Cyclic set: Difference between generator and unit

[smithnya]

Hello everyone,

I've just begun a lesson on cyclic sets, but I am having problems determining a few concepts. One question will ask me to find the generators and the units of a cyclic set Z₈. I have become confused and realized that I did not understand the difference between a generator and a unit in a cyclic set. Could someone explain such difference?

 

[Deveno]

perhaps you mean cyclic groups, or more generally a cyclic ring?

in Z₈ there are two operations: addition modulo 8, and multiplication modulo 8. now 8 is not prime, and one consequence of this is that Z₈ - {0} is not a group under multiplication.

nevertheless, certain elements of (Z₈)* do indeed have inverses:

(3)(3) = 9 = 1 (mod 8), so 3 is its own inverse in Z₈ (under multiplication mod 8).

the elements that possesses inverses under multiplication modulo 8 are called units, and form a group, the group of units of Z₈ (sometimes denoted U(8)).

one can show that U(8) is NOT cyclic, so it does not have a single generator (it takes at least two).

for certain integers n, U(n) IS cyclic, and one can speak of a generator, or "primitive element".

as far as the ADDITIVE group of Z₈, that is (of course) cyclic, and it just so happens that the generators of Z₈ as a cyclic group (under addition modulo 8) are also the units (elements of U(8)).

there is, in fact, this theorem:

k is a unit of Z_n iff gcd(k,n) = ?

(see if you can guess the answer).

 

[smithnya]

Is an element only a unit if it possesses a multiplicative inverse then?

 

[Deveno]

Is an element only a unit if it possesses a multiplicative inverse then?

in a ring, yes.

ok, the cyclic group Z₈ is {0,1,2,3,4,5,6,7}, where the group operation is addition modulo 8. technically, i should write [k] instead of k, since these are NOT integers, but equivalence classes of integers, but this abuse of notation is common-place.

as a(n additive) group the generators of Z₈ are 1,3,5 and 7 (these are the only elements of additive order 8).

but as a ring, Z₈ is not a domain, because it has zero-divisors: for example, 2 and 4 are zero-divisors, since (2)(4) = 0 (mod 8). this also means 2 and 4 cannot possibly have inverses, because if (for example again) 4 had an inverse a:

((2)(4))a = 0a
2(4a) = 0
2(1) = 0
2 = 0, a contradiction.

this happens precisely because 4 and 8 have a common divisor.