Explore BrainMass

Explore BrainMass

    Check whether the following sets are groups

    Not what you're looking for? Search our solutions OR ask your own Custom question.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    Determine if the following sets with the given operations are groups. Give a proof or counterexample. Please see attached image for further details.

    © BrainMass Inc. brainmass.com December 24, 2021, 9:04 pm ad1c9bdddf
    https://brainmass.com/math/discrete-math/check-whether-the-following-sets-are-groups-340794

    Attachments

    SOLUTION This solution is FREE courtesy of BrainMass!

    To determine whether a set is a group we need to check the following axioms:

    1) Closure under the operation;
    2) Associativity of the operation, that is, for any a,b,c in the set (a.b).c = a.(b.c) (here . is the operation)
    3) Existence of the identity element e, such that for any a a.e = e.a = a
    4) Existence of the inverses, that is for any a there is a^{-1} such that a.a^{-1} = a^{-1}.a = e

    a) Z is closed under subtraction. But subtraction is not associative: (a-b)-c is not a-(b-c). For example, (3-2)-1 = 0, while 3-(2-1) = 2.
    (Z,-) is NOT a group.

    b) G is obviously closed under multiplication. (9n) x (9k) = 9 (9nk); multiplication is associative; but there is no identity. There is no number of the form 9n, with integer n, that would satisfy the property (9n)x(9k) = 9k. G is NOT a group.

    c) An element of G can be written as e^{it} with real t. Then, multiplication e^{it} x e^{is} = e^{i(t+s)} is reduced to addition.
    The set is closed under multiplication; the operation is obviously associative (check!), there is an identity e^{i 0} = e^0 = 1;
    for every e^{it} there is an inverse e^{-it}. G is a group.

    d) The set is obviously closed under the operation (the results are only a, b, c, d, e and f); a is the identity; associativity is extremely hard to check, because we will need to check all possible triples of elements. But notice: (f.b).d = e.d = d, while f.(b.d) = f.a = f. This is NOT a group. Associativity fails.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 9:04 pm ad1c9bdddf>
    https://brainmass.com/math/discrete-math/check-whether-the-following-sets-are-groups-340794

    ADVERTISEMENT