EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 10, No. 5, 2017, 967-980
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

On weak graded rings

Najla Al-Subaie1, M. M. Al-Shomrani 2,∗

1 Department of Mathematics, Taif University, Taif, Saudi Arabia
2 Department of Mathematics, King Abdulaziz University, Jeddah, Saudi Arabia

Abstract. In this paper, we investigate some properties of weak graded rings that are rings graded
by a set G of coset representatives for the left action of a subgroup H on a group X. Moreover, a
graded rings by using the product H ×G are also discussed. A detailed example is given.

2010 Mathematics Subject Classifications: 16W50, 13A02, 16D25.

Key Words and Phrases: Weak graded rings, fully graded rings, left coset representatives.

1. Introduction

Let R be an associative ring and G be a semigroup. Recall that R is called G-graded
if there is an additive subgroup Rg of R, for each g ∈ G, such that R =

⊕
g∈GRg and the

inclusion property RgRh ⊆ Rgh is satisfied for all g, h ∈ G.
Semigroup graded rings and modules as well as a lot of their properties were investi-

gated by many mathematicians, see for example [1], [7], [10], [11], [12], [13] and [15].
The construction of group graded rings and their modules were deeply studied by Dade

in [6] that enriched and extended the concept of the classical stable Clifford theory.
The work of Dade was an initial source for many researchers who were interested in

the field of group graded rings and their modules, see for exampl [5], [8] and [9].

In [4], Beggs form a set G of left coset representatives for the left action of a subgroup
H on a group X and defined a binary operation on G which has a left identity and the
right division property. This binary operation is not associative, but the associativity can
be obtained by a ”cocycle” f : G×G −→ H.

In [2], a new concept named the weak graded rings and modules were introduced. In
more details, a graded ring R were constructed using a set G of left coset representatives
and some results were proved in the new setting. In addition, some properties of these
graded rings and their modules were derived.

In this paper, some properties of weak graded rings, that are rings graded by a set G
of coset representatives for the left action of a subgroup H on a group X, are investigated.

∗Corresponding author.

Email addresses: njlalsubaie@gmail.com (N. Al-Subaie), malshomrani@hotmail.com (M. Al-Shomrani)

http://www.ejpam.com 967 c© 2017 EJPAM All rights reserved.



N. Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 10 (5) (2017), 967-980 968

Moreover a graded rings by using the product H ×G are also discussed. A detailed
example is given.

Throughout this paper, for the sake of simplicity, it is assumed that all rings are
associative, commutative and with unities, although for many results associative rings are
only required.

2. Preliminaries

In this section some definitions and required results from [4] are presented.

Definition 1. For a group X and a subgroup H, we call G ⊂ X a set of left coset
representatives if for every x ∈ X there is a unique s ∈ G such that x ∈ Hs. The
decomposition x = us for u ∈ H and s ∈ G is called the unique factorization of x.

In what follows, it will be assumed that G ⊂ X is a fixed set of left coset representatives
for the subgroup H ⊂ X. Also, the identity in X will be denoted by e.

Definition 2. For elements s, t ∈ G we define f(s, t) ∈ H and s ∗ t ∈ G by the unique
factorization st = f(s, t)(s ∗ t) in X, where f is the cocycle map. Also, the functions
. : G × H → H and / : G × H → G are also defined by the unique factorization su =
(s . u)(s / u) for s, s / u ∈ G and u, s . u ∈ H.

It was proved that the binary operation ∗ on G has a unique left identity eG ∈ G and
also has the right division property which means that, there is a unique solution p ∈ G
satisfies the equation p ∗ s = t for all s, t ∈ G. It is noted that if e ∈ G then eG = e is also
a right identity.

Proposition 1. For s, t, p ∈ G and u, v ∈ H, the following identities between (G, ∗) and
f hold:

s . (t . u) = f(s, t)
(
(s ∗ t

)
. u)f

(
s / (t . u), t / u

)−1
(s ∗ t) / u =

(
s / (t . u)

)
∗ (t / u)

s . uv = (s . u)
(
(s / u) . v

)
s / uv = (s / u) / v

f(p, s)f(p ∗ s, t) =
(
p . f(s, t)

)
f
(
p / f(s, t), s ∗ t

)(
p / f(s, t)

)
∗ (s ∗ t) = (p ∗ s) ∗ t.

Proposition 2. For t ∈ G and v ∈ H, the following identities between (G, ∗) and f hold:

eG / v = eG, eG . v = eGve
−1
G , t . e = e, t / e = t,

f(eG, t) = eG, t . e−1G = f
(
t / e−1G , eG

)−1
,

(
t / e−1G

)
∗ eG = t.



N. Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 10 (5) (2017), 967-980 969

3. G-weak Graded Rings

Definition 3. [2] Let X be a group, H be a subgroup of X and (G, ∗) be a fixed set of left
coset representatives for the subgroup H with the binary operation ∗ which is defined as in
2. A ring R is called a G-weak graded ring if

R =
⊕
s∈G

Rs (1)

and
RsRt ⊆ Rs∗t for all s, t ∈ G, (2)

where Rs is an additive subgroup for each s ∈ G. If (2) is replaced by

RsRt = Rs∗t for all s, t ∈ G, (3)

then R is called a fully (or strongly) G-weak graded ring.

It can be noted that any ring R can be put into a G-weak graded ring by placing
R = ReG and Rs = 0 for all eG 6= s ∈ G which is called the trivial G-weak graded ring.

Example 1. Consider the Morita ring

T =

{(
r m
n s

)
: r ∈ R,m ∈M,n ∈ Nand s ∈ S

}
,

with a Morita contex (R,S,RMS , SNR, φ, ϕ) where the bimodule homomorphisms

φ : M ⊗S N −→ R

ϕ : N ⊗R M −→ S

satisfy (mn)m′ = m(nm′) as φ(m,n) = mn and ϕ(n,m) = nm, i.e. φ(m ⊗ n)m′ =
mϕ(n ⊗ m′) and ϕ(n ⊗ m)n′ = nφ(m ⊗ n′) for all m,m′ ∈ M and n, n′ ∈ N . It is
well known that T with the usual matrix addition and multiplication forms a ring. Now

to put the Morita ring T =

(
R M
N S

)
into a weak graded ring we consider the group

X = Z2 × Z3 under addition with the additive subgroup H = {(0, 0), (1, 0)} of X. Take
the set of left coset representatives to be G = {(1, 0), (0, 1), (1, 2)}. Then the ∗ and f
operations as well as the actions /, . are given by the following tables:

Table 1: ∗ and f operations.

∗ (1, 0) (0, 1) (1, 2)

(1, 0) (1, 0) (0, 1) (1, 2)
(0, 1) (0, 1) (1, 2) (1, 0)
(1, 2) (1, 2) (1, 0) (0, 1)

f (1, 0) (0, 1) (1, 2)

(1, 0) (1, 0) (1, 0) (1, 0)
(0, 1) (1, 0) (1, 0) (0, 0)
(1, 2) (1, 0) (0, 0) (0, 0)



N. Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 10 (5) (2017), 967-980 970

Table 2: / and . actions.

s . u (0, 0) (1, 0)

(1, 0) (0, 0) (1, 0)
(0, 1) (0, 0) (1, 0)
(1, 2) (0, 0) (1, 0)

s / u (0, 0) (1, 0)

(1, 0) (1, 0) (1, 0)
(0, 1) (0, 1) (0, 1)
(1, 2) (1, 2) (1, 2)

Hence the ring T can be written as T = T(1,0) ⊕ T(0,1) ⊕ T(1,2) where,

T(1,0) =

(
R 0
0 S

)
=

{(
r 0
0 s

)
: r ∈ R, ands ∈ S

}

T(0,1) =

(
0 M
0 0

)
=

{(
0 m
0 0

)
: m ∈M

}
T(1,2) =

(
0 0
N 0

)
=

{(
0 0
n 0

)
: n ∈ N

}
Next, to ensure that the inclusion property is satisfied, the following calculations are

needed:

(1) T(1,0)T(1,0) ⊆ T(1,0)∗(1,0) as for all

(
r1 0
0 s1

)
,

(
r2 0
0 s2

)
∈ T(1,0) we have :(

r1 0
0 s1

) (
r2 0
0 s2

)
=

(
r1r2 0

0 s1s2

)
∈ T(1,0) = T(1,0)∗(1,0).

(2) T(1,0)T(0,1) ⊆ T(1,0)∗(0,1) as for all

(
r 0
0 s

)
∈ T(1,0) ,

(
0 m
0 0

)
∈ T(0,1) we have :(

r 0
0 s

)(
0 m
0 0

)
=

(
0 rm
0 0

)
∈ T(0,1) = T(1,0)∗(0,1).

It can be noted that rm ∈M as M is a left R-module.

(3) T(1,0)T(1,2) ⊆ T(1,0)∗(1,2) as for all

(
r 0
0 s

)
∈ T(1,0) ,

(
0 0
n 0

)
∈ T(1,2) we have :

T(1,0)T(1,2) =

(
r 0
0 s

)(
0 0
n 0

)
=

(
0 0
sn 0

)
∈ T(1,2) = T(1,0)∗(1,2).

It can be noted that sn ∈ N as N is a left S-module.

(4) T(0,1)T(1,0) ⊆ T(0,1)∗(1,0) for all

(
0 m
0 0

)
∈ T(0,1) ,

(
r 0
0 s

)
∈ T(1,0) we have :

T(0,1)T(1,0) =

(
0 m
0 0

)(
r 0
0 s

)
=

(
0 ms
0 0

)
∈ T(0,1) = T(0,1)∗(1,0).

It can be noted that ms ∈M as M is a right S-module.



N. Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 10 (5) (2017), 967-980 971

(5) T(0,1)T(0,1) ⊆ T(0,1)∗(0,1) as for all

(
0 m1

0 0

)
,

(
0 m2

0 0

)
∈ T(0,1) we have :

T(0,1)T(0,1) =

(
0 m1

0 0

)(
0 m2

0 0

)
=

(
0 0
0 0

)
∈ T(1,2) = T(0,1)∗(0,1).

(6) T(0,1)T(1,2) ⊆ T(0,1)∗(1,2) as for all

(
0 m
0 0

)
∈ T(0,1) ,

(
0 0
n 0

)
∈ T(1,2) we have :

T(0,1)T(1,2) =

(
0 m
0 0

)(
0 0
n 0

)
=

(
mn 0
0 0

)
∈ T(1,0) = T(0,1)∗(1,2).

(7) T(1,2)T(1,0) ⊆ T(1,2)∗(1,0) as for all

(
0 0
n 0

)
∈ T(1,2),

(
r 0
0 s

)
∈ T(1,0) we have :

T(1,2)T(1,0) =

(
0 0
n 0

)(
r 0
0 s

)
=

(
0 0
nr 0

)
∈ T(1,2) = T(1,2)∗(1,0).

It can be noted that nr ∈ N as N is a left R-module

(8) T(1,2)T(0,1) ⊆ T(1,2)∗(0,1) as for all

(
0 0
n 0

)
∈ T(1,2) ,

(
0 m
0 0

)
∈ T(0,1) we have :

T(1,2)T(0,1) =

(
0 0
n 0

)(
0 m
0 0

)
=

(
0 0
0 nm

)
∈ T(1,0) = T(1,2)∗(0,1).

(9) T(1,2)T(1,2) ⊆ T(1,2)∗(1,2) as for all

(
0 0
n1 0

)
,

(
0 0
n2 0

)
∈ T(1,2) we have:

T(1,2)T(1,2) =

(
0 0
n1 0

)(
0 0
n2 0

)
=

(
0 0
0 0

)
∈ T(0,1) = T(1,2)∗(1,2).

Thus, T is a G-weak graded ring. However, it is not a fully (strongly) G-weak graded ring.
For instance, T(1,2)T(1,2) 6= T(1,2)∗(1,2) as T(0,1) = T(1,2)∗(1,2) * T(1,2)T(1,2).

4. Some properties of G-weak Graded Rings

In this section, in the light of [4], some properties of G-weak graded rings are proved.

Proposition 3. Let X be a group, H be a subgroup of X, G ⊂ X be a set of left coset
representatives and R be a G-weak graded ring. Then for any s, t, p ∈ G and u, v ∈ H, the
following properties are satisfied:

(i) Rs.(t.u) = R
f(s,t)

(
(s∗t).u

)
f
(
s/(t.u),t/u

)−1.

(ii) R(s∗t)/u = R(
s/(t.u)

)
∗(t/u)

.

(iii) Rs.uv = R
(s.u)

(
(s/u).v

).



N. Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 10 (5) (2017), 967-980 972

(iv) Rs/uv = R(s/u)/v.

(v) Rf(p,s)f(p∗s,t) = R(
p.f(s,t)

)
f
(
p/f(s,t),s∗t

).
(vi) R(

p/f(s,t)
)
∗(s∗t)

= R(p∗s)∗t.

Proof. The associativity of X implies that R(st)u = Rs(tu) which is used to prove
relations (i) and (ii) as follows:

R(st)u = Rf(s,t)(s∗t)u = R
f(s,t)

(
(s∗t).u

)(
(s∗t)/u

).
On the other hand,

Rs(tu) = Rs(t.u)(t/u)

= R(
s.(t.u)

)(
s/(t.u)

)
(t/u)

= R(
s.(t.u)

)
f
(
s/(t.u),t/u

)(
s/(t.u)∗(t/u)

).
Thus,

R
f(s,t)

(
(s∗t).u

)(
(s∗t)/u

) = R(
s.(t.u)

)
f
(
s/(t.u),t/u

)(
s/(t.u)∗(t/u)

).
As the factorization is unique, we get:

R(
s.(t.u)

)
f
(
s/(t.u),t/u

) = R
f(s,t)

(
(s∗t).u

),
or equivalently,

Rs.(t.u) = R
f(s,t)

(
(s∗t).u

)
f
(
s/(t.u),t/u

)−1 .

Also,
R(

(s∗t)/u
) = R(

s/(t.u)∗(t/u)
).

Next, to prove relations (iii) and (iv), we consider Rs(uv) = R(su)v, which is true by
the associativity of X, as follows:

Rs(uv) = R(s.uv)(s/uv).

On the other hand,

R(su)v = R(s.u)(s/u)v = R
(s.u)

(
(s/u).v

)(
(s/u)/v

).
Thus,

R(s.uv)(s/uv) = R
(s.u)

(
(s/u).v

)(
(s/u)/v

).
Again, by the uniqueness of factorization, we get:

R(s.uv) = R
(s.u)

(
(s/u).v

),



N. Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 10 (5) (2017), 967-980 973

and,
R(s/uv) = R(s/u)/v.

Finally, as before, the associativity of X yields Rp(st) = R(ps)t which is used to prove
relations (v) and (vi) as follows:

Rp(st) = Rpf(s,t)(s∗t)

= R(
p.f(s,t)

)(
p/f(s,t)

)
(s∗t)

= R(
p.f(s,t)

)
f
(
p/f(s,t),(s∗t)

)(
(p/f(s,t))∗(s∗t)

).
On the other hand,

R(ps)t = Rf(p,s)(p∗s)t = R
f(p,s)f

(
(p∗s),t

)(
(p∗s)∗t

).
Thus,

R
f(p,s)f

(
(p∗s),t

)(
(p∗s)∗t

) = R(
p.f(s,t)

)
f
(
p/f(s,t),(s∗t)

)(
(p/f(s,t))∗(s∗t)

).
The uniqueness of factorization yields:

R
f(p,s)f

(
(p∗s),t

) = R(
p.f(s,t)

)
f
(
p/f(s,t),(s∗t)

),
and

R(
(p∗s)∗t

) = R(
p/f(s,t)

)
∗(s∗t)

.

Proposition 4. Let X be a group, H be a subgroup of X, G ⊂ X be a set of left coset
representatives and R be a G-weak graded ring. Then for any t ∈ G and v ∈ H, the
following properties are satisfied:

(i) ReG/v = ReG , and ReG.v = ReGve−1
G
.

(ii) Rt.e = Re, and Rt/e = Rt.

(iii) Rf(eG,t) = ReG.

(iv) Rt.e−1
G

= R
f
(
t/e−1

G ,eG

)−1 , and R(
t/e−1

G

)
∗eG

= Rt.

Proof. To prove (i) we consider ReGv = R(eGv)e−1
G eG

= R(eGve−1
G )eG

, by the associativity

of X, where eG ∈ G and eGve
−1
G ∈ H. But ReGv = R(eG.v)(eG/v). As the factorization is

unique we get:
ReG/v = ReG , and ReG.v = ReGve−1

G
.

Now for (ii), as X is a group, we have Rte = Ret where e ∈ H is the identity and t ∈ G.
Also, Rt = Rte = R(t.e)(t/e) which can be written as Ret = R(t.e)(t/e). The uniqueness of
factorization implies

Re = R(t.e), and Rt = R(t/e).



N. Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 10 (5) (2017), 967-980 974

Next, for (iii) consider ReGt = Rf(eG,t)(eG∗t) = Rf(eG,t)t which is true by the definitions
of ∗ and f . Hence, the uniqueness of factorization gives

ReG = Rf(eG,t).

Finally, to prove (iv) we consider:

Ret = Rt = Rte−1
G eG

= R(t.e−1
G )(t/e−1

G )eG
= R

(t.e−1
G )f

(
(t/e−1

G ),eG

)(
(t/e−1

G )∗eG
),

which implies, Re = R
(t.e−1

G )f
(
(t/e−1

G ),eG

), or equivalently,

R(t.e−1
G ) = R

f
(
(t/e−1

G ),eG

)−1 ,

and,
Rt = R(t/e−1

G )∗eG .

Example 2. Consider the Morita ring T =

(
R M
N S

)
mentioned in example 1 with

X = Z2×Z3, H = {(0, 0), (1, 0)} and a set of left coset representatives G = {(1, 0), (0, 1), (1, 2)}.
As before T = T(1,0) ⊕ T(0,1) ⊕ T(1,2) is a G-weak graded ring where,

T(1,0) =

(
R 0
0 S

)
=

{(
r 0
0 s

)
: r ∈ R, and s ∈ S

}
,

T(0,1) =

(
0 M
0 0

)
=

{(
0 m
0 0

)
: m ∈M

}
and

T(1,2) =

(
0 0
N 0

)
=

{(
0 0
n 0

)
: n ∈ N

}
.

Now put s = (0, 1), t = (1, 2), p = (1, 0) in G and u = (0, 0), v = (1, 0) in H, then
the above properties can be illustrated one by one as follows:

(i) T(s∗t)/u = T(
s/(t.u)

)
∗(t/u)

.

We start with the left hand side as follows :

T(s∗t)/u = T(
(0,1)∗(1,2)

)
/(0,0)

= T(1,0)/(0,0) = T(1,0).

On the other hand,

T(
s/(t.u)

)
∗(t/u)

= T(
(0,1)/

(
(1,2).(0,0)

))
∗
(
(1,2)/(0,0)

)
= T(

(0,1)/(0,0)
)
∗(12)

= T(0,1)∗(1,2) = T(1,0).

Hence, the equality is satisfied.



N. Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 10 (5) (2017), 967-980 975

(ii) Ts.(t.u) = T
f(s,t)

(
(s∗t).u

)
f
(
s/(t.u),t/u

)−1.

We start with the left hand side as follows :

Ts.(t.u) = T
(0,1).

(
(1,2).(0,0)

) = T(0,1).(0,0) = T(0,0).

On the other hand,

T
f(s,t)

(
(s∗t).u

)
f
(
s/(t.u),t/u

)−1 = T
f
(
(0,1),(1,2)

)((
(0,1)∗(1,2)

)
.(0,0)

)
f
(
(0,1)/

(
(1,2).(0,0)

)
, (1,2)/(0,0)

)−1

= T
(0,0)

(
(1,0).(0,0)

)
f
(
(0,1)/(0,0) (1,2)

)−1

= T
(0,0)f

(
(0,1) (1,2)

)−1

= T
(0,0)

(
(1,0).(0,0)

)
f
(
(0,1)/(0,0) (1,2)

)−1

= T
f
(
(0,1) , (1,2)

)−1 = T(0,0)−1 = T(0,0).

Hence, the equality is satisfied.

(iii) Ts.uv = T
(s.u)

(
(s/u).v

).
We start with the left hand side as follows :

Ts.uv = T
(0,1).

(
(0,0)+(1,0)

) = T(0,1).(1,0) = T(1,0).

On the other hand,

T
(s.u)

(
(s/u).v

) = T(
(0,1).(0,0)

)((
(0,1)/(0,0)

)
.(1,0)

)
= T

(0,0)
(
(0,1)/(1,0)

) = T(0,0)+(1,0) = T(1,0).

Hence, the equality is satisfied.

(iv) Ts/uv = T(s/u)/v.
We start with the left hand side as follows :

Ts/uv = T
(0,1)/

(
(0,0)+(1,0)

) = T(0,1)/(1,0) = T(0,1).

On the other hand,

T(s/u)/v = T(
(0,1)/(0,0)

)
/(1,0)

= T(0,1)/(1,0) = T(0,1).

Hence, the equality is satisfied.

(v) Tf(p,s)f(p∗s,t) = T(
p.f(s,t)

)
f
(
p/f(s,t),s∗t

).
We start with the left hand side as follows :

Tf(p,s)f(p∗s,t) = T
f
(
(1,0),(0,1)

)
f
(
(1,0)∗(0,1),(1,2)

) = T
(1,0)f

(
(0,1),(1,2)

) = T(1,0)(0,0) = T(1,0).



N. Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 10 (5) (2017), 967-980 976

On the other hand,

T(
p.f(s,t)

)
f
(
p/f(s,t),s∗t

) = T(
(1,0).f

(
(0,1),(1,2)

))
f
(
(1,0)/f

(
(0,1),(1,2)

)
,(0,1)∗(1,2)

)
= T(

(1,0).(0,0)
)
f
(
(1,0)/(0,0),(1,0)

)
= T

(0,0)f
(
(1,0),(1,0)

) = T(1,0).

Hence, the equality is satisfied.

(vi) T(
p/f(s,t)

)
∗(s∗t)

= T(p∗s)∗t.

We start with the left hand side as follows :

T(
p/f(s,t)

)
∗(s∗t)

= T(
(1,0)/f

(
(0,1),(1,2)

))
∗
(
(0,1)∗(1,2)

)
= T(

(1,0)/(0,0)
)
∗(1,0)

= T(1,0)∗(1,0) = T(1,0).

On the other hand,

T(p∗s)∗t = T(
(1,0)∗(0,1)

)
∗(1,2)

= T(0,1)∗(1,2) = T(1,0).

Hence, the equality is satisfied.

In addition, to illustrate the proposition in 4, let t = (1, 2) ∈ G and v = (0, 0) ∈ H where
eG = (1, 0) ∈ H ∩G. Then

(i) TeG/v = T(1,0)/(0,0) = T(1,0) = TeG, as required.

(ii) Tt/e = T(1,2)/e = T(1,2) = Tt, as required.

(iii) T(t/e−1
G )∗eG = T(

(1,2)/(1,0)−1
)
∗(1,0)

= T(1,2)∗(1,0) = T(1,2) = Tt, as required.

5. Grading by H ×G

Definition 4. Let X be a group, H be a subgroup of X and (G, ∗) be a fixed set of left
coset representatives for the subgroup H with the binary operation ∗ which is defined as in
2. A ring R is called a H ×G-graded ring if

R =
⊕

(u,s)∈H×G

R(u,s) (4)

and

R(u,s)R(v,t) ⊆ R(u,s)(v,t) for all (u, s), (v, t) ∈ H ×G, where u, v ∈ H and s, t ∈ G, (5)

where R(u,s) is an additive subgroup for each (u, s) ∈ H ×G. If (5) is replaced by

R(u,s)R(v,t) = R(u,s)(v,t) for all (u, s), (v, t) ∈ H ×G, where u, v ∈ H and s, t ∈ G, (6)

then R is called a fully (or strongly) H ×G-graded ring.



N. Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 10 (5) (2017), 967-980 977

Theorem 1. Let X be a group, H be a subgroup of X and G ⊂ X be a set of left coset
representatives. Then the equality R(u,s)(v,t) = R(

u(s.v)f((s/v),t),(s/v)∗t
) for s, t ∈ G and

u, v ∈ H, makes R into a H ×G-graded ring where the functions . : G × H → H ,
/ : G×H → G and f : G×G→ H are supposed to satisfy the identities in proposition 1.

Proof. The proof of the theorem follows from the next three lemmas.

Lemma 1. Let X be a group, H be a subgroup of X, G ⊂ X be a set of left coset rep-
resentatives and R(u,s)(v,t) = R(

u(s.v)f((s/v),t),(s/v)∗t
). Then the equality R

(u,s)
(
(v,t)(w,p)

) =

R(
(u,s)(v,t)

)
(w,p)

is satisfied for all (u, s), (v, t) and (w, p) in H × G where the functions

. : G×H → H , / : G×H → G and f : G×G→ H are supposed to satisfy the identities
in proposition 1.

Proof. For s, t, p ∈ G and u, v, w ∈ H, we start with the left hand side as follows:

R
(u,s)

(
(v,t)(w,p)

) = R
(u,s)

(
v(t.w)f

(
(t/w),p

)
,(t/w)∗p

)
= R((

u
(
s.(v(t.w)f((t/w),p))

)
f
(
s/(v(t.w)f((t/w),p)),(t/w)∗p

))
,
(
(s/(v(t.w)f((t/w),p)))∗((t/w)∗p)

)).
(7)

Now, we simplify (7) using the proposition in 1 as follows:

u
(
s . (v(t . w)f ((t / w), p))

)
f
(
s /
(
v(t . w)f ((t / w), p)

)
, (t / w) ∗ p

)
= u

(
s . (v(t . w))

)(
s / (v(t . w))

)
. f
(
(t / w), p

)
f
(
s /
(
v(t . w)f ((t / w), p)

)
, (t / w) ∗ p

)
= u

(
s . (v(t . w))

)(
s / (v(t . w))

)
. f
(
(t / w), p

)
f
((
s / (v(t . w))

)
/ f
(
(t / w), p

)
, (t / w) ∗ p

)
= u

(
s . (v(t . w))

)
f
(
s / (v(t . w)), (t / w)

)
f
(
(s / (v(t . w))) ∗ (t / w), p

)
= u

(
s . (v(t . w))

)
f
((

(s / v) / (t . w)
)
, (t / w)

)
f
((

(s / v) / (t . w)
)
∗ (t / w), p

)
= u

(
s . (v(t . w))

)
f
((

(s / v) / (t . w)
)
, (t / w)

)
f
(
((s / v) ∗ t) / w, p

)
= u

(
(s . v)

(
(s / v) . (t . w)

))
f
((

(s / v) / (t . w)
)
, (t / w)

)
f
(
((s / v) ∗ t) / w, p

)
= u

(
(s . v)f((s / v), t)

(
((s / v) ∗ t) . w

))
f
(
((s / v) ∗ t) / w, p

)
.

Also, we have:(
s /
(
v(t . w)f ((t / w), p)

))
∗
(
(t / w) ∗ p

)
=
((
s / (v(t . w))

)
/ f
(
(t / w), p

))
∗
(
(t / w) ∗ p

)
=
((
s / (v(t . w))

)
∗ (t / w)

)
∗ p

=
((

(s / v) / (t . w)
)
∗ (t / w)

)
∗ p

=
((

(s / v) ∗ t
)
/ w
)
∗ p.

Hence, equation (7) can be rewritten as

R
(u,s)

(
(v,t)(w,p)

) = R
(u,s)

(
v(t.w)f

(
(t/w),p

)
,(t/w)∗p

)
= R((

u(s.v)f((s/v),t)
)(

((s/v)∗t).w
)
f
(
((s/v)∗t)/w,p

)
,
(
((s/v)∗t)/w

)
∗p
). (8)



N. Al-Subaie, M. M. Al-Shomrani / Eur. J. Pure Appl. Math, 10 (5) (2017), 967-980 978

On the other hand,

R(
(u,s)(v,t)

)
(w,p)

= R(
u(s.v)f((s/v),t),(s/v)∗t

)
(w,p)

= R((
u(s.v)f((s/v),t)

)(
((s/v)∗t).w

)
f
(
((s/v)∗t)/w,p

)
,
(
((s/v)∗t)/w

)
∗p
). (9)

Thus, equations (8) and (9) show that the equality is satisfied.

Lemma 2. Let X be a group, H be a subgroup of X, G ⊂ X be a set of left coset
representatives and R(u,s)(v,t) = R(

u(s.v)f((s/v),t),(s/v)∗t
) for all (u, s) and (v, t) in H ×G.

Suppose that there is an element eH ∈ H such that for all s ∈ G and u ∈ H, we have

eG / u = eG , eG . u = eHue
−1
H , s . e = e , s / e = s ,

f(eG, s) = eH , s . e−1H = f(s / e−1H , eG)−1 , (s / e−1H ) ∗ eG = s.

Then R(e−1
H ,eG) is the identity component for RH×G, where eG is the left identity in G and

the functions . : G×H → H , / : G×H → G and f : G×G→ H are supposed to satisfy
the identities in proposition 1.

Proof. We start with

R(u,s)(e−1
H ,eG) = R(

u(s.e−1
H )f(s/e−1

H ,eG), (s/e−1
H )∗eG

)
= R(

uf(s/e−1
H ,eG)−1f(s/e−1

H ,eG), (s/e−1
H )∗eG

)
= R(u, s).

(10)

On the other hand,

R(e−1
H ,eG)(u,s) = R(

e−1
H (eG.u) f(eG/u,s),(eG/u)∗s

)
= R(

e−1
H (eG.u) f(eG, s), eG∗s

)
= R(

e−1
H (eG.u)eH , s

)
= R(

e−1
H (eHue−1

H ) eH , s
)

= R(u, s).

(11)

Thus, equations (10) and (11) show that R(e−1
H ,eG) is the identity component for RH×G.

Equivalently, we can say that (e−1H , eG) is a 2-sided identity of H ×G.

It can be noted that since (G, ∗) has the right division property, i.e., for all s, t ∈ G
there is a unique solution p ∈ G to the equation p∗s = t, then there is a unique left inverse
sL for all s ∈ G by putting t = eG, the left identity in G. Consequently, we can define a
left inverse for all (u, s) in H ×G as we see in the next lemma.



REFERENCES 979

Lemma 3. Let X be a group, H be a subgroup of X, G ⊂ X be a set of left coset represen-
tatives and R(u,s)(v,t) = R(

u(s.v)f((s/v),t),(s/v)∗t
) for all (u, s), (v, t) in H ×G. Suppose that

the properties in lemmas 1 and 2 are satisfied, then H ×G has a left inverse satisfying
the following equality:

R(v,t)L = R(
e−1
H f(tL,t)−1(tL.v−1), tL/v−1

).
Proof. To show that R(v,t)L(v,t) = R(e−1

H ,eG), we start with the left hand side as follows:

R(v,t)L(v,t) = R(
e−1
H f(tL,t)−1(tL.v−1), tL/v−1

)
(v,t)

= R(
e−1
H f(tL,t)−1(tL.v−1)

(
(tL/v−1).v

)
f
(
(tL/v−1)/v,t

)
,
(
(tL/v−1)/v

)
∗t
)

= R(
e−1
H f(tL,t)−1(tL.v−1v)f

(
tL/v−1v,t

)
, (tL/v−1v)∗t

)
= R(

e−1
H f(tL,t)−1(tL.e)f

(
tL/e,t

)
, (tL/e)∗t

)
= R(

e−1
H f(tL,t)−1f(tL,t)

)
, tL∗t

)
= R(

e−1
H , eG

).
As required.

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