Abstract Algebra...please help?

Question

Is U(p^2), for p a prime, cyclic? How do I show this is true or not?

Answer 1

The answer is yes. In fact if g is a generator for Z_p*
(a primitive root of p)
and g^(p-1) .ne. 1(mod p²) then g is also a generator
for U(p²). If g^(p-1) = 1(mod p²) then g+p is a
generator for U(p²).
Here is the proof.
We need some preliminary facts.
I'll let you work them out, but I'll give a hint for the second one.
1). The order of U(p²) = p(p-1) = φ(p²).
2). Suppose a^k = 1(mod p). Then a^(kp) = 1(mod p²).
Proof: a^k = 1+ mp
a^(kp) = (1+mp)^p. Now expand this by the binomial
theorem and you will see that every term after the
first is divisible by p².
3). The order of a primitive root g mod p is p-1.
So by part 2, g^(p)(p-1) = 1(mod p²).
Now we must find the order of g mod p².
The order is a divisor of p(p-1).
Since g^(p-1) .ne. 1(mod p²) it can't be p-1.
It certainly can't be p. Also it can't be a proper
divisor of p-1 because then we would have
g^k = 1(mod p) with k < p-1, contradicting the
fact that g is a primitive root mod p.
So the order of g mod p² is p(p-1) and U(p²) is cyclic.
For the second statement, we have g^(p-1) = 1(mod p²).
Look at (g+p)^(p-1) = 1 + (p-1 C_2)g^(p-2)*p (mod p²)
which is not 1 because the second term is only
divisible by p and not by p².
Whew! Enough theory. How about an example?
Let p = 5. Then we can take g = 2.
Since 2^4 = 16(mod 25), we know that 2 generates
U(25).
Let's compute the powers of 2 mod 25 and show
that we get all the units mod 25.
2 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 7
2^6 = 14
2^7 = 3
2^8 = 6
2^9 = 12
2^10 = 24 = -1
2^11 = 23
2^12 = 21
2^13 = 17
2^14 = 9
2^15 = 18
2^16 = 11
2^17 = 22
2^18 = 19
2^19 = 13
2^20 = 1.
So 2 is indeed a generator for U(25).
Hope that helps!