56 I The Expanding Universe of Numbers
group. For supposea,b,care distinct elements ofA,letf:A→Abe the bijective
map defined by
f(a)=b, f(b)=a, f(x)=x ifx=a,b,and letg:A→Abe the bijective map defined by
g(a)=c, g(c)=a, g(x)=x ifx=a,c.Thenf◦g=g◦f,since(f◦g)(a)=cand(g◦f)(a)=b.
For arbitrary groups, instead ofφ(a,b)we usually writea·bor simplyab.For
commutative groups, instead ofφ(a,b)we often writea+b.
Since, by the associative law,
(ab)c=a(bc),we will usually dispense with brackets.
We now derive some simple properties possessed by all groups. By (iii) we have
a−^1 a=e. In fact alsoaa−^1 =e. This may be seen by multiplying on the left, by the
inverse ofa−^1 , the relation
a−^1 aa−^1 =ea−^1 =a−^1.By (ii) we haveea=a. It now follows that alsoae=a,since
ae=aa−^1 a=ea.For all elementsa,bof the groupG, the equationax = bhas the solution
x =a−^1 band the equationya=bhas the solutiony =ba−^1. Moreover, these
solutions are unique. For fromax=ax′we obtainx=x′by multiplying on the left
bya−^1 , and fromya=y′awe obtainy=y′by multiplying on the right bya−^1.
In particular, the identity elementeis unique, since it is the solution ofea=a,
and the inversea−^1 ofais unique, since it is the solution ofa−^1 a=e. It follows that
the inverse ofa−^1 isaand the inverse ofabisb−^1 a−^1.
As the preceding argument suggests, in the definition of a group we could have re-
placed left identity and left inverse by right identity and right inverse, i.e. we could have
requiredae=aandaa−^1 =e, instead ofea=aanda−^1 a=e. (However, left iden-
tity and right inverse, or right identity and left inverse, would not give the same result.)
IfH,Kare nonempty subsets of a groupG, we denote byHKthe subset ofG
consisting of all elementshk,whereh∈Handk∈K.IfLis also a nonempty subset
ofG, then evidently
(HK)L=H(KL).A subsetHof a groupGis said to be asubgroupofGif it is a group under the
same group operation asGitself. A nonempty subsetHis a subgroup ofGif and only
ifa,b∈Himpliesab−^1 ∈H. Indeed the necessity of the condition is obvious. It is
also sufficient, since it implies firste=aa−^1 ∈Hand thenb−^1 =eb−^1 ∈H.(The
associative law inHis inherited fromG.)