# 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.

## Proof

### Identity

 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 $i\in \lbrace k\in K\;|\;\exists \;g\in G:f(g)=k\rbrace$ 3. $i\in K$ 2. 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)

### Inverse

 0. Choose ${\color {OliveGreen}i}\in \lbrace k\in K\;|\;\exists \;g\in G:f(g)=k\rbrace$ 1. $\exists \;{\color {OliveGreen}g}\in G:f({\color {OliveGreen}g})={\color {OliveGreen}i}$ 0. 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. ${\color {OliveGreen}i}\circledast f({\color {BrickRed}g^{-1}})=f({\color {BrickRed}g^{-1}})\circledast {\color {OliveGreen}i}=e_{K}$ 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 $i_{1},i_{2}\in \lbrace k\in K\;|\;\exists \;g\in G:f(g)=k\rbrace$ 1. $\exists \;g_{1},g_{2}\in G:f(g_{1})=i_{1},f(g_{2})=i_{2}$ 0. 2. $g_{1}\ast g_{2}\in G$ Closure in G 3. $f(g_{1}\ast g_{2})\in \lbrace k\in K\;|\;\exists \;g\in G:f(g)=k\rbrace$ 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. $i_{1}\circledast i_{2}\in imf$ 3. and 4.

### Associativity

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