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Chapter 10: Problem 5

Let \(a\) be any element of a group \(G\). Prove the following: If \(a\) has order \(n\), and \(a^{r}=a^{s}\), then \(n\) is a factor of \(r-s\)

Short Answer

Expert verified
If \(a^r = a^s\), then \(n | (r-s)\), so \(n\) divides \(r-s\).

Step by step solution

01

Understand the Order of an Element

The order of an element \(a\) in a group \(G\) is the smallest positive integer \(n\) such that \(a^n = e\), where \(e\) is the identity element of \(G\). It means raising \(a\) to the power \(n\) returns the identity element.
02

Use Given Equation

We are given that \(a^r = a^s\). To solve for the relationship between \(r\) and \(s\), we manipulate this equation. By multiplying both sides by \((a^s)^{-1}\), we obtain:\[a^r \cdot (a^s)^{-1} = e\].
03

Simplify the Equation

Using the properties of exponents, \(a^r \cdot (a^s)^{-1} = a^{r-s}\). Thus, the equation from Step 2 becomes \(a^{r-s} = e\).
04

Apply Definition of Order

Since \(a^{r-s} = e\) and \(a^n = e\) (with \(n\) as the minimal such integer), it follows from the definition of order that \(n\) divides \(r-s\). This is because \(r-s\) is another integer for which raising \(a\) to its power returns the identity, hence \(r-s\) must be a multiple of \(n\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Order of an Element
The order of an element in group theory is a fundamental concept that defines the "cyclic pattern" of a group element under the operation of the group. For an element \(a\) in a group \(G\), the order is the smallest positive integer \(n\) such that raising \(a\) to the exponent \(n\) results in the identity element of the group, notated as \(a^n = e\). This means:
  • If \(a^n = e\), \(a\) cycles back to the identity after \(n\) operations.
  • If no smaller positive integer achieves this, then \(n\) is indeed the order of \(a\).
  • The identity element \(e\) acts as the 'reset' point, so reaching \(e\) confirms the completion of a full cycle.
This property helps in understanding the structure of the group as it provides insight into how elements interact through repetition.
Exponential Equations in Groups
In group theory, exponential equations are equations where elements are raised to exponents. These are essential for understanding relationships between group elements. Take the equation \(a^r = a^s\) in a group \(G\). This implies a relationship between the exponents \(r\) and \(s\). We solve this by:
  • Recognizing both expressions represent the same group element.
  • Rewriting the equation by canceling common terms through multiplication by inverses, like \(a^r \cdot (a^s)^{-1} = e\).
  • Simplifying as \(a^{r-s} = e\), revealing that raising \(a\) to the power of \(r-s\) results in the identity.
This satisfies the order condition of the group element: \(r-s\) must be a multiple of the order \(n\), since raising \(a\) to a multiple of its order always results in the identity.
Identity Element in Groups
The identity element is a cornerstone of group structure. In any group \(G\), the identity element, denoted as \(e\), is the unique element that leaves other elements unchanged when combined with them. Formally, for any element \(a\) in \(G\), the following holds:
  • \(a \cdot e = a\)
  • \(e \cdot a = a\)
In the context of the original exercise, the identity plays a crucial role.
  • The order of an element is determined by when the element, under repetition of the group operation, returns to the identity \(e\).
  • When an exponent equation like \(a^{r-s} = e\) is established, it confirms that \(r-s\) forms a complete cycle, recognizable by returning to the identity.
This reinforces the pervasive role of the identity element in bridging group operations and the cyclic properties of its elements.

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Most popular questions from this chapter

Let \(a\) denote an element of a group \(G\). Let \(a\) have order \(n .\) Let \(k\) be an integer such that every prime factor of \(k\) is a factor of \(n\). Prove: If \(a\) has a \(k\) th root \(b\), then ord \((b)=n k\).

From elementary arithmetic we know that every integer may be written uniquely as a product of prime numbers. Two integers \(m\) and \(n\) are said to be relatively prime if they have no prime factors in common. (For example, 15 and 8 are relatively prime.) Here is a useful fact: If \(m\) and \(n\) are relatively prime, and \(m\) is a factor of \(n k\), then \(m\) is a factor of \(k\). (Indeed, all the prime factors of \(m\) are factors of \(n k\) but not of \(n\), hence are factors of \(k\).) Let \(a\) be an element of order \(n\) in a group \(G\). Prove the following: If \(a^{m}\) has order \(n\), then \(m\) and \(n\) are relatively prime. [HINT: Assume \(m\) and \(n\) have a common factor \(q>1\), hence we can write \(m=m^{\prime} q\) and \(n=n^{\prime} q\). Explain why \(\left(a^{m}\right)^{n^{\prime}}=e\), and proceed from there.]

In \(\mathbb{Z}_{24}\), list all the elements \((a)\) of order \(2 ;(b)\) of order \(3 ;(c)\) of order \(4 ;(d)\) of order 6

Let \(a\) and \(b\) be elements of a group \(G .\) Let ord \((a)=m\) and \(\operatorname{ord}(b)=n ; \operatorname{lcm}(m, n)\) denotes the least common multiple of \(m\) and \(n .\) Prove the following: If \(a\) and \(b\) commute, then ord \((a b)\) is a divisor of \(\operatorname{lcm}(m, n)\).

Prove that \(a^{m} a^{n}=a^{m+n}\) in the following cases: (i) \(m=0\) (ii) \(m<0\) and \(n>0\) (iii) \(m<0\) and \(n<0\)
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