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Let $a,b,p\in\mathbb Z$ and $n\in\mathbb N$

(a) Show that the relation defined by $a\sim b$ if $a\equiv b\pmod n$ is an equivalence relation. For the purposes of this exercise use Definition 2.1.3 of $a\equiv b\pmod n$.

(b) Suppose $p$ is prime, and suppose $a\in\mathbb Z_p\setminus\{0\}$. Show that $a^{-1}$ exists. Then show $|\mathbb Z_p^\times|=p-1$.

(c) Find $\mathbb Z^\times_{105}$

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I was just wondering if anyone could help me with part a on this question as i seem to be stuck, would the relation be reflexive, and how would one go about showing this as i haven't yet came across an example like this before.

For part b to show that $a^{-1}$ would you need to show that $gcd(a,p)=1$ (or is there an easier way) and how would you then show that the magnitude of the set is p-1.

I think I have done part c correctly it is just calculating the gcd of the individual elements up to 105 with 105 for example if $gcd(y,105)=1$ then $y$ is an element in the set we are trying to construct thanks for taking time to read this looking forward to the responses.

(from duplicate question):

for showing that the magnitude of the set is p-1 is this just to do with the definition of a prime number as the only number in that set that is not coprime with p is p itself

for part C i got 48 elements in the set but it would take to long to list them here.

Clyde Kertzer
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    Note that $105 = 3\cdot 5\cdot 7$. – Arthur Dec 04 '16 at 18:23
  • 3,5,7 are all prime, which part of the question are you referring to or is this just a general point –  Dec 04 '16 at 18:33
  • Part (c). Have you heard about the Chinese remainder theorem? – Arthur Dec 04 '16 at 18:41
  • Not really sure what "Definition 2.1.3 of $a \equiv b \mod n$ is, but $a \equiv b \mod n \iff a-b \in n \mathbb{Z}$ for some arbitrary integer $n$. – bthmas Dec 04 '16 at 18:42
  • @Arthur yes i have but it hasnt been taught on this course, we have been given the definition for a is invertible mod n and the general definition for a set like it is asking us to find in part c, in this case n=105. –  Dec 04 '16 at 19:25
  • @bthmas $\ $ Correct is $\ a\equiv b \pmod n,\ $ not $\ a\equiv b\mod n\ \ $ – Bill Dubuque Dec 04 '16 at 19:27
  • @bthmas would we then be able to say the relation in part a is symmetric then from your point?, how would we go about showing that it is also reflexive and transitive –  Dec 04 '16 at 19:27
  • @mathdeds Try proving it is symmetric and tell us where get stuck, i.e. you need to prove $,a\sim b,\Rightarrow,b\sim a,,$ i.e. $\ a\equiv b,\Rightarrow, b\equiv a\pmod m\ \ $ – Bill Dubuque Dec 04 '16 at 19:29
  • @BillDubuque we must show that a~b and b~a for (would we write the sign here or (mod n) ) to be symmetric. then if a=b (modn) then n|a-b thus there exists a k in the integers such that a-b=nk which means that a~b then b-a=n(-k) thus b~a and hence it is symmetric?.. is this right? –  Dec 04 '16 at 20:18
  • @BillDubuque for reflexive would we just need to write that a~a as a-a=mn=0 thus n|a-a=0 and hence we have shown it is reflexive? –  Dec 04 '16 at 20:19
  • @mathdeds Yes, that is correct, symmetry follows because $,n\Bbb Z = $ multiples of $n$ are closed under negation, and transitivity follows because $,n\Bbb Z,$ is closed under addition, and reflexivity because $,0\in n\Bbb Z,\ $. Equivalently, all are true because $,n\Bbb Z,$ is closed under subtraction (i.e. is an additive subgroup of $,\Bbb Z,$ by the subgroup test). $\ \ $ – Bill Dubuque Dec 04 '16 at 20:21
  • @BillDubuque thanks for your help i have shown that it is also transitive now thanks again and a stupid question maybe as i have now done the question but what part of it was we showing is an equivalence relation is it the congruent sign or what? and could you help me with part b? –  Dec 04 '16 at 20:25
  • By definition the binary relation $,a\sim b,$ is an equivalence relation if it is reflexive, symmetric and transitive, which you proved. The congruence relation is ternary but if we fix the modulus then we get a binary relation $, a\equiv b,$ (where $n$ is now fixed, so we don't need to explicitly mention it). So the exercise shows that congruence to a fixed modular is an equivalence relation. One should think of equivalence relations as generalized equality relations. – Bill Dubuque Dec 04 '16 at 20:33
  • There are many ways to prove that inverses exist, e.g. see this post. for a handful. You can safely ignore the 3 ridiculous downvotes there (political nonsense). – Bill Dubuque Dec 04 '16 at 20:35
  • regarding my comment above is it true that Every positive integer less than p is relatively prime to p, so φ(p) = p − 1 which is what it is asking me to show? –  Dec 04 '16 at 21:18
  • and since we know p is prime gcd(a,p)=1 since a does not equal 0 or p and a and p share no common factors? thus gcd(a,p)=1 is this a proof? or is it incorrect? and since p has no factors other than 1 and p it's self we get that |pZ|=p-1? –  Dec 04 '16 at 21:38
  • For some basic information about writing math at this site see e.g. here, here, here and here. – Martin Sleziak Dec 08 '16 at 17:34

1 Answers1

1

Let $a,b,p\in\mathbb Z$ and $n\in\mathbb N$

(a) Show that the relation defined by $a\sim b$ if $a\equiv b\pmod n$ is an equivalence relation.

We need to show that the relation is reflexive, symmetric and transitive. I'm using the definition you see on the first line of the symmetric argument.

  • Reflexive $: \\a=a \implies a\equiv a \bmod n \therefore a\sim a \quad \checkmark$

  • Symmetric $: \\ a\equiv b \bmod n \iff \exists k\in \mathbb Z: a = b +kn \\ \implies b=a+(-k)n \implies b\equiv a \bmod n \therefore a\sim b \implies b\sim a \quad \checkmark$

  • Transitive $: \\ a\equiv b \bmod n \text{ and }b\equiv c \bmod n \implies \exists k_1,k_2\in \mathbb Z: a = b +k_1n, b = c +k_2n\\ \implies a=c+(k_1+k_2)n \implies a\equiv c \bmod n \\ \therefore a\sim b \ \land\ b\sim c \implies a\sim c \quad \checkmark$

(b) Suppose $p$ is prime, and suppose $a\in\mathbb Z_p\setminus\{0\}$. Show that $a^{-1}$ exists. Then show $|\mathbb Z_p^\times|=p-1$.

Consider two numbers $b,c \in\mathbb Z_p\setminus\{0\}$ . Then $b\ne c \implies b-c \ne 0 \implies a(b-c) \ne 0$ $ \implies ab-ac \ne 0 \implies ab \ne ac$ . So the $p\mathord{-}1$ members of $\mathbb Z_p\setminus\{0\}$ give $p\mathord{-}1$ different results multiplied to $a$, that is, the full set. In particular one such $d$ must give $ad =1$ so $a$ has an inverse, and since $a$ was a general choice, all members of the set are units and $|\mathbb Z_p^\times|=p-1$

(c) Find $\mathbb Z^\times_{105}$

As above all coprime members of the set will form a closed set under multiplication since factors of $105 = 3\cdot 5\cdot 7$ will not appear by multiplying numbers that do not contain them. There are $(3\mathord{-}1)(5\mathord{-}1)(7\mathord{-}1) = 48$ such numbers $<105$ so $|\mathbb Z^\times_{105}|=48$, consisting of the powers of two, primes $11$ and upwards and $2,4, 8$ times those primes as approriate.

Joffan
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