Precursor to Algebra: Herstein shows a way!

Reference: Abstract Algebra, 3rd edition, I. N. Herstein, Prentice Hall International Edition.

Problems:

  1. Let S be a set having an operation * which assigns an element a*b of S for any a,b \in S. Let us assume that the following two rules hold:

i) If a, b are any objects in S, then a*b = a

ii) If a, b are any objects in S, then a*b=b*a

Show that S can have at most one object.

II. Let S be the set of all integers 0, \pm 1, \pm 2, \pm 3, \ldots, \pm n, \ldots. For a, b in S, define * by a*b=a-b. Verify the following:

(i) a*b \neq b*a unless a=b.

(ii) (a*b)*c \neq a*(b*c) in general. Under what conditions on a, b, c is (a*b)*c=a*(b*c)?

(iii) the integer 0 has the property that a*0=a for every a in S.

(iv) For a in S, a*a=0.

III) Let S consist oif f the two objects \square and \triangle. We define the operation * on S by subjecting \square and \triangle to the following conditions:

i) \square * \triangle = \triangle = \triangle * \squarei

ii) \square * \square =\square

iii) \triangle * \triangle=\square

of verify by explicit calculations that if a, b, c are any elements of S (that is, a, b and c can be any of \square or \triangle then

i) a*b is in S

ii) (a*b)*c=a*(b*c)

iii) a*b=b*a

iv) There is a particular a in S such that $la=latex a*b=b*a=b$ for all b in S.

,v) Given b in S, then b*b=a where a is the parituclar element in part “iv” above.

Cheers,

Nalin Pithwa

 

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