Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

14
ADVANCED ABSTRACT ALGEBRA
Thus
is a group.
Theorem3:
Theorem.
Let G be a group and let Z (G) be the center of G. If
is cyclic, then G is abelian.
Proof:
we claim
−
we show that g Z G g ⊆ Z G ∀ g ∈G
let x
1
bg bg .
∈Zbg,
G then
= b g= b gc
Θ ∈ bgh
−1
= dg gix = ex = x ∈Zbg
G
−1 ∈ bg∀ ∈ , ∀ ∈ bg
−1 , bg ⊆ bg∀ ∈ .
−1 −1 −1
g xg g x g g g x x Z G
Hence g xg Z G g G x Z G
Therefore g Z G g Z G g G
We can now form a factor group
Let G x / g G
Zbg bg Θ is cyclic
G
Z bg G
=
F H G I K J
Let a, b ∈ G.
To show ab=ba
Z
aZ ab =
G
hence
bg bg n n
c h bg
bg bg m m
c h bg
n
bg
m
= for some ∈bg
aZ G = xZ G = x Z G
and bZ G = xZ G = x Z G , where n, m are integers.
bg
Thus a ∈aZ G a = x y for some y ∈Z G
and b x t t G
Now
= b a
We often use it as: If G is not abelian, then
is not cyclic.