Proving uniqueness of inverse by identity (Groups)

In summary, The conversation discussed finding the identity element, inverse, and other properties of a group. The group is defined on the set of real numbers with a complex multiplication rule. The identity axiom and finding inverses proved to be challenging, but it was suggested to solve for e1 and e2 to find the identity element, and then use it to find the inverse.
  • #1
YABSSOR
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1. Which of the following is a group?
To find the identity element, which in these problems is an ordered pair (e1, e2) of real numbers, solve the equation (a,b)*(e1, e2)=(a,b) for e1 and e2.




2. (a,b)*(c,d)=(ac-bd,ad+bc), on the set ℝxℝ with the origin deleted.



3. The question also asks for you to find the inverse and I think implicitly for associativity and then for commutativity. I've got the other three down, but the identity axiom is giving me trouble.

(a,b)*(a,b)-1=(aa-1-bb-1,ab-1+ba-1)
Unless I'm really screwing up here, I think this implies that e1-e2 is the first value, which would be equivalent to e1. However, I don't know how to resolve the ab-1+ba-1) part.


I think I've been dropped from this class for a lack of prerequisites, but I think I'm still going to try and finish the class nonetheless.
 
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  • #2
YABSSOR said:
1. Which of the following is a group?
To find the identity element, which in these problems is an ordered pair (e1, e2) of real numbers, solve the equation (a,b)*(e1, e2)=(a,b) for e1 and e2.




2. (a,b)*(c,d)=(ac-bd,ad+bc), on the set ℝxℝ with the origin deleted.

That certainly is a complex multiplication rule.

3. The question also asks for you to find the inverse and I think implicitly for associativity and then for commutativity. I've got the other three down, but the identity axiom is giving me trouble.

(a,b)*(a,b)-1=(aa-1-bb-1,ab-1+ba-1)

That won't work; it's not the case here that [itex](a,b)^{-1} = (a^{-1},b^{-1})[/itex] (what inverses did you assign to [itex](1,0)[/itex] and [itex](0, 1)[/itex], both of which are members of this alleged group?).

It is a consequence of the definitions that you cannot assign an inverse to an element until you have found the identity.

As suggested, you need to set [itex](a,b)*(e_1,e_2) = (ae_1 - be_2, ae_2 + be_1) = (a,b)[/itex] and solve for [itex]e_1[/itex] and [itex]e_2[/itex].

Then, to find [itex](a,b)^{-1} = (c,d)[/itex], set [itex](a,b)*(c,d) = (e_1,e_2)[/itex] and solve for [itex]c[/itex] and [itex]d[/itex].
 
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1. What is the inverse of an element in a group?

The inverse of an element in a group is the element that, when combined with the original element using the group's operation, results in the identity element.

2. How can we prove the uniqueness of the inverse in a group?

To prove the uniqueness of the inverse in a group, we can use the identity property, which states that the identity element multiplied by any element results in that element. By using this property and the definition of an inverse, we can show that there can only be one inverse for each element in a group.

3. What is the identity element in a group?

The identity element in a group is the element that, when combined with any other element using the group's operation, results in that same element.

4. Can a group have more than one identity element?

No, a group can only have one identity element. This is because if there were more than one identity element, it would violate the definition of a group, which requires that there be a unique identity element for the group's operation.

5. How does proving the uniqueness of the inverse in a group relate to the group's operation?

The uniqueness of the inverse in a group is directly related to the group's operation, as it is defined as the element that, when combined with another element using the group's operation, results in the identity element. By proving the uniqueness of the inverse, we are also showing that the group's operation is well-defined and follows the properties of a group.

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