Abstract Algebra: Properties of the Group U(n)

  1. 1. The problem statement, all variables and given/known data
    (This is an example of a group in my text).

    An integer 'a' has a multiplicative inverse modulo n iff 'a' and 'n' are relatively prime. So for each n > 1, we define U(n) to be the set of all positive integers less than 'n' and relatively prime to 'n'. Then U(n) is a group under multiplication modulo n.

    2. Relevant equations

    3. The attempt at a solution

    1) I am unable to wrap my mind around "An integer 'a' has a multiplicative inverse modulo n iff 'a' and 'n' are relatively prime". I know what "relatively prime" is. I just can't seem to get a grip on this, and could use a nudge in the right direction to fully grasp this group.

    2) I am unsure of the identity for the group. I understand that an identity element 'e' is the element which, when multiplied by any element 'a' in the group = 'a'. So under ordinary multiplication, my identity would be 1. But this group is under multiplication modulo n, and I am unsure if the identity is something other than 1. It is the modular arithmetic component that is buggering me up.
  2. jcsd
  3. Dick

    Dick 25,735
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    1 is still the identity. Take U(9). 2 and 9 are relatively prime. So 2 must have an inverse in U(9). 2*5=10=1 mod 9. So 5 is the inverse of 2. Just try doing some more examples like this until you get used to the idea.
  4. Or better yet, prove it! When I took my first abstract algebra class, one of the problems we did was:

    a) Show that [itex] a \in U(n) [/itex] is a unit if and only if a and n are relatively prime.
    (Hint: Use Bézout's identity (http://en.wikipedia.org/wiki/Bézout's_identity).)

    b) Every element of U(n) is either a unit or a zero-divisor.

    N.B. U(n) is often called [itex](\mathbb{Z}/n\mathbb{Z})^{\times}[/itex] since it is the group of elements of [itex] \mathbb{Z}/n\mathbb{Z}[/itex] which have multiplicative inverses.
  5. I'm not sure if anyone has cleared this up for you, but I will try. As you said, your confusion seems to be the modular arithmetic. If you already know this, then I apologize, but I will go over some modular arithmetic stuff.

    First, what is mod arithmetic? The equation a + b = c (mod n) or a+b (mod n) are examples of equations/statements in modular arithmetic. a+b (mod c) means to normally add a and b, divide by c, and take the remainder. In other words, add a and b normally, then see how far away they are from the last multiple of c.

    Example: 5 + 4 (mod 4) = 5 (mod 4), which is usually written like this:
    [tex] 5 + 4 \equiv 5 (mod 4) [\tex]

    Second, if a = b (mod n) then we say that a is congruent to b, mod n. It is of utmost importance to know that congruence modulo n is an equivalence relation (look up that this means).

  6. Are you sure about that? I think that [itex]\mathbb{Z}_n[/itex] is what you mean. [itex](\mathbb{Z}/n\mathbb{Z})^{\times}[/itex] contains sets, but it is isomorphic to U(n).
  7. SammyS

    SammyS 8,048
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    The number, 1, is considered to be coprime (relatively prime) to every positive integer, so U(n) has an identity element for all n.

    Generate some examples of U(n) for various values of n, to see how things work out.
  8. No, [itex](\mathbb{Z}/n\mathbb{Z})^{\times}[/itex] is what I meant. First of all, the notation [itex] \mathbb{Z}_n[/itex] is ambiguous, since it might also refer to the p-adic numbers: http://en.wikipedia.org/wiki/P-adic_integer#p-adic_expansions.

    Secondly, as you pointed out the two groups are isomorphic, so they are fundamentally the same: talking about [itex] \mathbb{Z}_n[/itex] and [itex](\mathbb{Z}/n\mathbb{Z})^[/itex] as if they were different things is misleading. But more to the point, the notation often used for modular arithmetic is very indicative of a quotient group. [itex] \overline{a} [/itex] is often used to denote the residue of a mod n, and similarly, for any quotient group G/N, the coset a+N is often written as [itex] \overline{a}[/itex] (cf. Dummit and Foote, for example).
    Last edited: Feb 20, 2011
  9. Yeah, I understand what you're saying. Though, there are lots of ambiguous notations. I guess that the point of my comment was to make it clear the the Z/nZ are, in fact, different groups, and while ther are isomorphic the differences should be noted.
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