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

TheoremEdit

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.

ProofEdit

IdentityEdit

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)

InverseEdit

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.

ClosureEdit

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.

AssociativityEdit

0. im f is a subset of K
1. \circledast is associative in K
2. \circledast is associative in im f 1 and 2
Last modified on 14 October 2012, at 08:35