# 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

 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   ${\text{ker}}f=\lbrace g\in G\;|\;f(g)={\color {OliveGreen}e_{K}}\rbrace$ 3. $k\in G$ 2. 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

 0. Choose ${\color {OliveGreen}k}\in \lbrace g\in G\;|\;f(g)=e_{K}\rbrace$ 1. $f({\color {OliveGreen}k})=e_{K}$ 0. 2. ${\color {OliveGreen}k}\ast {\color {BrickRed}k^{-1}}={\color {BrickRed}k^{-1}}\ast {\color {OliveGreen}k}=e_{G}$ definition of inverse in G (usage 3) 3. $f({\color {BrickRed}k^{-1}})=[e_{K}]^{-1}=e_{K}$ 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 $x,y\in \lbrace g\in G\;|\;f(g)=e_{K}\rbrace$ 1. $f(x)=f(y)=e_{K}$ 0. 2. $f(x\ast y)=f(x)\circledast f(y)$ f is a homomorphism 3. $f(x\ast y)=e_{K}\circledast e_{K}=e_{K}$ 1. and eK is identity of K 4. $x\ast y\in \lbrace g\in G\;|\;f(g)=e_{K}\rbrace$ ## Associativity

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