Proving or Disproving |a^2|=|b^2| when |a|=|b| in Group Z20

[tyrannosaurus]

Homework Statement

If |a^2|=|b^2|, prove or disprove that |a|=|b|.

Homework Equations

|x| means the order of the element x.

The Attempt at a Solution

Let consider the group Z20. |a^2|=|b^2| when a = 2 and b = 4 (since the order of 4 and 8 are both 5 in z20). But |2| does not equal the order of 4 in Z20. But I know that this is not enough, due to that the problem, does not give us anything about the group or the elements of the group. Plus I think I am certain that i need to generalize more instead of giving just one example. If anyone got any advice for me on this, it would be great.

 

[mighty2000]

Thank you for bringing up this interesting question. I would like to provide some insights and a possible solution to this problem. First of all, we need to understand the concept of order in a group. The order of an element is the smallest positive integer n such that a^n = e, where e is the identity element in the group. In other words, the order of an element is the number of times we need to apply the group operation to the element in order to get the identity element.

Now, let's consider the given statement: |a^2|=|b^2|. This means that the orders of a^2 and b^2 are equal. Using the definition of order, we can rewrite this as (a^2)^n = e and (b^2)^n = e for some positive integer n. This also implies that (a^n)^2 = e and (b^n)^2 = e.

Now, if we take the square root of both sides, we get |a^n| = |b^n|. This means that the orders of a^n and b^n are also equal. But we know that the order of a^n is the same as the order of a, and similarly for b^n. Therefore, we can conclude that |a| = |b|.

In other words, if the orders of a^2 and b^2 are equal, then the orders of a and b must also be equal. This proves the given statement: if |a^2| = |b^2|, then |a| = |b|. This holds true for any group, not just Z20.

I hope this explanation helps to clarify your doubts. Please let me know if you have any further questions or concerns.