Question

Given a group of order \(p q,(p>q)\), with \(S^{q}=1\) and \(T^{p}=\) 1, show that if \(S^{-1} T S=T^{r}\), then \(r^{q} \equiv 1(\bmod p)\).

Short answer

Based on the step-by-step solution provided, answer the following question: Question: Given a group of order pq where p>q, and two elements S and T such that \(S^q=1\) and \(T^p=1\). If \(S^{-1}TS=T^r\), prove that \(r^q \equiv 1 (\bmod p)\). Answer: To prove that \(r^q \equiv 1 (\bmod p)\), we followed the steps outlined in the provided solution, which involved rewriting the given equation, raising both sides to appropriate powers, applying group properties, and using Fermat's Little Theorem. As a result, we successfully derived the congruence \(r^q \equiv 1 (\bmod p)\), completing the proof.

Step-by-step solution

Rewrite Equation

To begin, let's rewrite the given equation \(S^{-1}TS=T^r\) by multiplying both sides by \(S\) on the left: \(STS^{-1} = ST^r\). This will help us get rid of the inverse, and make the equation easier to work with.

Raise Both Sides to the Power q

In this step, we will raise both sides of the equation \(STS^{-1} = ST^r\) to the power of \(q\): \((STS^{-1})^q = (ST^r)^q\).

Apply Binomial Theorem on the Left Side

We can now expand the left side of the equation using the binomial theorem: \((S^qT^qS^{-q}) = (ST^r)^q\). Since \(S^q = 1\), we are left with \(T^q = (ST^r)^q\).

Raise Both Sides to the Power p

In this step, we will raise both sides of the equation \(T^q = (ST^r)^q\) to the power of \(p\): \((T^q)^p = ((ST^r)^q)^p\).

Apply Group Property on the Left Side

Using the group property, we have \((T^{qp}) = ((ST^r)^{qp})\). Given that \(T^p = 1\), we have that \(1 = ((ST^r)^{qp})\).

Apply Conjugation Property on the Right Side

Using the conjugation property, we have that \(1=((S^{-1}T^rS)^{qp})\) and so \((S^{-1}T^rS)^{qp}=1\).

Rewrite Equation

Recall that our original equation was \(S^{-1}TS=T^r\). Now we have \((S^{-1}T^rS)^{qp}=1\). Thus, raising both sides to \(qp\), we get \((T^r)^{qp}=T^{rqp}=1\).

Use Fermat's Little Theorem

Now we apply Fermat's Little Theorem: \(r^q \equiv 1(\bmod p)\). Since \(T^p=1\), by Fermat's Little Theorem, we know that: \((T^{r^q})^p=1\). Comparing this to the equation \(T^{rqp}=1\), we notice that these two equations are equivalent if \(r^q \equiv 1(\bmod p)\). Therefore, we have proven that \(r^q \equiv 1(\bmod p)\).

Key concepts

Fermat's Little Theorem

Fermat's Little Theorem is a foundational principle in number theory, providing insights into powers modulo prime numbers. It states that if \( p \) is a prime number and \( a \) is any integer not divisible by \( p \), then \( a^{p-1} \equiv 1 (\bmod \; p) \). This means, if you raise \( a \) to the power \( p-1 \), and divide it by \( p \), the remainder will be 1.
This theorem is particularly useful in problems involving modulus operations, as it provides a way to verify the equivalency of expressions. In the context of group theory, it assists in understanding the behavior of elements when raised to certain powers.
For the exercise given, Fermat's Little Theorem is applied to conclude that if raising a number \( r \) to the power \( q \) results in 1 when taken modulo \( p \), then \( r^q \equiv 1 (\bmod \; p) \). This step is crucial as it links the behavior of the group elements under transformation and powers.

Binomial Theorem

The Binomial Theorem is a powerful mathematical tool, crucial for expanding expressions raised to any power. It allows us to expand \((x + y)^n\) into a series using combinations, mathematically represented as:

• \( (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \)

Here, \( \binom{n}{k} \) represents the binomial coefficient, indicating the number of ways to choose \( k \) elements from a set of \( n \) elements.
In the context of the original problem, we utilize this theorem to expand group elements raised to powers. Specifically, it helps us simplify terms by splitting powers across elements and subsequently leverage known group properties, like \( S^q = 1 \).
Thus, in the solution, we use this theorem to transform the expression and systematically break it down into more manageable parts, leading us to subsequent simplifications and analytical steps.

Conjugation in Groups

Conjugation in groups revolves around a fundamental operation where an element \( x \) is taken to another element \( x' \) by conjugating it with a group element \( g \). This is represented as \( x' = g^{-1}xg \). This action reveals important structural properties of the group.
Conjugation doesn't alter the order of elements in a group, meaning the structural properties remain intact despite transformations. In simpler terms, elements that are conjugates of each other are similar in both behavior and structure within the group.
For the problem at hand, understanding how conjugation impacts the elements \( S \) and \( T \), and how it alters or preserves their properties, is essential. The equation \( S^{-1}TS = T^r \) uses conjugation to transform \( T \) into another element of the group, offering insights or confirmations about the element \( r \) and its positioning within the group.
This concept supports the solution's emphasis on using conjugation to validate and simplify equivalency relationships within the calculations.