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