# 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 eK 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