ρ e
ð Þρ e
ð Þ ¼ ρ ee
ð Þ ¼ ρ e
ð Þ, i:e:, ρ e
ð Þ ¼ e
0 ,
where e
0 is an identity element of ℊ
0 . Also, we have
ρ x
ð Þρ x
À1
À Á ¼ ρ xx
À1
À
Á ¼ ρ e
ð Þ ¼ e
0
:
Therefore,
ρ x
ð Þ
½
À1 ¼ ρ x
À1
À Á :
The two groups can be either a finite group or an infinite group. Note that in the
above the mapping is not injective. The mapping may or may not be surjective.
Regarding the identity and inverse elements, we have the same relations as (16.18)
and (16.20). From Definitions 16.1 and 16.2, we say that the bijective homomorphism is the isomorphism.
Let us introduce an important notion of a kernel of a mapping. In this regard, we
have a following definition.
Definition 16.3 Let ℊ ¼ {e, x, y, Á Á Á} and ℊ
0
¼ {e
0 , x
0 , y
0 , Á Á Á} be groups and let
e and e
0 be the identity elements. Suppose that there exists a homomorphic mapping
ρ: ℊ ! ℊ
0 . Also let F be a subset of ℊ such that
ρ F
ð Þ ¼ e
0
:
ð16:22Þ
Then, F is said to be a kernel of ρ.
Regarding the kernel, we have following important theorems.
Theorem 16.2 Let ℊ ¼ {e, x, y, Á Á Á} and ℊ
0
¼ {e
0
, x
0
, y
0
, Á Á Á} be groups, where e and
e
0 are identity elements. A necessary and sufficient condition for a surjective and
homomorphic mapping ρ : ℊ ! ℊ
0 to be isomorphic is that a kernel F ¼ e
f g.
Proof We assume that F ¼ e
f g. Suppose that ρ(x) ¼ ρ( y). Then, we have
ρ x
ð Þ ρ y
ð Þ
½
À1 ¼ ρ x
ð Þρ y
À1
À Á ¼ ρ xy
À1
À
Á ¼ e
0
:
ð16:23Þ
The first and second equalities result from the homomorphism of ρ. Since F ¼ e
f g,
xy
À1
¼ e, i.e., x ¼ y. Therefore, ρ is injective (i.e., one-to-one correspondence). As ρ
is surjective from the assumption, ρ is bijective. The mapping ρ is isomorphic
accordingly.
Conversely, suppose that ρ is isomorphic. Also suppose for
∃
x 2 ℊ ρ(x) ¼ e
0 .
From (16.18), ρ(e) ¼ e
0
. We have ρ(x) ¼ ρ(e) ¼ e
0
⟹ x ¼ e due to the isomorphism
of ρ (i.e., one-to-one correspondence). This implies F ¼ e
f g. This completes the
proof.
∎
We become aware of close relationship between Theorem 16.1 and linear transformation versus kernel already mentioned in Sect. 11.2 of Part III. Figure 16.1
16.4 Isomorphism and Homomorphism
629
ð Þρ e
ð Þ ¼ ρ ee
ð Þ ¼ ρ e
ð Þ, i:e:, ρ e
ð Þ ¼ e
0 ,
where e
0 is an identity element of ℊ
0 . Also, we have
ρ x
ð Þρ x
À1
À Á ¼ ρ xx
À1
À
Á ¼ ρ e
ð Þ ¼ e
0
:
Therefore,
ρ x
ð Þ
½
À1 ¼ ρ x
À1
À Á :
The two groups can be either a finite group or an infinite group. Note that in the
above the mapping is not injective. The mapping may or may not be surjective.
Regarding the identity and inverse elements, we have the same relations as (16.18)
and (16.20). From Definitions 16.1 and 16.2, we say that the bijective homomorphism is the isomorphism.
Let us introduce an important notion of a kernel of a mapping. In this regard, we
have a following definition.
Definition 16.3 Let ℊ ¼ {e, x, y, Á Á Á} and ℊ
0
¼ {e
0 , x
0 , y
0 , Á Á Á} be groups and let
e and e
0 be the identity elements. Suppose that there exists a homomorphic mapping
ρ: ℊ ! ℊ
0 . Also let F be a subset of ℊ such that
ρ F
ð Þ ¼ e
0
:
ð16:22Þ
Then, F is said to be a kernel of ρ.
Regarding the kernel, we have following important theorems.
Theorem 16.2 Let ℊ ¼ {e, x, y, Á Á Á} and ℊ
0
¼ {e
0
, x
0
, y
0
, Á Á Á} be groups, where e and
e
0 are identity elements. A necessary and sufficient condition for a surjective and
homomorphic mapping ρ : ℊ ! ℊ
0 to be isomorphic is that a kernel F ¼ e
f g.
Proof We assume that F ¼ e
f g. Suppose that ρ(x) ¼ ρ( y). Then, we have
ρ x
ð Þ ρ y
ð Þ
½
À1 ¼ ρ x
ð Þρ y
À1
À Á ¼ ρ xy
À1
À
Á ¼ e
0
:
ð16:23Þ
The first and second equalities result from the homomorphism of ρ. Since F ¼ e
f g,
xy
À1
¼ e, i.e., x ¼ y. Therefore, ρ is injective (i.e., one-to-one correspondence). As ρ
is surjective from the assumption, ρ is bijective. The mapping ρ is isomorphic
accordingly.
Conversely, suppose that ρ is isomorphic. Also suppose for
∃
x 2 ℊ ρ(x) ¼ e
0 .
From (16.18), ρ(e) ¼ e
0
. We have ρ(x) ¼ ρ(e) ¼ e
0
⟹ x ¼ e due to the isomorphism
of ρ (i.e., one-to-one correspondence). This implies F ¼ e
f g. This completes the
proof.
∎
We become aware of close relationship between Theorem 16.1 and linear transformation versus kernel already mentioned in Sect. 11.2 of Part III. Figure 16.1
16.4 Isomorphism and Homomorphism
629