Commutative finite ring and the Euler-Lagrange Theorem

[Hugheberdt]

Homework Statement

We are given the ring Z/1026Z with the ordinary addition and multiplication operations. We define G as the group of units of Z/1026Z. We are to show that g^{18}=1.

Homework Equations

The Euler-phi (totient) function, here denoted \varphi(n)

The Attempt at a Solution

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

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

So the element (g^{18})^{18}) is necessarily the identity or of order 2. A simple check shows that there are no integers h\inG between 2 and 1025 such that h=(g^{18})^{18})=\sqrt{1026n+1}, n some positive integer.
Thus (g^{18})^{18})=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!

 

[morphism]

$\phi(1026)=18^2$ and not $18^2 \cdot 2$.

 

[Hugheberdt]

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\neq1. For instance, can't there be any elements of order 12 or 27?

 

[morphism]

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.