shows this relationship. Figure 16.1a represents homomorphic mapping ρ: ℊ ! ℊ
0
between two groups, whereas Fig. 16.1b shows linear transformation (endomorphism) A: V
n
! V
n in a vector space V
n .
Theorem 16.3 Suppose that there exists a homomorphic mapping ρ: ℊ ! ℊ
0 ,
where ℊ and ℊ
0 are groups. Then, a kernel F of ρ is an invariant subgroup of ℊ.
Proof Let k i and k j be any two arbitrarily chosen elements of F . Then,
ρ k i
ð Þ ¼ ρ k j
À Á ¼ e
0 ,
ð16:24Þ
where e
0 is the identity element of ℊ
0
. From (16.21) we have
ρ k i k j
À Á ¼ ρ k i
ð Þρ k j
À Á ¼ e
0
e
0
¼ e
0
:
ð16:25Þ
Therefore, k i k j 2 F . Meanwhile, from (16.20) we have
ρ k
À1
i
À Á ¼ ρ k i
ð Þ
½
Š
À1 ¼ e
0 À1 ¼ e
0
:
ð16:26Þ
Then, k
À1
i 2 F . Thus, F is a subgroup of ℊ.
Next, for
8 g 2 ℊ we have
ρ gk i g
À1
À
Á ¼ ρ g
ð Þρ k i
ð Þρ g
À1
À Á ¼ ρ g
ð Þe
0
ρ g
À1
À Á ¼ e
0
:
ð16:27Þ
Accordingly we have gk i g
À1
2 F . Thus, gF g
À1
⊂ F . Since g is chosen arbitrarily,
replacing it with g
À1 we have g
À1
F g ⊂ F. Multiplying g and g
À1 on both sides from
the left and right, respectively, we get F ⊂ gF g
À1 . Consequently, we get
gF g
À1
¼ F :
ð16:28Þ
This implies that F of ρ is an invariant subgroup of ℊ.
∎
(a)
(b)
ℊ
ℊ
⟷ : isomorphism
: inverƟble (bijecƟve)
Fig. 16.1 Mapping in a
group and vector space. (a)
Homomorphic mapping ρ:
ℊ ! ℊ
0 between two
groups. (b) Linear
transformation
(endomorphism) A: V
n ! V
n
in a vector space V
n
630
16 Introductory Group Theory