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

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Theorem[edit]

Let f be a homomorphism from group G to group K. Let eK be identity of K.


 {\text{ker}}~ f = \lbrace g \in G \; | \;  f(g) = {\color{OliveGreen}e_{K}} \rbrace is a subgroup of G.

Proof[edit]

Identity[edit]

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

Inverse[edit]

0. Choose  {\color{OliveGreen}k} \in \lbrace g \in G \; | \;  f(g) = e_{K} \rbrace
1.  f({\color{OliveGreen}k}) = e_{K}
0.
2.  {\color{OliveGreen}k} \ast {\color{BrickRed}k^{-1}} = {\color{BrickRed}k^{-1}} \ast {\color{OliveGreen}k} = e_{G}
definition of inverse in G (usage 3)
3.  f({\color{BrickRed}k^{-1}}) = [e_{K}]^{-1} = e_{K}
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[edit]

0. Choose  x, y \in \lbrace g \in G \; | \;  f(g) = e_{K} \rbrace
1.  f(x) = f(y) = e_{K}
0.
2.  f(x \ast y) = f(x) \circledast f(y)
f is a homomorphism
3.  f(x \ast y) = e_{K} \circledast e_{K} = e_{K}
1. and eK is identity of K
4.  x \ast y \in \lbrace g \in G \; | \;  f(g) = e_{K} \rbrace

Associativity[edit]

0. ker f is a subset of G
1. \ast is associative in G
2. \ast is associative in ker f 1 and 2