The latter part of the question asks to find |U(750)|, and suggests using HW3 from Ch3. I'm reasonably sure this is supposed to suggest HW38 from Ch3, which can be found here.
Does anyone know how to do the first part describing U(p^n)? Is p prime or what is the question asking? -Neely
It states at the beginning of the problem that p is a prime.
__ I worked with my guess from the previous homework and applied it to the U(p^n). I tested it with the first two primes (2 and 3). This is what I have: |U(p^n)| = |U(1)|*|U(p^n)| if p is even, n>=0 |U(p^n)| = |U(p)|*|U(p^n-1)| if p is odd, n>=0
I'm not sure if it's right, though. -Kristie
__ I think the order of all the elements of U(p^n) is p
-Matt
Are we supposed to use the guess from problem 38, or the revised guess from problem 40? I'm going to use the one from 38, because that's closest to '3'. -Tim
I think the answer to the first part is that $ U(p^n) $ contains all values less that $ p^n $ except multiples of p.
Finding the order is another problem. I said $ |U(p^n)|=p^n-1-(p-1)=p^n-p $, which makes sense because there will be p-1 multiples of p and you only check to see if all numbers up to $ p^n-1 $ are in the group, so at most the group has $ p^n-1 $ members and you just subtract how many multiples you have. This equation works for $ |U(3^2)| $.
For the last part, Is this true, $ |U(750)|=|U(5^3)||U(2)||U(3)| $? if it is, then $ |U(750)|=|U(5^3)||U(2)||U(3)|=(125-5)(1)(2)=240 $.
Let me know what you guys think?
I think the equation for $ |U(p^n)| $ should be $ |U(p^n)|=(p^n-1)-(n-1)=p^n-n $. I agreed with the above that $ U(p^n) $ contains all values less than $ p^n $ minus the multiples of p. $ p^n-1 $ takes care of the part "all values less than $ p^n $." Suppose n=3, the mulpile of p are $ p^0 $ , $ p^1 $, $ p^2 $, and $ p^3 $; however, $ p^3 $ is not less than itself and was not counted at the beginning, and $ p^0 $=1 has been counted once when finding all values less than $ p^n $. Therefore, the number of multiples of p is n-1.
take U(25), this is $ U(5^2) $. there are more than n-1=2-1=1 multiples of 5. there are 4. 5,10,15,20, so p-1 fits.
I'm not sure if my equation is right, but I think it's
$ |U(p)| * p^(n-1) $
Of course this is assuming that we know |U(p)|. But I think we should be able to assume we know |U(p)| since if we did not know it, we can use our new found generalized equation to calculate what |U(p^1)| is instead of writing the values for it. I tried this for $ U(2^n), U(3^n) and U(5^n) $. The above equations seems to fail if we try for n > 3.
Edit: I just realized that the question assumes p is a prime number. In that case |U(p)| = p-1
