about abstract algebra?

Question

let * be defined on the set of complex numbers by a*b = labl. determine whether the binary operation * gives a group structure on the given set.

please show the properties how you satisfied them.

Answer 1

A group (G, *) is a set G with a binary operation * that satisfies the following four axioms:

* Closure : For all a, b in G, the result of a * b is also in G.
* Associativity: For all a, b and c in G, (a * b) * c = a * (b * c).
* Identity element: There exists an element e in G such that for all a in G, e * a = a * e = a.
* Inverse element: For each a in G, there exists an element b in G such that a * b = b * a = e, where e is an identity element.

Some texts omit the explicit requirement of closure, since the closure of the group follows from the fact that the operation * is a binary operation.

The absolute value of the product of complex numbers is a real number and a real number is in the set of complex numbers. So, there is closure.

Let a = d + ei, b = f + gi, and c = p + qi where d, e, f, g, p, q are real numbers and i is the imaginary unit.

(a*b)*c
= ((d + ei)*(f + gi))*(p + qi)
= | df + dgi + efi - eg |*(p + qi)

If df + dgi + efi - eg > 0, then
( df + dgi + efi - eg )*(p + qi)
= | dfp + dgpi + efpi - egp + dfqi - dgq - efq - egqi |
= | (d + ei)( fp + gpi + fqi - gq) |
= | (d + ei)(f + gi)(p + qi) |
= | abc |
= a*|bc|
= a*(b*c)

So, there is associativity.

The absolute value of the product of complex numbers is a real number. Since not every complex number is a real number, there is NO identity element for all complex numbers under this binary operation.

Answer: No.

Answer 2

Well let's see. Does it meet the 4 requirements of a group:

1. Closure: is |ab| a member of the set. Well |ab| is always real, but that is a special case of complex number, so I would say yes to this

2. Associativity: is a*b = b*a. I'd say so.

3. Identity element: an element I such that any member of the set times I gives the same member. The real number 1 is such an element.

4: Inverse element: for every member of the set, there is another member that when operated on by the set definition results in the identity element. I think so. For a complex number x + iy, the inverse is 1/(x + iy), which is itself a complex number and part of the set. The set operation results in the identity element.

EDIT: maybe math is different from physics, but in quantum theory the state vector components are complex numbers, and real numbers are considered group elements. The question said complex numbers, not imaginary numbers. On the other hand, if we allow a zero imaginary component, then a zero real component should also be allowed, and so should both, in which case zero is included in the set, and there is no inverse component. Therefore the set under the given definition is not a group.

EDIT again: I withdraw my response: only ksoileau has the right answer.

Answer 3

In a group, multiplication of an element by the unique identity returns that element, yet in this case multiplication always returns a real number. Thus no number can operate in this way on the non real complex numbers, hence there is no identity element, hence this is not a group.

Answer 4

No, it doesn´t, even if you exclude zero, because it simply has no unity element: if z is a unity element for C under the operation *, then it must be that for all complex a we´d get a*z = |az| = a, but then:
|az| = a <==> |a||z| = a <==> Sqrt(a.a') |z| = a -- with a' = complex conjugate of a --, and then:

|az| = a <==> |z| = Sqrt(a)/Sqrt(a') = Sqrt(a/a') , and this last expression depends on a.

For example, for a = 1 then z would be such that |z| = Sqr(1/1) = 1, but for a = 1 we'd get Sqrt(i/-i) = Sqrt(-1), which isn't even defined in the reals. (remember! |z| is ALWAYS real and non-negative for all z in C).

Regards
Tonio

Answer 5

You would have to exclude 0 which has no inverse.