Question about Automorphisms

[daon]

I am to prove that the mapping from U(16) (the set of non-negative integers less-than and relatively prime to 16) to itself given by x->x³ is an automorphism.

Well, this isn't very hard. I have listed out each element, cubed them to show it is indeed closed, and have shown it is operation preserving using two cayley tables. I basically showed that the mapping is a permutation so that it must also be bijective... hence it is an isomorphism.

But, is there a better way? If I had been given a much larger number such as U(2502).. listing them would be kind of rediculous.

I ask this because I am supposed to generalize that x->x^k is an automorphism if k is odd.

Thanks,
-Daon

 

[pka]

Please give more information about U(16) or U(N)!
Are these groups? What is the operation?
I am not sure what 13<SUP>3</SUP> would equal.
How is it in U(16)?

 

[daon]

I'm sorry, my professor lead me to believe the group U(n) had standardized notation. Here is everything in more detail:

The group U(n) consists of all positive integers less-than or equal to n that are relatively prime to n.

So, U(16)={1, 3, 5, 7, 9, 11, 13, 15}

It is a group under multiplication modulo n. So in your example [13³]₁₆ = [5]₁₆.

So, again, what I did was create a cayley table of multiplication modulo n for U(16) and showed that it was closed. Second, I made a map of each element to its cube in U(16). I suppose that is enough information to verify that for all x,y in U(16), (xy)³ = x³y³, so it is operation preserving.

And my question is: Is there a way to determine that the mapping for \(\displaystyle x \rightarrow x^k\) where k is odd (for my example when k=3) is an Automorphism? I'm sure there is, as my way does not seem practical at all, and it may not even be correct.

Thanks,
-Daon

 

[pka]

I am is a bit of a rush, so cannot help much.
But I am sure that it has to do with the prime factorization theorem.
Being relatively prime, the third power will not add any more prime factors.

 

[daon]

Thank you pka.

I think it was my job to realize that for every element x in the group, x⁴=e.

Therefore, since x->x and x->x³ are Automorphisms, any odd power larger than 3 can be written as 4r + 1 or 4r+3. And:
..... x^4r+1 = x^4rx = 1*x = x
..... x^4r+3 = x^4rx³ = 1*x³ = x³

which are both automorphisms.

-Daon