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. $i_{1} \circledast i_{2} \in im f$	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