SECTION 4.2 Basics of Group Theory 229
(b) If|H|<∞andHis closed under∗, thenH is a subgroup ofG.Proof. Notice first that we don’t have to check the associativity of∗,
as this is already inherited from the “parent” groupG. Now assume
condition (a). SinceHis non-empty, we may choose an elementh∈H.
By condition (a), we know thate=h−^1 h∈H, and soH contains the
identity element ofG (which is therefore also the identity element of
H). Next, given h ∈ H we appeal again to condition (a) to obtain
(sincee∈H)h−^1 =h−^1 ∗e∈H. It follows thatHis a subgroup ofG.
Next, assume condition (b), and leth ∈H. Since H is closed un-
der∗, we conclude that all of the products h, h^2 , h^3 , ...are all inH.
SinceH is a finite set, it is impossible for all of these elements to be
distinct, meaning that there must be powers m < nwith hm = hn.
This implies thate=hn−m∈H, forcingH to contain the identity of
G. Furthermore, the same equation above shows thate=hn−m−^1 ∗h,
wheren−m− 1 ≥0. Therefore,hn−m−^1 ∈Hande=hn−m−^1 ∗himplies
thath−^1 =hn−m−^1 ∈H. Therefore, we have shown that H contains
both the identity and the inverses of all of its elements, forcingHagain
to be a subgroup ofG.
I can’t overestate how useful the above result is!One very easy way to obtain a subgroup of a given group (G,∗) is
start with an elementx∈Gand form the setH ={xk|k∈Z}; that
is, H contains all the positive and negative powers of x. Clearly H
satisfies condition (a) of the above proposition sincexm, xn ∈ H ⇒
x−mxn =xm−n ∈H. ThereforeH is a subgroup ofG; asH is cyclic,
we say thatH is thecyclic subgroup of Ggenerated byx. This
cyclic subgroup generated byxis often denoted〈x〉.
Exercises
- Let Gbe a group, and let H 1 andH 2 be subgroups. Prove that
the intersectionH 1 ∩H 2 is also a subgroup ofG. - Let Gbe a group, and let H 1 andH 2 be subgroups. Prove that
unlessH 1 ⊆H 2 orH 2 ⊆H 1 , thenH 1 ∪H 2 isnota subgroup ofG.