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Question: prove that the set of all Rn (relatively prime with respect to any n) is a group

... there is a theorem that states Rn is a group for n > 0, but i dont know where that came from...

(Just to be clear, for example, the set of all Rn and n = 15 is {1,2,4,7,8,11,13,14}).

i found out the definition of a group. a set A, with an operation * is a group if:

  • CLOSURE: For all a, b in A, the result of the operation a * b is also in A.
  • ASSOCIATIVITY: For all a, b and c in A, the equation (a * b) * c = a * (b * c) holds.
  • IDENTITY ELEMENT: There exists an element e in A, such that for all elements a in A, the equation e * a = a * e = a holds.
  • INVERSE ELEMENT: For each a in A, there exists an element b in A such that a * b = b * a = e, where e is the identity element.

What I have done so far: I was able to prove the "identity element" part. since all the elements in Rn are less than n, and 1 is always in Rn for any integer n > 0 (because 1 is relatively prime to any integer n >= 1)... and any element in the set of Rn multiplied by 1 will cause no changes, thus the equation e * a = a * e = a, where e = 1, holds and proves the identity element part...

but im having trouble trying to prove the other properties... i have an idea that for multiplying 2 or more numbers, associativity always holds, but im not sure if that proof is enough? am i even in the right track?

JennyMelody
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2 Answers2

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You haven't said what the multiplication is. Conventionally in this context one reduces modulo $n$; thus if $n=15$ and one multiplies $11$ by $13$, one gets the remainder on division of $1\times13=143$ by $15$, and this is $8$, since $143=15\times 9 + 8$. If you're convinced that a number cannot have more than one prime factorization, then you conclude that when you multiply two numbers that have no prime factors in common with $15$, you get a product that has no prime factors in common with $15$. The set of all common factors of $15$ and $x$ is the same as the set of all common factors of $15$ and the remainder on division of $x$ by $15$ (that's a routine algebra exercise). Thus the only common factor between $15$ and the remainder is $1$, so you've got a number in the set of those less than $15$ that are coprime to $15$. That proves closure.

You're on the right track with associativity.

As for inverses, that's more involved. See this question. (That question assumes the modulus is prime, so not everything is exactly the same.)

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Michael Hardy
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I'd like to add an alternate proof, which basically involves proving that each element of the set maps to a different element, under multiplication mod n.

For this, we show two things (associativity already follows from the associativity of multiplication):

(a.) If $a, b \in \mathbb{U}_n \implies ab \in \mathbb{U}_n$. This follows since if $a$ and $b$ have no common divisors with $n$, then $ab$ can't have a common divisor either, because if $ab$ has a common divisor, then it has atleast one common prime divisor $p | ab, p| n$. Thus, either $p | a$ or $p | b$, which means that either $a$ or $b$ has a common divisor $p$ with n, which is a contradiction.

(b.) $a, b, c \in \mathbb{U}_n \implies a\cdot c \ne a\cdot b \;\;\text{if}\;\; b \ne c$. This follows because if $$a\cdot b = a\cdot c \implies a\cdot b \equiv a\cdot c\mod{n} \implies n \mid ab - ac$$ $\implies n | a(b - c) \implies n | b - c,$ since $gcd(a, n) = 1$

Since $b, c < n \;\;\text{and}\;\;b \ne c$, this is not possible, and so, $a\cdot b \ne a\cdot c$.

Thus, each element in the set $\mathbb{U}_n$ maps to a unique element under mult. mod n, and hence there is some element that maps to 1, since $1 \in \mathbb{U}_n$. So, there exists an inverse of each of the elements in $\mathbb{U}_n$. This also proves that $\mathbb{U}_n$ is a group under multiplication mod n.

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