SECTION 4.2 Basics of Group Theory 237
bijective. Note that in this case, the inverse mappingf−^1 :H→Gis
also a homomorphism. The argument is as follows: ifh, h′ ∈H then
watch this:
f(f−^1 (h)∗f−^1 (h′)) = f(f−^1 (h))? ff−^1 (h′))
(
sincefis a homo-
morphism
)
= h? h′
Ü
since f and f−^1
are inverse func-
tions
ê
= f(f−^1 (h? h′)) (same reason!)
However, sincef is one-to-one, we infer from the above that
f−^1 (h)∗f−^1 (h′) =f−^1 (h? h′),
i.e., thatf−^1 is a homomorphism.
Before going any further, a few comments about homomorphisms
are needed here. Namely, let G 1 and G 2 be groups (we don’t need
to emphasize the operations here), and assume thate 1 ande 2 are the
identity elements ofG 1 andG 2 , respectively Assume thatf:G 1 →G 2
is a homomorphism. Then,
f(e 1 ) =e 2. This is because f(e 1 )^2 = f(e 1 )f(e 1 ) = f(e 1 e 1 ) =f(e 1 ).
Now multiply both sides byf(e 1 )−^1 and getf(e 1 ) =e 2.
Ifx∈G 1 , thenf(x−^1 ) =f(x)−^1. Note that by what we just proved,
e 2 =f(e 1 ) =f(xx−^1 ) =f(x)f(x−^1 ). Now multiply both sides by
f(x)−^1 and get f(x)−^1 = f(x)−^1 e 2 =
f(x)−^1 (f(x)f(x−^1 )) = (f(x)f(x)−^1 )f(x−^1 ) =e 2 f(x−^1 ) =f(x−^1 ).
Theorem.Let(G,∗)and(H,?)be cyclic groups of the same ordern.
Then(G,∗)and(H,?) are isomorphic.
Proof. Let G have generator x and letH have generator y. Define
the mappingf:G→Hby settingf(xk) =xk, k= 0, 1 , 2 , ..., n−1.
Note thatf is obviously onto. But since |G|=|H|=n it is obvious