description/proof of that range of group homomorphism is subgroup of codomain
Topics
About: group
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of %structure kind name% homomorphism.
Target Context
- The reader will have a description and a proof of the proposition that for any group homomorphism, the range of the homomorphism is a subgroup of the codomain.
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: Structured Description
Here is the rules of Structured Description.
Entities:
\(G_1\): \(\in \{\text{ the groups }\}\)
\(G_2\): \(\in \{\text{ the groups }\}\)
\(f\): \(: G_1 \to G_2\), \(\in \{\text{ the group homomorphisms }\}\)
\(ran f\): \(= \text{ the range of } f\)
//
Statements:
\(ran f \in \{\text{ the subgroups of } G_2\}\).
//
2: Natural Language Description
For any groups, \(G_1, G_2\), and any group homomorphism, \(f: G_1 \to G_2\), the range of \(f\), \(ran f\), is a subgroup of \(G_2\).
3: Proof
The identity element of any group is denoted as \(e\).
For any elements, \(g_1, g_2 \in ran f\), \(g_1 g_2 \in ran f\), because \(g_j = f (g'_j)\) for a \(g'_j \in G_1\), \(g_1 g_2 = f (g'_1) f (g'_2) = f (g'_1 g'_2)\).
\(e \in ran f\), because \(f (e) = e\).
For any element, \(g \in ran f\), \(g^{-1} \in ran f\), because \(g = f (g')\) for a \(g' \in G_1\), \(g^{-1} = {f (g')}^{-1} = f (g'^{-1})\).
\(ran f\) satisfies the associative law, because the elements are elements of \(G_2\) and the operation is inherited from that of \(G_2\).
