Home
Random
Log in
Settings
Donations
About Wikibooks
Disclaimers
Search
Abstract Algebra/Group Theory/Homomorphism/Homomorphism Maps Identity to Identity
Language
Watch
Edit
<
Abstract Algebra
|
Group Theory
|
Homomorphism
Theorem
edit
Let
f
be a homomorphism from group
G
to group
K
.
Let
e
G
and
e
K
be identities of
G
and
K
.
f
(
e
G
) =
e
K
Proof
edit
0.
f
(
e
G
)
∈
K
{\displaystyle f({\color {Blue}e_{G}})\in K}
f
maps to K
1.
[
f
(
e
G
)
]
−
1
{\displaystyle {\color {BrickRed}[}f({\color {Blue}e_{G}}){\color {BrickRed}]^{-1}}}
inverse in K
.
2.
f
(
e
G
∗
e
G
)
=
f
(
e
G
)
⊛
f
(
e
G
)
{\displaystyle f({\color {Blue}e_{G}}\ast {\color {Blue}e_{G}})=f({\color {Blue}e_{G}})\circledast f({\color {Blue}e_{G}})}
f
is a homomorphism
3.
f
(
e
G
)
=
f
(
e
G
)
⊛
f
(
e
G
)
{\displaystyle f({\color {Blue}e_{G}})=f({\color {Blue}e_{G}})\circledast f({\color {Blue}e_{G}})}
identity
e
G
.
4.
[
f
(
e
G
)
]
−
1
⊛
f
(
e
G
)
=
[
f
(
e
G
)
]
−
1
⊛
f
(
e
G
)
⊛
f
(
e
G
)
{\displaystyle {\color {BrickRed}[}f({\color {Blue}e_{G}}){\color {BrickRed}]^{-1}}\circledast f({\color {Blue}e_{G}})={\color {BrickRed}[}f({\color {Blue}e_{G}}){\color {BrickRed}]^{-1}}\circledast f({\color {Blue}e_{G}})\circledast f({\color {Blue}e_{G}})}
1.
.
5.
e
K
=
e
K
⊛
f
(
e
G
)
{\displaystyle {\color {OliveGreen}e_{K}}={\color {OliveGreen}e_{K}}\circledast f({\color {Blue}e_{G}})}
identity
e
K
, definition of inverse
6.
e
K
=
f
(
e
G
)
{\displaystyle {\color {OliveGreen}e_{K}}=f({\color {Blue}e_{G}})}
identity
e
K
![{\displaystyle {\color {BrickRed}[}f({\color {Blue}e_{G}}){\color {BrickRed}]^{-1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0109bb676f4b1376a243634ddc2021f552d4699e)
![{\displaystyle {\color {BrickRed}[}f({\color {Blue}e_{G}}){\color {BrickRed}]^{-1}}\circledast f({\color {Blue}e_{G}})={\color {BrickRed}[}f({\color {Blue}e_{G}}){\color {BrickRed}]^{-1}}\circledast f({\color {Blue}e_{G}})\circledast f({\color {Blue}e_{G}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/86562c2f5cda1d9f6bfb1415f56840c3ca4e91a5)
