Commutative finite ring and the Euler-Lagrange Theorem


by Hugheberdt
Tags: commutative, eulerlagrange, finite, ring, theorem
Hugheberdt
Hugheberdt is offline
#1
Mar26-12, 10:25 AM
P: 6
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!
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morphism
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#2
Mar28-12, 09:30 PM
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##\phi(1026)=18^2## and not ##18^2 \cdot 2##.
Hugheberdt
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#3
Mar29-12, 09:54 AM
P: 6
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?

morphism
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#4
Mar29-12, 06:58 PM
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Commutative finite ring and the Euler-Lagrange Theorem


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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