## Commutative finite ring and the Euler-Lagrange Theorem

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

2. Relevant equations
The Euler-phi (totient) function, here denoted $\varphi$(n)

3. The attempt at a solution

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

I also know from the Euler-Lagrange theorem that since every g$\in$G 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$\in$G 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!
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 Recognitions: Homework Help Science Advisor ##\phi(1026)=18^2## and not ##18^2 \cdot 2##.
 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$\neq$1. For instance, can't there be any elements of order 12 or 27?

Recognitions:
Homework Help
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.