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# Check whether the following sets are groups

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

https://brainmass.com/math/discrete-math/check-whether-the-following-sets-are-groups-340794

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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.

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