Abstract Algebra/Group Theory/Homomorphism/Kernel of a Homomorphism is a Subgroup

=Theorem= Let f be a homomorphism from group G to group K. Let eK be identity of K.


 * $ {\text{ker}}~ f = \lbrace g \in G \; | \; f(g) = {\color{OliveGreen}e_{K}} \rbrace$ is a subgroup of G.

=Proof=

Identity

 * {|Style = "width:75%;boarder:1px"

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 * 0. $$f({\color{Blue}e_{G}}) = {\color{OliveGreen}e_{K}}$$
 * homomorphism maps identity to identity
 * 1. $$ {\color{Blue}e_{G}} \in \lbrace g \in G \; | \; f(g) = {\color{OliveGreen}e_{K}} \rbrace $$
 * 0. and $ {\color{Blue}e_{G}} \in G$
 * 2. Choose $$ k \in \text{ker} f$$  where   $$\text{ker} f = \lbrace g \in G \; | \;  f(g) = {\color{OliveGreen}e_{K}} \rbrace $$||
 * 0. and $ {\color{Blue}e_{G}} \in G$
 * 2. Choose $$ k \in \text{ker} f$$  where   $$\text{ker} f = \lbrace g \in G \; | \;  f(g) = {\color{OliveGreen}e_{K}} \rbrace $$||
 * 2. Choose $$ k \in \text{ker} f$$  where   $$\text{ker} f = \lbrace g \in G \; | \;  f(g) = {\color{OliveGreen}e_{K}} \rbrace $$||
 * 2. Choose $$ k \in \text{ker} f$$  where   $$\text{ker} f = \lbrace g \in G \; | \;  f(g) = {\color{OliveGreen}e_{K}} \rbrace $$||
 * 2. Choose $$ k \in \text{ker} f$$  where   $$\text{ker} f = \lbrace g \in G \; | \;  f(g) = {\color{OliveGreen}e_{K}} \rbrace $$||
 * 2. Choose $$ k \in \text{ker} f$$  where   $$\text{ker} f = \lbrace g \in G \; | \;  f(g) = {\color{OliveGreen}e_{K}} \rbrace $$||
 * 2. Choose $$ k \in \text{ker} f$$  where   $$\text{ker} f = \lbrace g \in G \; | \;  f(g) = {\color{OliveGreen}e_{K}} \rbrace $$||
 * 3. $$ k \in G$$
 * 3. $$ k \in G$$

|-
 * 2.
 * 4. $$ {\color{Blue}e_{G}} \ast k = k \ast {\color{Blue}e_{G}} = k$$
 * 4. $$ {\color{Blue}e_{G}} \ast k = k \ast {\color{Blue}e_{G}} = k$$

|-    |-     |-
 * k is in G and eG is identity of G (usage3)
 * 5. $$ \forall \; g \in G: e_{G} \ast g = g \ast e_{G} = g$$
 * 2, 3, and 4.
 * 6. $$ {\color{Blue}e_{G}}$$ is identity of $$ \text{ker} \; f$$
 * definition of identity (usage 4)
 * }

Inverse

 * {|Style = "width:75%;boarder:1px"

|-    |-
 * 0. Choose $$ {\color{OliveGreen}k} \in \lbrace g \in G \; | \; f(g) = e_{K} \rbrace $$||
 * 1. $$ f({\color{OliveGreen}k}) = e_{K} $$
 * 1. $$ f({\color{OliveGreen}k}) = e_{K} $$

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 * 0.
 * 2. $$ {\color{OliveGreen}k} \ast {\color{BrickRed}k^{-1}} = {\color{BrickRed}k^{-1}} \ast {\color{OliveGreen}k} = e_{G} $$
 * 2. $$ {\color{OliveGreen}k} \ast {\color{BrickRed}k^{-1}} = {\color{BrickRed}k^{-1}} \ast {\color{OliveGreen}k} = e_{G} $$

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 * definition of inverse in G (usage 3)
 * 3. $$ f({\color{BrickRed}k^{-1}}) = [e_{K}]^{-1} = e_{K} $$
 * 3. $$ f({\color{BrickRed}k^{-1}}) = [e_{K}]^{-1} = e_{K} $$

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 * homomorphism maps inverse to inverse
 * 4. k has inverse  k-1 in ker f
 * 4. k has inverse  k-1 in ker f

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 * 2, 3, and eG is identity of ker f
 * 5. Every element of ker f has an inverse.
 * }

Closure

 * {|Style = "width:75%;boarder:1px"

|-    |-
 * 0. Choose $$ x, y \in \lbrace g \in G \; | \; f(g) = e_{K} \rbrace $$||
 * 1. $$ f(x) = f(y) = e_{K}$$
 * 1. $$ f(x) = f(y) = e_{K}$$

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 * 0.
 * 2. $$ f(x \ast y) = f(x) \circledast f(y)$$
 * 2. $$ f(x \ast y) = f(x) \circledast f(y)$$

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 * f is a homomorphism
 * 3. $$ f(x \ast y) = e_{K} \circledast e_{K} = e_{K}$$
 * 3. $$ f(x \ast y) = e_{K} \circledast e_{K} = e_{K}$$

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 * 1. and e K is identity of K
 * 4. $$ x \ast y \in \lbrace g \in G \; | \; f(g) = e_{K} \rbrace$$
 * 4. $$ x \ast y \in \lbrace g \in G \; | \; f(g) = e_{K} \rbrace$$


 * }
 * }

Associativity

 * {| Style ="width:60%"

|-    |-     |-
 * 0. ker f is a subset of G||
 * 1. $$\ast$$ is associative in G||
 * 2. $$\ast$$ is associative in ker f || 1 and 2
 * }