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

Theorem

Let f be a [[../Definition of Homomorphism, Kernel, and Image|homomorphism]] from [[../../Group/Definition of a Group|group]] G to [[../../Group/Definition of a Group|group]] K. Let e_K be [[../../Group/Definition of a Group/Definition of Identity|identity]] of K.

[[../Definition of Homomorphism, Kernel, and Image|$\text{ker}f=\{g\in G|f(g)=e_K\}$]] is a [[../../Subgroup/Definition of a Subgroup|subgroup]] of G.

Proof

Identity

0. $f(e_G)=e_K$	[[../Homomorphism Maps Identity to Identity|homomorphism maps identity to identity]]
1. $e_G\in \{g\in G|f(g)=e_K\}$	0. and [[../../Group/Definition of a Group/Definition of Identity|$e_G\in G$]]
.
	2. Choose $k\in \text{ker}f$   where   $\text{ker}f=\{g\in G|f(g)=e_K\}$
3. $k\in G$	2.
4. $e_G\ast k=k\ast e_G=k$	[[../../Group/Definition of a Group/Definition of Identity#Usage3|k is in G and eG is identity of G(usage3)]]
.
5. $\mathrm{\forall }g\in G:e_G\ast g=g\ast e_G=g$	2, 3, and 4.
6. $e_G$ is identity of $\text{ker}f$	[[../../Group/Definition of a Group/Definition of Identity#Usage4|definition of identity(usage 4)]]

Inverse

0. Choose $k\in \{g\in G|f(g)=e_K\}$
1. $f(k)=e_K$	0.
2. $k\ast k^{-1}=k^{-1}\ast k=e_G$	[[../../Group/Definition of a Group/Definition of Inverse#Usage3|definition of inverse in G (usage 3)]]
3. $f(k^{-1})=[e_K{]}^{-1}=e_K$	[[../Homomorphism Maps Inverse to Inverse|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 $x,y\in \{g\in G|f(g)=e_K\}$
1. $f(x)=f(y)=e_K$	0.
2. $f(x\ast y)=f(x)⊛f(y)$	f is a homomorphism
3. $f(x\ast y)=e_K⊛e_K=e_K$	1. and eK is identity of K
4. $x\ast y\in \{g\in G|f(g)=e_K\}$

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