Proving A^n = e for Finite G with Identity e

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In summary, for any element a belonging to a finite group G with identity e, there exists a positive n such that a^n = e, thus proving that every element of a finite group has finite order. This can be shown by considering the sequence of elements a, a^2, a^3, a^4, ... which are all in G but since G is finite, they must eventually repeat, leading to a^n = a^j for some j, k where j < k, and thus a^(k-j) = e. Setting n = k-j, we have a^n = e and n > 0.
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rsa58
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Homework Statement


Show that if a belongs to G where G is finite with identity e, then there exists a positive n such that a^n = e


Homework Equations





The Attempt at a Solution



Someone tell me if i am reasoning correctly.

Suppose a^n is not equal to e.

By finiteness of G, a^n = b for some b belonging to G. Then b^t = (a^n)^t is not equal to e for any positive t. so G is infinite? i don't know i am stuck.
 
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  • #2
since a is in G, then the elements a, a^2, a^3, a^4, ... are also in G, but G is finite, so a^k = a^j for some j, k, say j < k, so a^(k-j) = e, set n = k - j, so a^n = e and n > 0.

realize what you proved, you proved that every element of a finite group has finite order.
 
  • #3
In your attempt, you assume b^t =/= e for all t, which isn't a great assumption, considering you're supposed to prove that (in this case given b) there IS t for which b^t=e
 

1. What is the definition of A^n?

A^n refers to the nth power of matrix A, calculated by multiplying A by itself n times.

2. What is a finite group with identity e?

A finite group with identity e is a group consisting of a finite number of elements that follows the properties of a group, including having an identity element e that when combined with any other element in the group, results in that element.

3. How do you prove that A^n = e for a finite group with identity e?

To prove that A^n = e for a finite group with identity e, you must show that multiplying A by itself n times results in the identity element e. This can be done through mathematical proof or by providing specific examples of A^n and showing that they all equal e.

4. What is the significance of proving A^n = e for a finite group with identity e?

Proving A^n = e for a finite group with identity e is significant because it shows that the group has a property known as finite order, meaning that the group has a finite number of elements and that raising any element to a certain power will result in the identity element. This is a fundamental concept in group theory and has many real-world applications in fields such as physics and cryptography.

5. Are there any exceptions to the proof of A^n = e for a finite group with identity e?

Yes, there are a few exceptions to this proof. One exception is the group of real numbers under addition, where the identity element is 0 and raising any number to the power of 0 will result in 1, not 0. Another exception is the group of nonzero real numbers under multiplication, where the identity element is 1 and raising any number to the power of 0 will result in that same number, not 1. These exceptions show that the proof of A^n = e for a finite group with identity e is not always applicable to all groups.

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