Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 0 down vote favorite

This problem is from Herstein's Topics in Algebra.

I have to use the property that if a finite set is closed under an associative product and that both cancellation laws hold in $G$, then $G$ is a group to prove the following:

  1. Non-zero integers modulo $p$, a prime number, form a group under multiplication $\mod p$

  2. Non zero integers relatively prime to $n$, form a group under multiplication $\mod n$.

I am very new to modular arithmetic and group theory, and stuck badly. I am trying to do the above, by assuming some numbers $a= pq + r$, where $q$ is an integer, so $a=r (mod\, p)$ but I don't know how to incorporate the facts prime and relatively prime, to solve the problem.

share|improve this question

1 Answer

up vote 0 down vote accepted

Hint: If $a$ and $b$ are relatively prime and $a$ divides $bc$, then $a$ divides $c$.

share|improve this answer
:Sorry, This is not helping. I have proved closure, and associativity. But I am getting that the above theorem is true wrt the mod of any number, not necessarily prime or relatively prime. I show it below. Let $\bar{a}$ be the congruence class of a modulo p ($a \lt p$). TPT: $\bar{a}\bar{b}=\bar{a}\bar{c}$ implies $\bar{b}=\bar{c}$. I write $\bar{a}=pn_1+a$, and similarly for others, with labels of the n different. I substitute into the given statement, and get $$(pn_1+a)(pn_3+c)=(pn_1+a)(pn_2+b)$$. Rearranging I get $$(c-b)(pn_1+a)=p(n_2-n_3)(pn_1+a)$$Which proves b congruent to c mod p – ramanujan_dirac Jan 11 at 12:24
@ramanujan_dirac Here is a counterexample: Let $p = 4$, $c = 3$, $b = 1$ and $a = 2$. So there must be something wrong with your argument. – Serkan Jan 11 at 15:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.