Let p be a prime number

This problem has been solved! You'll get a detailed solution from a subject matter expert that helps you learn core concepts. See Answer See Answer See Answer done loading Question:Let p be a prime number. Show that the equation x2 = [1]p has just two solutions in Zp. Let p be a prime number. Show that (p − 1)!= -1 modulo p.

The correct option is D 0 Explanation for the correct option: Finding the number of primes in the list n + 1 , n + 2 , n + 3 , . . . . , n + p – 1 is Given: p be a prime number such that p ≥ 3 Also: x = p ! + 1 For 1 ≤ k ≤ p - 1 x + k = p ! + k + 1 2 ≤ k + 1 ≤ p ∵ 1 ≤ k ≤ p - 1 ⇒ x + k = ( 1 , 2 , 3 , . . . . . , k + 1 , . . . p ) + ( k + 1 ) Since there are no prime number in the above equation Hence the correct option is (D)

Let $p$ be a prime number. Prove that $ord_p$ has the following property. $ord_p(ab)=ord_p(a)+ord_p(b)$. (Thus $ord_p$ resembles the logarithm function, since it converts multiplication into addition!) In my book, they describe the order as the exponent of the term in the prime factorization. But how do I use this to proof this statement? Hint $\ \ \overbrace.\ $ If $\ p\nmid c,d\ $ then $\ p\nmid cd\ $ (by $\,p\,$ prime) thus $\ v_p(a)+v_p(b)\, =\, j+k\,=\, v_p(ab)$ Remark $\ $ This can also be deduced directly from the Fundamental Theorem of Arithmetic (existence and uniqueness of prime factorizations). Above is essentially the same but for a single prime $\,p,\,$ i.e. the existence and uniqueness of factorizations of the form $\ n\,p^j\,$ where $\ p\nmid n.$ Note that, by the Fundamental Theorem of Arithmetic, there exists unique prime factorizations for $a$ and $b$. If we let $p_\left(b\right)$, as required.

• العربية • Български • Català • Čeština • Dansk • Deutsch • Español • Euskara • فارسی • Français • 한국어 • Hrvatski • Bahasa Indonesia • Italiano • עברית • Қазақша • Kreyòl ayisyen • Latviešu • Lietuvių • Magyar • Nederlands • 日本語 • Polski • Português • Română • Русский • Simple English • Slovenčina • Slovenščina • کوردی • Suomi • Svenska • ไทย • Українська • Tiếng Việt • 中文 Theorem on prime numbers In Wilson's theorem states that a n> 1 is a n is one less than a multiple of n. That is (using the notations of ( n − 1 ) ! = 1 × 2 × 3 × ⋯ × ( n − 1 ) which shows (n-1)! being congruent to -1 (mod q). But ( n−1)!≡0(mod q) by the fact that q is a term in (n-1)! making (n-1)! a multiple of q. A contradiction is now reached. In fact, more is true. With the sole exception of 4, where 3!=6≡2(mod4), if n is composite then ( n−1)! is congruent to 0 (mod n). The proof is divided into two cases: First, if n can be factored as the product of two unequal numbers, n = ab, where 2≤ a2. But then 2 q< q 2= n, both q and 2 q will be factors of ( n−1)!, and again n divides ( n−1)!. Prime modulus [ ] Elementary proof [ ] The result is trivial when p = 2, so assume p is an odd prime, p ≥ 3. Since the residue classes (mod p) are a field, every non-zero a has a unique multiplicative inverse, a −1. a for which a ≡ a −1 (mod p) are a ≡ ±1 (mod p) (because the congruence a 2 ≡ 1 can have at most two roots (mod p)). Therefore, with the exception of ±1, the factors of ( p − 1)! can be arranged in disjo...

• • • • Question:5. Let p be a prime number. The goal of this problem is to prove that any group G of order p2 is abelian. (a) Let the group G act on itself by the action defined in question 4. Prove that h∈Z(G) if and only if the orbit of h has exactly 1 element. (b) Suppose that ∣Z(G)∣=1. By the orbit stabilizer theorem, the size of the orbit of any element g∈G must divide 5. Let p be a prime number. The goal of this problem is to prove that any group G of order p 2 is abelian. (a) Let the group G act on itself by the action defined in question 4. Prove that h ∈ Z ( G ) if and only if the orbit of h has exactly 1 element. (b) Suppose that ∣ Z ( G ) ∣ = 1. By the orbit stabilizer theorem, the size of the orbit of any element g ∈ G must divide p 2. Use part b and the fact that orbits partition G into disjoint equivalence classes to come to a contradiction. (c) Suppose that ∣ Z ( G ) ∣ = p and let g ∈ G with g ∈ / Z ( G ). Define ⟨ Z ( G ) , g ⟩ to be the group generated by g and every element of Z ( G ). Show that ⟨ Z ( G ) , g ⟩ is abelian. (d) Suppose that ∣ Z ( G ) ∣ = p and let g ∈ G with g ∈ / Z ( G ). Define ⟨ Z ( G ) , g ⟩ to be the group generated by g and every element of Z ( G ). Show that ⟨ Z ( G ) , g ⟩ is all G. (e) Deduce that G is abelian. (f) Give an example of a group with p 3 elements that is not abelian. Previous question Next question

\( \newcommand\) No headers Definition: Prime Numbers - integers greater than \(1\) with exactly \(2\) positive divisors: \(1\) and itself. Let \(n\) be a positive integer greater than \(1\). Then \(n\) is called a prime number if \(n\) has exactly two positive divisors, \(1\) and \(n.\) Composite Numbers - integers greater than 1 which are not prime. Note that: \(1\) is neither prime nor composite. There are infinitely many primes, which was proved by Euclid in 100BC. Theorem \(\PageIndex\). Since \(m \mid n\) and \(1