# Theorem

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

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

# Proof

## Identity

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

## Inverse

 0. Choose ${\displaystyle {\color {OliveGreen}k}\in \lbrace g\in G\;|\;f(g)=e_{K}\rbrace }$ 1. ${\displaystyle f({\color {OliveGreen}k})=e_{K}}$ 0. 2. ${\displaystyle {\color {OliveGreen}k}\ast {\color {BrickRed}k^{-1}}={\color {BrickRed}k^{-1}}\ast {\color {OliveGreen}k}=e_{G}}$ definition of inverse in G (usage 3) 3. ${\displaystyle f({\color {BrickRed}k^{-1}})=[e_{K}]^{-1}=e_{K}}$ homomorphism maps inverse to inverse 4. k has inverse k-1 in ker f 2, 3, and eG is identity of ker f 5. Every element of ker f has an inverse.

## Closure

 0. Choose ${\displaystyle x,y\in \lbrace g\in G\;|\;f(g)=e_{K}\rbrace }$ 1. ${\displaystyle f(x)=f(y)=e_{K}}$ 0. 2. ${\displaystyle f(x\ast y)=f(x)\circledast f(y)}$ f is a homomorphism 3. ${\displaystyle f(x\ast y)=e_{K}\circledast e_{K}=e_{K}}$ 1. and eK is identity of K 4. ${\displaystyle x\ast y\in \lbrace g\in G\;|\;f(g)=e_{K}\rbrace }$

## Associativity

 0. ker f is a subset of G 1. ${\displaystyle \ast }$ is associative in G 2. ${\displaystyle \ast }$ is associative in ker f 1 and 2