L = Z5[x]/(x^2+2x+4) is a field and the multiplicative group L* is generated as a cyclic group by s+3, where s is the congruence class of x modulo x^2+2x+4. Prove the following statements about the orders of elements in L*.
(a) s has order 12.
(b) s+1 has order 8.
(c) s+2 has order 12.
(d) 2s has order 6.
I thought this problem wouldn't be so bad but I am having difficulties. Any help would be appreciated!
Addendum which somehow got left out.
For (c), from the list of powers of s, s+2=s^{11}. Since 11 is prime to 12, s+2 has order s, that is 12.
For (d), 2s=s^{10}. So 2s has order 12/gcd(10,12)=6
(a) remember |L*| = 24, so we only need to check orders which are divisors of 24.
s^{2} = -2s - 4 = 3s + 1 ≠ 1, so s does not have order 2.
s^{3} = s(s^{2}) = s(3s + 1) = 3s^{2} + s = 3(3s + 1) + s = 4s + 3 + s = 3 (cubes do not always behave as expected in this strange type of field).
anyway, s^{3} ≠ 1, so s does not have order 3.
s^{4} = (s^{2})^{2} = (3s + 1)^{2} = 4s^{2} + s + 1 = 4(3s + 1) + s + 1 = 2s + 4 + s + 1 = 3s ≠ 1
(hmm....it would have easier to calculate s^{4} = s(s^{3}) = s(3) = 3s. oh well, s doesn't have order 4).
s^{6} = (s^{3})^{2} = 3^{2} = 4 ≠ 1, s does not have order 6.
s^{8} = (s^{4})^{2} = (3s)^{2} = 4s^{2} = 4(3s + 1) = 2s + 4 ≠ 1, s does not have order 8.
s^{12} = (s^{6})^{2} = 4^{2} = 1 <---bingo!
of course you could have calculated ALL the powers of s+3, and made a "discrete log table" (that is: a homomorphism L*-->Z_{24}).
for example, (s+3)^{2} = s^{2} + s + 4 = 3s + 1 + s + 4 = 4s, so:
log_{(s+3)}(4s) = 2 (our logarithm "base" is s+3, and products in L* get turned into sums in Z_{24}).
this would allow you to realize any element of L* as (s+3)^{k} (for k = 0,1,2,...,23)
and then the order of (s+3)^{k} is: 24/gcd(k,24). this puts all the work "up-front" (creating the log table is the most time-consuming), but when finished the order of any element is computed easily:
|g| = 24/gcd(log_{(s+3)}(g),24)