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

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Theorem

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

${\displaystyle {\text{im}}~f=\lbrace k\in K\;|\;\exists \;g\in G:f(g)=k\rbrace }$ is a subgroup of K.

Proof

Identity

 0. ${\displaystyle f({\color {Blue}e_{G}})={\color {OliveGreen}e_{K}}}$ homomorphism maps identity to identity 1. ${\displaystyle {\color {OliveGreen}e_{K}}\in \lbrace k\in K\;|\;\exists \;g\in G:f(g)=k\rbrace }$ 0. and ${\displaystyle {\color {Blue}e_{G}}\in G}$ 2. Choose ${\displaystyle i\in \lbrace k\in K\;|\;\exists \;g\in G:f(g)=k\rbrace }$ 3. ${\displaystyle i\in K}$ 2. 4. ${\displaystyle {\color {OliveGreen}e_{K}}\ast i=i\ast {\color {OliveGreen}e_{K}}=i}$ i is in K and eK is identity of K(usage3) 5. ${\displaystyle \forall \;i\in {\text{im}}f:{\color {OliveGreen}e_{K}}\ast i=i\ast {\color {OliveGreen}e_{K}}=i}$ 2, 3, and 4. 6. ${\displaystyle {\color {OliveGreen}e_{K}}}$ is identity of ${\displaystyle {\text{im}}\;f}$ definition of identity(usage 4)

Inverse

 0. Choose ${\displaystyle {\color {OliveGreen}i}\in \lbrace k\in K\;|\;\exists \;g\in G:f(g)=k\rbrace }$ 1. ${\displaystyle \exists \;{\color {OliveGreen}g}\in G:f({\color {OliveGreen}g})={\color {OliveGreen}i}}$ 0. 2. ${\displaystyle f({\color {OliveGreen}g})\circledast f({\color {BrickRed}g^{-1}})=f({\color {BrickRed}g^{-1}})\circledast f({\color {OliveGreen}g})=e_{K}}$ homomorphism maps inverse to inverse between G and K 3. ${\displaystyle {\color {OliveGreen}i}\circledast f({\color {BrickRed}g^{-1}})=f({\color {BrickRed}g^{-1}})\circledast {\color {OliveGreen}i}=e_{K}}$ homomorphism maps inverse to inverse 4. i has inverse f( k-1) in im f 2, 3, and eK is identity of im f 5. Every element of im f has an inverse.

Closure

 0. Choose ${\displaystyle i_{1},i_{2}\in \lbrace k\in K\;|\;\exists \;g\in G:f(g)=k\rbrace }$ 1. ${\displaystyle \exists \;g_{1},g_{2}\in G:f(g_{1})=i_{1},f(g_{2})=i_{2}}$ 0. 2. ${\displaystyle g_{1}\ast g_{2}\in G}$ Closure in G 3. ${\displaystyle f(g_{1}\ast g_{2})\in \lbrace k\in K\;|\;\exists \;g\in G:f(g)=k\rbrace }$ 4. ${\displaystyle i_{1}\circledast i_{2}=f(g_{1})\circledast f(g_{2})=f(g_{1}\ast g_{2})}$ f is a homomorphism, 0. 5. ${\displaystyle i_{1}\circledast i_{2}\in imf}$ 3. and 4.

Associativity

 0. im f is a subset of K 1. ${\displaystyle \circledast }$ is associative in K 2. ${\displaystyle \circledast }$ is associative in im f 1 and 2