T /T is an isomorphism. By Theorem 4, T = {(a,b) ∈ T ×T |f(a+T ) =
22 21 12 22 11
b+T } is an additive subgroup of S ×S .
21 1 2
SUBGROUPS OF DIRECT PRODUCTS OF GROUPS 99
(a) Suppose that S T ⊆T . Let (t ,t )∈T, that is, f(t +T )=t +T . Let
i i2 i1 1 2 1 11 2 21
(s ,s ) ∈ S ×S . Then s t ∈ T . So f(s t +T ) = f(0+T ) = 0+T =
1 2 1 2 i i i1 1 1 11 11 21
s t +T ; so (s ,s )(t ,t ) = (s t ,s t ) ∈ T. Hence T is a left ideal of S ×S .
2 2 21 1 2 1 2 1 1 2 2 1 2
The other cases are similar.
(b) Let (t ,t(cid:2)),(t ,t(cid:2))∈T. So f(t +T )=t(cid:2) +T . Then f a ring isomorphism
1 1 2 2 i 11 i 21
givesf(t t +T )=t(cid:2)t(cid:2) +T ; so(t t ,t(cid:2)t(cid:2))∈T. HenceT isasubring ofS ×S .
1 2 11 1 2 21 1 2 1 2 1 2
The “moreover” statement is immediate.
(3) Clear. (cid:2)
ThereadermayfinditoddthatinTheorem11(2)forsubringstheisomorphism
f:T /T → T /T is required to be a ring isomorphism, but for ideals (which
12 11 22 21
are after all subrings), f is only required to be an abelian group isomorphism. The
reason is that in the ideal case T /T has the zero product since S T ⊆T .
i2 i1 i i2 i1
Asin the case of subgroups, the subproduct T =T ×T of left idealsT of the
1 2 i
rings S corresponds to T = T = T with f the trivial map. Suppose that S
i i1 i2 i T 1
is a left e-ring and T is a left ideal of S ×S . Then we have T = S T ⊆ T ;
1 2 12 1 12 11
so T = T . Since f is an isomorphism, we must also have T = T . Thus
11 12 21 22
T = T ×T . So every left ideal of S ×S is a subproduct. This gives another
12 22 1 2
proof ofProposition9andpart ofTheorem10. Conversely, suppose that everyleft
ideal of S ×S is a subproduct of left ideals. Suppose that T is a left ideal of S
1 1 1 1
with S T (cid:1) T . Then the identity map f:T /S T → T /S T gives rise to the
1 1 1 1 1 1 1 1 1
left ideal T = {(t ,t ) ∈ S ×S |t +S T = t +S T } of S ×S that is not a
1 2 1 1 1 1 1 2 1 1 1 1
subproduct. Indeed, if x ∈ T −S T , then (x,x) ∈ T, but (x,0) (cid:9)∈ T. Hence we
1 1 1
must have S T =T for each left ideal T of S . Hence S must be a left e-ring.
1 1 1 1 1 1
WenextillustrateTheorem11byfindingtheidealsandsubringsofn Z×n Z.
1 2
Example 12. The ideals and subrings of n Z×n Z. Note that for the
1 2
rings n Z, the notions of ideal, subring, and additive subgroup coincide and have
i
the form n (cid:2)Zwhere (cid:2)≥0. We first find the ideals of n Z×n Z. So in accordance
i 1 2
to Theorem 11 we consider pairs of ideals I ⊆ I ⊆ n Zwith n ZI ⊆ I and
i1 i2 i i i2 i1
determine the isomorphisms I /I → I /I , if any. Note that if say I = 0,
12 11 22 21 1j
then I = I = 0, so an isomorphism I /I → I /I forces I = I . This
11 12 12 11 22 21 21 22
case gives the ideal 0×I . Likewise, the case I = 0 gives the ideal I ×0. So
22 2j 12
let I = n k Zwhere k ≥ 1. Now the condition I = n k Z ⊇ I ⊇ n ZI =
i2 i i i i2 i i i1 i i2
n2k Ztranslates to I = n k (cid:2) Zwhere 1 ≤ (cid:2) ≤ n and (cid:2) |n . Now I /I =
i i i1 i i i i i i i i2 i1
n k Z/n k (cid:2) Zis isomorphic to Z/(cid:2) Zvia n k a + n (cid:2) k Z → a + (cid:2) Z. So there
i i i i i i i i i i i i
is an isomorphism I /I → I /I ⇔ (cid:2) = (cid:2) . Let (cid:2) = (cid:2) = (cid:2) ; then there
12 11 22 21 1 2 1 2
are φ((cid:2)) isomorphisms f :Z/(cid:2)Z → Z/(cid:2)Z given by f (a+(cid:2)Z) = ja+(cid:2)Zwhere
(cid:1),j (cid:1),j
1 ≤ j ≤ (cid:2) with gcd(j,(cid:2)) = 1. These isomorphisms give rise to the family of ideals
(n k Z×n k Z) = {(n k a ,n k a ) ∈ n k Z×n k Z|ja ≡ a mod(cid:2)} where
1 1 2 2 (cid:1),j 1 1 1 2 2 2 1 1 2 2 1 2
(cid:2)≥1 with (cid:2)|gcd(n ,n ) and 1≤j ≤(cid:2) with gcd(j,(cid:2))=1. Note that the case (cid:2)=1
1 2
(and hence j = 1)corresponds to the subproducts n k Z×n k Z. In particular,
1 1 2 2
every ideal of n Z×n Z is a subproduct if and only if gcd(n ,n )=1.
1 2 2 1 2
For the case of subrings of n Z×n Z, we just drop the requirement that I ⊇
1 2 i1
n ZI ; so we can take I = n k (cid:2) Z for any (cid:2) ≥ 0. As before, the case I = 0
i i2 i1 i i i i 12
or I = 0 leads to the subring 0×I or I ×0. So assume that I (cid:9)= 0. For
22 22 12 i2
(cid:2) = (cid:2) = (cid:2) ≥ 1, the group isomorphisms n k Z/n k (cid:2)Z → n k Z/n k (cid:2)Z have
1 2 1 1 1 1 2 2 2 2
the form n k a+n k (cid:2)Z → n k ja+n k (cid:2)Z where 1 ≤ j ≤ (cid:2) with gcd(j,(cid:2)) = 1.
1 1 1 1 2 2 2 2
Denote this map by f. For f to be a ring homomorphism we need n k jn k ab+
2 2 1 1
1100 D.D.ANDERSONANDV.CAMILLO
n k (cid:2)Z = f(n k n k ab+n k (cid:2)Z) = f(n k a+n k (cid:2)Z)·f(n k b+n k (cid:2)Z) =
2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1
(n k ja+n k (cid:2)Z)(n k jb+n k (cid:2)Z)=n2k2j2ab+n k (cid:2)Zorn k (cid:2)|(n k n k jab−
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1
n2k2j2ab)=n k jab(n k −n k j). Takinga=b=1andrecallingthatgcd(j,(cid:2))=
2 2 2 2 1 1 2 2
1,weneed(cid:2)|(n k −n k j). (Notethatinthecaseofidealswealwayshave(cid:2)|(n k −
1 1 2 2 1 1
n k j) since (cid:2)|n .)With these conditions on j and (cid:2) we get the subrings (n k Z×
2 2 i 1 1
n k Z) ={(n k a ,n k a )∈n k Z×n k Z|ja ≡a mod(cid:2)}. Finally, suppose
2 2 (cid:1),j 1 1 1 2 2 2 1 1 2 2 1 2
that (cid:2) = 0. We have a ring isomorphism n k Z → n k Z precisely when n k =
1 1 2 2 1 1
n k and then the isomorphism is the identity map. In this case {(n k a,n k a)∈
2 2 1 1 2 2
n k Z×n k Z|a∈Z} is the subring.
1 1 2 2
We next characterize the pairs of rings S ,S for which every ideal of S ×S
1 2 1 2
is a subproduct.
Theorem 13. Let S and S be rings. Then the following are equivalent.
1 2
(1)Every left ideal of S ×S is a subproduct.
1 2
(2)There do not exist left ideals I (cid:1) I with S I ⊆ I such that (I /I ,+)
i1 i2 i i2 i1 12 11
and (I /I ,+) are isomorphic.
22 21
(3)One of the following two conditions holds.
(a)S or S is a left e-ring.
1 2
(b)For each left ideal I of S , (I /S I ,+) is torsion and for g ∈I /S I ,
i i i i i i i i i
gcd(o(g ),o(g ))=1.
1 2
Proof. (1)⇔(2)ThisfollowsfromTheorem11. (2)⇒(3)SupposethatS and
1
S arenotlefte-rings. LetI bealeftidealofS withS I (cid:1)I . Suppose, say,that
2 i i i i i
I /S I is not torsion. So let i ∈ I with o(i +S I ) = ∞. Let i ∈ I −I S .
1 1 1 1 1 1 1 1 2 2 2 2
If o(i + S I ) = ∞, then there is a group isomorphism ((i ) + S I )/S I →
2 2 2 1 1 1 1 1
((i )+S I )/S I . Put I(cid:2) = (i )+S I and I(cid:2) = S I . So I(cid:2) (cid:1) I(cid:2) are left
2 2 2 2 2 i2 i i i i1 i i i1 i2
ideals of S with S I(cid:2) ⊆ I(cid:2) and I(cid:2) /I(cid:2) is isomorphic to I(cid:2) /I(cid:2) , a contradiction.
i i i2 i2 12 11 22 21
So assume that o(i + S I ) = n > 1. Put I(cid:2) = (ni ) + S I , I(cid:2) = S I
2 2 2 12 1 1 1 11 1 1,
I(cid:2) = (i )+S I , and I(cid:2) = S I . Again I(cid:2) (cid:1) I(cid:2) are left ideals with S I(cid:2) ⊆ I(cid:2)
22 2 2 2 21 2 2 i1 i2 i i2 i1
and I(cid:2) /I(cid:2) is isomorphic to I(cid:2) /I(cid:2) , a contradiction. So each I /S I must be
12 11 22 21 i i i
torsion. Butasintheproof of(2)⇒(3)of Theorem7 thisleads toacontradiction.
(3)⇒(2) Suppose that there exist left ideals I (cid:1) I with S I ⊆ I and an
i1 i2 i i1 i1
isomorphism f:I /I → I /I . Now I /S I is torsion, so its quotient group
12 11 22 21 i2 i i2
I /I is also torsion. Moreover, I /I and I /I inherit the property that for
i2 i1 12 11 22 12
g ∈I /I , we have gcd(o(g ), o(g ))=1. But if 0(cid:9)=g ∈I /I and g =f(g ),
i i2 i1 1 2 1 12 11 2 1
then o(g )=o(g ), so gcd(o(g ),o(g ))(cid:9)=1, a contradiction. (cid:2)
1 2 1 2
We end this section by stating a module version of Goursat’s Lemma.
Theorem 14. Let R be a ring and A and A R-modules.
1 2
(1)Let A be an R-submodule of A ×A . Let A = {a ∈ A |(a,0) ∈ A}, A =
1 2 11 1 21
{b ∈ A |(0,b) ∈ A}, A = {a ∈ A |(a,b) ∈ A for some b ∈ A}, and A = {b ∈
2 12 1 22
A |(a,b) ∈ A for some a ∈ A }. Then A ⊆ A are R-submodules of A and the
2 1 i1 i2 i
map f :A /A →A /A given by f (a+A )=b+A where (a,b)∈A is an
A 12 11 22 21 A 11 21
R-module isomorphism.
(2)Let A ⊆ A be R-submodules of A with an R-module isomorphism
i1 i2 i
f:A /A → A /A . Then A = {(a,b) ∈ A ×A |f(a +A ) = b+A }
12 11 22 21 12 22 11 21
is an R-submodule of A ×A .
1 2
(3)The constructions given in (1)and (2)are inverses to each other.