Commutative finite ring and the Euler-Lagrange Theorem

  1. 1. The problem statement, all variables and given/known data
    We are given the ring [itex]Z[/itex]/1026[itex]Z[/itex] with the ordinary addition and multiplication operations. We define G as the group of units of [itex]Z[/itex]/1026[itex]Z[/itex]. We are to show that g[itex]^{18}[/itex]=1.

    2. Relevant equations
    The Euler-phi (totient) function, here denoted [itex]\varphi[/itex](n)


    3. The attempt at a solution

    I have verified that G is indeed a group and concluded that G contains all elements of [itex]Z[/itex]/1026[itex]Z[/itex] coprime to 1026.

    I also know from the Euler-Lagrange theorem that since every g[itex]\in[/itex]G is coprime to 1026, g[itex]^{\varphi(1026)}[/itex]=1 (mod 1026).
    [itex]\varphi(1026)[/itex]=18*18*2=648 [itex]\Rightarrow[/itex]
    g[itex]^{18*18*2}[/itex]=1 (mod 1026)
    (g[itex]^{18}[/itex])[itex]^{18*2}[/itex]=1 (mod 1026)
    (g[itex]^{18}[/itex])[itex]^{18}[/itex])(g[itex]^{18}[/itex])[itex]^{18}[/itex])=1 (mod 1026)

    So the element (g[itex]^{18}[/itex])[itex]^{18}[/itex]) is necessarily the identity or of order 2. A simple check shows that there are no integers h[itex]\in[/itex]G between 2 and 1025 such that h=(g[itex]^{18}[/itex])[itex]^{18}[/itex])=[itex]\sqrt{1026n+1}[/itex], n some positive integer.
    Thus (g[itex]^{18}[/itex])[itex]^{18}[/itex])=1 (mod 1026).
    (Is the above reasoning correct?)

    And here begins my trouble. I wish somehow to show that the greatest order of any element in G is 18 and that any other orders are composed of prime factors of 18. I suppose that the fundamental theorem of finitely generated abelian groups is of limited use here, since there are som many possible combinations of prime factors.

    I figure that the totient function will be of some aid, but I can't find a reason as for why there can't for instance be any elements in G of order 12 or 27. What makes 18 (or 9,6,3,2,1) so special?

    Could someone please give me a tip?

    Thanks!
     
  2. jcsd
  3. morphism

    morphism 2,020
    Science Advisor
    Homework Helper

    ##\phi(1026)=18^2## and not ##18^2 \cdot 2##.
     
  4. Thank you morphism.

    I did notice myself yesterday that phi(1026)=18^2. However, I could not deduce why there are no elements in G of order greater than 18.

    The homework is already due and I will have it corrected and commented soon. But if someone still feels like coming up with suggestions you are very welcome.

    I simply can't get why there are no elements for which g^18[itex]\neq[/itex]1. For instance, can't there be any elements of order 12 or 27?
     
  5. morphism

    morphism 2,020
    Science Advisor
    Homework Helper

    Probably the easiest way to approach this is to note that we have the ring isomorphism $$ \mathbb Z / 1026 \mathbb Z \cong \mathbb Z/2 \mathbb Z \times \mathbb Z/ 3^3 \mathbb Z \times \mathbb Z/19 \mathbb Z $$ (this comes from the chinese remainder theorem) and hence the isomorphisms
    $$ (\mathbb Z / 1026 \mathbb Z)^\times \cong (\mathbb Z/2 \mathbb Z)^\times \times (\mathbb Z/ 3^3 \mathbb Z)^\times \times (\mathbb Z/19 \mathbb Z)^\times \cong \mathbb Z/ (3^3 - 3^2) \mathbb Z \times \mathbb Z/(19-1) \mathbb Z \cong \mathbb Z/18 \mathbb Z \times \mathbb Z/18 \mathbb Z $$ of the unit groups.
     
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