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

Theorem

Let f be a homomorphism from group G to group K. Let e_K 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