5. Decide whether or not each set is a group with respect to the indicated opera

Question

5. Decide whether or not each set is a group with respect to the indicated operation. If it is a group, just say "yes it is a group", and check whether or not the group is Abelian. If it is not a group, find the first of the four group properties (closed, associative, identity, inverses) that does not hold, and show that it does not hold. (a) R -Q under addition. (b) the half-open interval (0, 1] under multiplication. (c) the set of positive divisors of 10 with respect to GCD (d) the set of surjective functions from Z to Z with respect to composition. (e) the set of polynomials in Rr] having constant term 1 under multiplication.

Explanation / Answer

(a) (R-Q , +) is not closed under addition , since, for 1+ 2 and 1-2 both belong to R-Q but their sum is 2 which is a rational,hence does not belong to R-Q. hence (R-Q,+) is not closed under addition and not a group.

(b) ( (0,1] , ) is closed under multiplication ,since for x,y in (0,1] , 0<x<=1 & 0<y<=1 => 0<xy<=1

the product is associative in R and hence being a heriditary property, is associative in (0,1] as well

the multiplicative identity 1 also belongs to (0,1].

(0,1] is abelian since is commutative in R and (0,1] is a subset of R

for x in (0,1] , 0<x<=1 => 0< 1/x <=1 such that x (1/x) = 1

hence ( (0,1] , ) is an abelian group

(c) the set of all positive divisors of 10 is S = {1,2,5,10}

a,b in (S,#) where # is defined by, a#b = gcd(a,b)

clearly S is closed under # , since, 1#2=1, 1#5=1, 1#10=1, 2#5=1, 2#10=2, 5#10=5

again,for the elements in S, it can be easily checked that , a#(b#c) = a#gcd (b,c ) = gcd (a, gcd(b,c) )

and (a#b)#c = gcd ( gcd(a,b) , c) = gcd (a, gcd(b,c) ) = a#(b#c)

hence # is associative,

if S has an identity e then, for all a in S we must have a#e = e#a = a

but gcd(a,e) = a => a divides e for all a in S.

i.e. e must be the lcm of all elements. hence e=10 is the identity of S

now, gcd(a,b) = gcd(b,a) => a#b = b#a => (S,# ) is abelian

now for the existence of inverses, 10 being the identity in S, 1 has no inverse in S, since there is no k in S such that gcd(k,1) = 10 ,since, gcd(k,1)=1.

hence S is not a group since the inverses of elements does not exist except for 10.

(d) let A be the set of surjective functions from Z to Z under composition of mapping

A is closed under composition since f : Z-->Z and g : Z-->Z are surjective then f g and g f are both surjective.

composition of mapping is associative i.e.,  f (g h) =  (f g) h

the identity mapping id is the identity element of (A, ) such that idf = fid = f

but A being the collection of only surjective functions, its elements have only one sided inverse. hence inverse does not exist in A. hence A is not a group under composition of mappings.

(e) the set of polynomials having constant term 1 is closed under multiplication ( since product of 2 such polynomials also has its constant term 1) , the product is associative, commutative, and the identity element is the polynomial f(x)=1. but the polynomials have no inverses except for the polynomial f(x)=1

hence, it is not a group under multiplication ( but a Ring)

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