# Cyclic set: Difference between generator and unit

1. Feb 3, 2012

### 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?

2. Feb 3, 2012

### 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 possess 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 Zn iff gcd(k,n) = ?

(see if you can guess the answer).

3. Feb 3, 2012

### smithnya

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

4. Feb 3, 2012

### Deveno

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