A description/proof of fundamental theorem for group homomorphism
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of group.
- The reader knows a definition of normal subgroup.
- The reader knows a definition of quotient group of group by normal subgroup.
- The reader knows a definition of %structure kind name% homomorphism.
- The reader admits the proposition that any groups map that maps the identity to the identity and maps any multiplication to the multiplication is a group homomorphism.
Target Context
- The reader will have a description and a proof of the fundamental theorem for group homomorphism: for any group (the 1st group), the quotient group of the group by any normal subgroup (the 2nd group), and any group homomorphism (the 1st group homomorphism) from the 1st group into any group (3rd group) whose kernel contains the 2nd group, there is the unique group homomorphism (the 2nd group homomorphism) from the quotient group into the 3rd group such that the 1st group homomorphism is the composition of the 2nd group homomorphism after the canonical homomorphism from 1st group onto the quotient group.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any group, \(G_1\), any normal subgroup, \(G_2 \subseteq G_1\), the quotient group of \(G_1\) by \(G_2\), \(G_1/G_2\), the canonical group homomorphism, \(\pi: G_1 \to G_1/G_2\), any group, \(G_3\), and any group homomorphism, \(f: G_1 \to G_3\), such that \(G_2 \subseteq ker f\), there is the unique group homomorphism, \(f': G_1/G_2 \to G_3\), such that \(f = f' \circ \pi\).
2: Proof
Let us define \(f' := g G_2 \mapsto f (g)\) for any \(g \in G_1\). Let us confirm that \(f'\) does not depend on the choice of \(g\) from \(g G_2\). For any 2 elements, \(g', g'' \in g G_2\), as \(g' G_2 = g'' G_2\), there are \(g'_2, g''_2 \in G_2\) such that \(g' g'_2 = g'' g''_2\). \(g'' = g' g'_2 {g''_2}^{-1}\). \(f (g'') = f (g' g'_2 {g''_2}^{-1}) = f (g') f (g'_2 {g''_2}^{-1}) = f (g')\), because \(f\) is a group homomorphism and \(G_2 \subseteq ker f\). So, \(f'\) does not depend on the choice. \(f' \circ \pi (g) = f' (g G_2) = f (g)\) for any \(g \in G_1\), so, \(f' \circ \pi = f\).
Let us confirm that \(f'\) is a group homomorphism. \(f' (e G_2) = f (e) = e\) where \(e\) is the identity elements. \(f' ((g' G_2) (g'' G_2)) = f' (g' g'' G_2) = f (g' g'') = f (g') f (g'') = f' (g' G_2) f' (g'' G_2)\). So, \(f'\) is a group homomorphism, by the proposition that any groups map that maps the identity to the identity and maps any multiplication to the multiplication is a group homomorphism.
Let us confirm that \(f'\) is unique. Let us suppose that there was another \(f''\). \(f' \circ \pi = f'' \circ \pi\), \(f' \circ \pi (g) = f' (g G_2) = f'' \circ \pi (g) = f'' (g G_2)\). So, \(f' = f''\).
