220 CHAPTER 4 Abstract Algebra
on page 208, we see that Gis closed under ◦. Since we already
know that◦is associative on Sym(X) it certainly continues to be
associative forG. Next, the identity permutation of the vertices
of the graph is clearly a graph automorphism. Finally, the same
proposition on page 208 shows that each element g ∈ G has as
inverse, proving thatGis a group.- There is one other group that is well worth mentioning, and is
a multiplicative version of (Zn,+). We start by writing Z∗n =
{ 1 , 2 , 3 , ..., n− 1 } (note, again, that we have dispensed with
writing the brackets ([·])). We would like to consider whether this
is a group relative to multiplication. Consider, for example, the
special case n = 10. Note that despite the fact that 2, 5 ∈ Z∗ 10
we have 2·5 = 0 6∈Z∗ 10. In other wordsZ∗ 10 is not closed under
multiplication and, hence, certainly cannot compose a group.
The problem here is pretty simple. If the integernisnot a prime
number, say,n=n 1 n 2 , where 1< n 1 , n 2 < nthen it’s clear that
whilen 1 , n 2 ∈Z∗nwe haven 1 n 2 = 06∈Z∗n. This says already that
(Z∗n,·) is not a group. Thus, in order for (Z∗n,·) to have any chance
at all of being a group, we must have that n = p, some prime
number. Next, we shall show that ifpis prime, thenZ∗p is closed
under multiplication. This is easy, ifa, b∈Zp*, then neitheranor
bis divisible byp. But thenabis not divisible bypwhich means
that ab 6 = 0 and so, in fact, ab ∈ Z∗p, proving that Z∗p is closed
under multiplication.Next, note that since multiplication is associative inZp, and since
Z∗p⊆Zp we have that multiplication is associative inZ∗p. Clearly
1 ∈Z∗pand is the multiplicative identity. It remains only to show
that every element ofZ∗phas a multiplicative inverse. There are a
number of ways to do this; perhaps the following argument is the
most elementary. Fix an elementa∈Z∗pand consider the elements1 ·a, 2 ·a, 3 ·a, ...,(p−1)·a∈Z∗p.