UNIT-6
Algebraic structures.
Q 1: Let S=
. Define * on S by a*b =a+ b+ a b
(a) Show that * is a binary operation on S.
(b) Show that [S, *
is a group.
(c) Find the solution of the equation 2*x*3= 7 in S
Solution: (a) We must show that S is closed under *, that is , that a+b+ab
for a,b
.Now a+b+ab=-1 if and only if 0=ab+a+b+1=(a+1)(b+1).This is the case if and only if either a=-1 or b = -1 , which is not the case for a,b
.
(b) We have a*(b*c) = a*(b+c+bc) = a+(b+c+bc)+a(b+c+bc)=a+b+c+ab+ac+bc+abc and (a*b)*c = (a+b+ab)*c =(a+b+ab)+c+(a+b+ab) = a+b+c+ab+ac+bc+abc. Thus * is associative. Note that 0 acts as identity element for *, since 0*a =a*0=a.
Also, 
acts as inverse of a , for a* 
Thus [s, *] is a group.
(c)Because the operation is commutative, 2*x*3=2*3*x=11*x. Now the inverse of 11 is 
,by part(b). from 11*x=7, we obtain
Q 2: Let G be a group with a finite number of elements. Show that for any 
Solution: Let G has m element, Then the elements e, a,
are not all different, since G has only m elements. If one of e, a,
is e, then we are done. If not, then we must have 
where i<j, Repeated left cancellation of a yields e=
Q 3: Any finite Abelian group is isomorphic to a direct sum of cyclic groups.
Proof: We need more than this, because two different direct sums may be isomorphic. For example, C2 ⊕C3 
C6.
(If a and b are generators of the summands, then 2a = 3b = 0, and successive multiplies of (a, b) are (a, b), (0,2b), (a,0), (0, b), (a,2b), and (0,0).)
There are two standard resolutions of this problem.
(a) An Abelian group is in Smith canonical form if it is written as Cn1 ⊕··· ⊕Cnr, where n1..., nr are integers greater than 1 and ni divides ni+1 for 1 ≤ i ≤ r –
(b) An Abelian group is in prime-power canonical form if it is written as
Cq1 ⊕··· ⊕Cqr, where q1, …., qr are rare prime powers greater than 1.
Q 4: (a) Any finite Abelian group can be written in Smith canonical form. If two groups in Smith canonical form are isomorphic, then the multisets of orders of the cyclic factors are equal.
Proof: To convert Smith into prime-power and vice versa, use the fact that if n =
where 
are powers of distinct prime numbers, then
Cn 
C
⊕··· ⊕C
. Thus, from Smith to prime-power, simply factorise the orders of the cyclic factors. From prime-power to Smith, gather up the largest power of each prime and multiply them; then repeat until nothing remains. For example, the group C4 ⊕C12 ⊕C36 is in Smith canonical form; the group C4 ⊕C4 ⊕C4 ⊕C3 ⊕C9 is in prime-power canonical form; and these two groups are isomorphic.
Q 5: let f and g be two permutation a X. then f=g if only if f(x) =g(x) for all x in X.
Where f = 1 2 3 4 and g = 3 2 1 4
Solution: we re write the values of f and g by using permutation combinations, we get f = 2 3 4 1 and g = 4 3 2 1
Evidently f (1) = 2 = g (1)
f (2) = 3=g (2)
f (3) = 4=g (3)
f (4) = 1=g (4)
thus, f(x) = g(x) for all x E = {1, 2, 3, 4}
which implies that f(x) =g(x).
Q 6: The group Z6 into cosets of the subgroup H = {0, 3}
solution: one coset is {0, 3} itself.
Q 7: Let H be the subgroup 
of 
. Find the partitions of 
.Into left cosets of H, and the partition into right cosets of H. (See the table for the symmetric group 
in the next slide.
Solution: For the partition into left cosets, we have
The partition into right cosets is
Since 
is not abelian, the partition into left cosets of H is different from the partition into right cosets.
Q 8: The following are three equivalent conditions for a subgroup H to be a normal subgroup of a group G.
1.
2.
3.
Proof : (1)
Suppose that H is a subgroup of G such that 
for all g
and all h
Then, 
We claim that actually 
=H.We must show that H
for all g
Let h
. Replacing g by 
in the relation 
, we obtain 
.
Consequently, 
(3)
Suppose that gH=Hg for all g
. Then 
, So 
for all g
And all 
By the preceding paragraph, this means that 
=H for all g
.
(2)
Conversely , if 
for all g
then 
=
gH
But also, 
giving 
,so that hg=g
and Hg
.
Q 9: If f: G
H is a homomorphism, define the kernel of f as
Ker f = 
;
Solution: Suppose that ker f is a normal sub-group of G. For if a 
and b 
ker f, we must show that aba-1 belongs to ker f.
But f (aba-1) = f(a). f (b). f (a-1) = f(a)(1)f(a)-1 = 1.
Conversely, every normal sub-group is the kern el of a homomorphism.
To see this, suppose that N 
G, and let H be the quotient group G/N. Define the map 
: G
G/N by 
(a) = aN;
Then 
is called the natural or canonical map.
Since,
is a homomorphism.
The kernel of 
is the set of all a 
G, such that,
aN = N(=1N), or equivalently, a 
N.
thus, ker
= N.
Q 10: The code with generator matrix:
G= 
Solution: The given generator matrix has code-words a follow,
And it is cyclic because the right shifts have the following impacts
Q 11: Solve the following generator matrix.
Solution: The given generator matrix can be written as,
The elements of this matrix are 2
2 matrices which are endomorphisms.
In this scenario, each code-word can be represented as,
where 
are the generators of G.
Q 12: The cancellation laws hold in a ring R if and only if R has no divisors of 0.
Proof : Assume that R is a ring in which the cancellation laws hold and let ab=0 for some a,b
.
If a
by right cancellation law. Thus, R has no divisors of 0, if the cancellation laws hold in R.
Conversely, suppose that R has no divisors of 0, and suppose that ab=ac, with a
. Since a
and since R has no divisors of 0, we have 0=ab-ac=a(c) 
b-c =0
.In similar way, b a=ca with a
. Thus, if R has no divisors of 0, The cancellation laws hold in R.
Q 13: Every field F is an integral domain.
Proof: Let 
Assume that ab=0.Since 
,the multiplicative we
inverse 
a exists, multiplying the above equation on both sides by 
, weget
his implies, 0=
b=eb=b, which shows that F has
no divisors of 0. Since F is a field, in particular F is a commutative ring with unity, and we showed that F has no divisors of 0. Hence F is an integral domain.
We know that Z is an integral domain, but not a field. We next prove that finite integral domains are fields.
Q 14: Consider the map det of 
into R where det(A) is the determinant of the matrix A for A
a ring homomorphism? Justify your answer.
Solution: Because det(A+B) need not equal det(A)+det(B) , we say that det is not a ring homomorphism .For example ,
