Abstract Algebra/Group Theory/Homomorphism/Image 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.

${\text{im}}~f=\lbrace k\in K\;|\;\exists \;g\in G:f(g)=k\rbrace$ is a subgroup of K.

Proof

Identity

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

Inverse

0. Choose ${\color {OliveGreen}i}\in \lbrace k\in K\;|\;\exists \;g\in G:f(g)=k\rbrace$
1. $\exists \;{\color {OliveGreen}g}\in G:f({\color {OliveGreen}g})={\color {OliveGreen}i}$	0.
2. $f({\color {OliveGreen}g})\circledast f({\color {BrickRed}g^{-1}})=f({\color {BrickRed}g^{-1}})\circledast f({\color {OliveGreen}g})=e_{K}$	homomorphism maps inverse to inverse between G and K
3. ${\color {OliveGreen}i}\circledast f({\color {BrickRed}g^{-1}})=f({\color {BrickRed}g^{-1}})\circledast {\color {OliveGreen}i}=e_{K}$	homomorphism maps inverse to inverse
4. i has inverse f( k-1) in im f	2, 3, and eK is identity of im f
5. Every element of im f has an inverse.

Closure

0. Choose $i_{1},i_{2}\in \lbrace k\in K\;|\;\exists \;g\in G:f(g)=k\rbrace$
1. $\exists \;g_{1},g_{2}\in G:f(g_{1})=i_{1},f(g_{2})=i_{2}$	0.
2. $g_{1}\ast g_{2}\in G$	Closure in G
3. $f(g_{1}\ast g_{2})\in \lbrace k\in K\;|\;\exists \;g\in G:f(g)=k\rbrace$
4. $i_{1}\circledast i_{2}=f(g_{1})\circledast f(g_{2})=f(g_{1}\ast g_{2})$	f is a homomorphism, 0.
5. $i_{1}\circledast i_{2}\in imf$	3. and 4.

Associativity

0. im f is a subset of K
1. $\circledast$ is associative in K
2. $\circledast$ is associative in im f	1 and 2