SECTION 4.2 Basics of Group Theory 225
proving thatGis abelian.
Let’s look at a few examples.- The infinite additive group (Z,+) is cyclic, with generator 1. Note,
however, in this context, we wouldn’t write 1nfor powers of 1 as
this notation is suggestive of multiplication and 1n = 1. Rather,
in this additive setting we write
n1 = 1 + 1 + 1 +︸ ︷︷ ···+ 1︸
nterms.
As any integer can be written as a positive or negative multiple
(“multiple” is the additive version of “power”), we conclude that
(Z,+) is an (infinite) cyclic group.- Ifnis a positive integer, then the additive group (Zn,+) is cyclic.
Notice here that we don’t really need any negative multiples of 1
to obtain all ofZn. One easy way to see this is that−1 = (n−1)1
and so if [a]∈Zn, then−[a] =a(n−1)1. - Ifpis prime, then the multiplicative group (Z∗p,·) is cyclic. While
not a deep fact, this is not easy to show using only what we’ve
learned up to this point.^11 As examples, note that Z∗ 5 is cyclic,
with generator 2, as 2^1 = 2, 22 = 4, 23 = 8 = 3,and 2^4 = 16 = 1.
Next,Z∗ 7 is cyclic, with generator 3, as
31 = 3, 32 = 9 = 2, 33 = 6, 34 = 4, 35 = 5, 36 = 1.Note, however, that while 2 is a generator ofZ∗ 5 , it isnota gener-
ator ofZ∗ 7.Related to the above is the following famous unsolved conjec-
ture: that the congruence class of the integer “2” a generator of(^11) Most proofs proceed along the following lines. One argues that ifZ∗pis not cyclic, then there will
have to exist a proper divisorkofp−1 such that every elementxofZ∗psatisfiesxk= 1. However,
this can be interpreted as a polynomial equation of degreekwhich has (p−1)> ksolutions. Since
(Zp,+,·) can be shown to be a “field,” one obtains a contradiction.