- #1

Hugheberdt

- 6

- 0

## Homework Statement

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.

## Homework Equations

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

## 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!