Determine if binary operation gives a group

**Problem:** Determine whether the binary operation gives a group structure on the given set.

Let be defined on by letting .

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My attempt:

The axioms for a group structure are:

1) Associativity

2) Existence of an identity element (which I will call )

3) Existence of an inverse

For the first axiom - associativity:

Since the entire expression (for each side) is enclosed by absolute value, the inner absolute values serve no purpose and can be removed. Therefore,

and it is shown that is associative.

For axiom 2 - existence of an identity element:

However,

since if is negative , the absolute value of will not equal .

Therefore, the binary operation does not have an identify element and fails axiom two, and so this does not form a group structure.

For axiom three - existence of an inverse:

And therefore an inverse does exist, so axiom three passes.

In conclusion, axiom 1 passes, axiom two fails and axiom three passes. Since the second failed, this is not a group structure.

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Did I do everything correctly? I have the nagging suspicion that I made an error. Any help is greatly appreciated.

Re: Determine if binary operation gives a group

first of all, on , |z| does not mean the absolute value of z, but rather the modulus (distance from the origin) of the complex number z.

so any of your arguments that use "absolute value" are null and void.

it IS true, however, that |zw| = |z||w| (you should PROVE this!), and this fact can be used to show associativity.

i don't think you understand what a group identity IS. it is an element e for which x*e = e*x = x for ALL x.

consider the complex number 1 = 1+0i. what is 1*z ( = |1z|)...? is this equal to z if z = -1? how about if z = √2/2 + i(√2/2)?

even as it stands, your argument on inverses is flawed. inverses must be UNIQUE.

for example, consider the complex number 4 = 4+0i. what is 4*(1/4), and 4*(-1/4)? is 1/4 a unique inverse for 4?

Re: Determine if binary operation gives a group

Okay, so, I still can't find an identity element. How do you go about proving that an identity exists or does not exist? Is it just math intuition, so people who are naturally good at math can just "see" it on observation? Is there some sort of method for those of us who are not so mathematically minded?

Re: Determine if binary operation gives a group

well, look: z*w = |zw| is always real and nonnegative. if z itself is not real and non-negative, than it doesn't matter what "w" is,

|zw| ≠ z. that is, there is NO identity for ANY complex number with any imaginary part, and no identity for any negative real.

so there is no identity. to prove no identity exists, all you have to do is find ONE z for which there is NO e with z*e = z.

we have LOTS of them. for example, suppose z = -1. then z*w = |-w| ≥ 0. since |-w| ≥ 0, it can't ever be -1, no matter what w is.

if we write w = r (cos(t) + i sin(t)), then |-w| = √(-rcos(t))^2 + (-rsin(t))^2) = √(r^2(cos^2(t) + sin^2(t))) = √r^2 = r.

"r" is the radius of w in polar form, it's "distance from the origin" which is 0, if w = 0, and positive otherwise. this can't be negative

so |-w| is nonnegative, but -1 IS negative, thus -1*w ≠ -1. ever.

it's hard to say, when you suspect (in general) that an operation does not have an identity WHICH elements to try.

sometimes the set itself gives you some clues: in the complex numbers, certain sets are "special", the real numbers, the negative reals,

0, and the unit circle all sometimes provide good "counter-examples". if your set that you are doing operations on is a power set

(the set of all subsets of a given set), good sets to test are the empty set, and the set that you are taking the power set of.

if you are considering operations on the integers, -1, 0 and 1 are good places to look for unusual behavior. if you are dealing with

the rational numbers, try -1/2, -1, 0, 1 and 1/2, as first places to look for "test cases"

but there's no "universal algorithm" for proving these kinds of things. you become acquainted with the character of certain structures over time.

and that's part of why you go to school, to benefit from the years of experience your instructor has, and to get "exposure" to the various examples.

and it's part of why self-study is so much harder, it's more hit-and-miss.