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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'. | 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 | -Tim | ||
| + | ---- | ||
| + | I think the answer to the first part is that <math>U(p^n)</math> contains all values less that <math>p^n</math> except multiples of p. | ||
| + | |||
| + | Finding the order is another problem. I said <math>|U(p^n)|=p^n-1-(p-1)=p^n-p</math>, which makes sense because there will be p-1 multiples of p and you only check to see if all numbers up to <math>p^n-1</math> are in the group, so at most the group has <math>p^n-1</math> members and you just subtract how many multiples you have. | ||
| + | This equation works for <math>|U(3^2)|</math>. | ||
| + | |||
| + | For the last part, Is this true, <math>|U(750)|=|U(5^3)||U(2)||U(3)|</math>? | ||
| + | if it is, then <math>|U(750)|=|U(5^3)||U(2)||U(3)|=(125-5)(1)(2)=240</math>. | ||
| + | |||
| + | Let me know what you guys think? | ||
Revision as of 18:17, 24 September 2008
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?
