Abstract Algebra/Group Theory/Homomorphism/Image 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{im}}~ f = \lbrace k \in K \; | \; \exists \; g \in G: f(g) = k \rbrace$ is a subgroup of K.

Identity

 * {|Style = "width:60%;boarder:0.5px"

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


 * 3. $$ i \in K$$
 * 3. $$ i \in K$$
 * 3. $$ i \in K$$


 * 2.
 * 4. $$ {\color{OliveGreen}e_{K}} \ast i = i \ast {\color{OliveGreen}e_{K}} = i$$
 * 4. $$ {\color{OliveGreen}e_{K}} \ast i = i \ast {\color{OliveGreen}e_{K}} = i$$
 * 4. $$ {\color{OliveGreen}e_{K}} \ast i = i \ast {\color{OliveGreen}e_{K}} = i$$


 * i is in K and eK is identity of K (usage3)
 * 5. $$ \forall \; i \in \text{im} f: {\color{OliveGreen}e_{K}} \ast i = i \ast {\color{OliveGreen}e_{K}} = i$$
 * 2, 3, and 4.
 * 6. $$ {\color{OliveGreen}e_{K}}$$ is identity of $$ \text{im} \; f$$
 * definition of identity (usage 4)
 * }
 * 2, 3, and 4.
 * 6. $$ {\color{OliveGreen}e_{K}}$$ is identity of $$ \text{im} \; f$$
 * definition of identity (usage 4)
 * }
 * }

Inverse

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

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

|-
 * 0.
 * 2. $$ f({\color{OliveGreen}g}) \circledast f({\color{BrickRed}g^{-1}}) = f({\color{BrickRed}g^{-1}}) \circledast f({\color{OliveGreen}g}) = e_{K} $$
 * 2. $$ f({\color{OliveGreen}g}) \circledast f({\color{BrickRed}g^{-1}}) = f({\color{BrickRed}g^{-1}}) \circledast f({\color{OliveGreen}g}) = e_{K} $$

|-
 * homomorphism maps inverse to inverse between G and K
 * 3. $$ {\color{OliveGreen}i} \circledast f({\color{BrickRed}g^{-1}}) = f({\color{BrickRed}g^{-1}}) \circledast {\color{OliveGreen}i} = e_{K} $$
 * 3. $$ {\color{OliveGreen}i} \circledast f({\color{BrickRed}g^{-1}}) = f({\color{BrickRed}g^{-1}}) \circledast {\color{OliveGreen}i} = e_{K} $$

|-
 * homomorphism maps inverse to inverse
 * 4. i has inverse f( k-1 ) in im f
 * 4. i has inverse f( k-1 ) in im f

|-
 * 2, 3, and eK is identity of im f
 * 5. Every element of im f has an inverse.
 * }

Closure

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

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 * 0. Choose $$ i_{1}, i_{2} \in \lbrace k \in K \; | \; \exists \; g \in G: f(g) = k \rbrace $$||
 * 1. $$ \exists \; g_{1},g_{2} \in G: f(g_{1}) = i_{1}, f(g_{2}) = i_{2}$$
 * 1. $$ \exists \; g_{1},g_{2} \in G: f(g_{1}) = i_{1}, f(g_{2}) = i_{2}$$

|-
 * 0.
 * 2. $$g_{1} \ast g_{2} \in G $$
 * 2. $$g_{1} \ast g_{2} \in G $$

|-
 * Closure in G
 * 3. $$f(g_{1} \ast g_{2}) \in \lbrace k \in K \; | \; \exists \; g \in G: f(g) = k \rbrace$$
 * 3. $$f(g_{1} \ast g_{2}) \in \lbrace k \in K \; | \; \exists \; g \in G: f(g) = k \rbrace$$

|-
 * 4. $$i_{1} \circledast i_{2} = f(g_{1}) \circledast f(g_{2}) = f(g_{1} \ast g_{2})$$
 * 4. $$i_{1} \circledast i_{2} = f(g_{1}) \circledast f(g_{2}) = f(g_{1} \ast g_{2})$$

|-
 * f is a homomorphism, 0.
 * 5. $$i_{1} \circledast i_{2} \in im f$$
 * 5. $$i_{1} \circledast i_{2} \in im f$$


 * 3. and 4.
 * }

Associativity

 * {| Style ="width:60%"

|-    |-     |-
 * 0. im f is a subset of K||
 * 1. $$\circledast$$ is associative in K||
 * 2. $$\circledast$$ is associative in im f || 1 and 2
 * }