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Abstract Algebra/Group Theory/Homomorphism/A Homomorphism with Trivial Kernel is Injective
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<
Abstract Algebra
|
Group Theory
|
Homomorphism
Theorem
edit
Let
f
be a
homomorphism
from
group
G
to
group
K
. Let
e
K
be
identity
of
K
.
ker
f
=
{
e
G
}
{\displaystyle {\text{ker}}~f={\lbrace e_{G}\rbrace }}
means
f
is injective.
Proof
edit
0. Choose
x
,
y
∈
G
{\displaystyle x,y\in G}
such that
f
(
x
)
=
f
(
y
)
{\displaystyle f(x)=f(y)}
1.
f
(
y
∗
x
−
1
)
=
f
(
y
)
⊛
f
(
x
−
1
)
{\displaystyle f(y\ast x^{-1})=f(y)\circledast f(x^{-1})}
f is a homomorphism
2.
=
f
(
x
)
⊛
f
(
x
−
1
)
{\displaystyle =f(x)\circledast f(x^{-1})}
0.
3.
=
f
(
x
∗
x
−
1
)
{\displaystyle =f(x\ast x^{-1})}
f is a homomorphism
4.
=
f
(
e
G
)
=
e
K
{\displaystyle =f(e_{G})=e_{K}}
homomorphism maps identity to identity
5.
y
∗
x
−
1
∈
ker
f
{\displaystyle y\ast x^{-1}\in {\text{ker}}~f}
1,2,3,4.
6.
y
∗
x
−
1
=
e
G
{\displaystyle y\ast x^{-1}=e_{G}}
given
ker
f
=
{
e
G
}
{\displaystyle {\text{ker}}~f={\lbrace e_{G}\rbrace }}
7.
y
=
x
{\displaystyle y=x}