description/proof of that group homomorphism is injective iff kernel is 1 subgroup
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 injection.
Target Context
- The reader will have a description and a proof of the proposition that any group homomorphism is injective if and only if the kernel is the 1 subgroup.
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:
\(G_1\): \(\in \{\text{ the groups }\}\)
\(G_2\): \(\in \{\text{ the groups }\}\)
\(f\): \(: G_1 \to G_2\), \(\in \{\text{ the group homomorphisms }\}\)
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Statements:
\(Ker (f) = \{1\}\)
\(\iff\)
\(f \in \{\text{ the injections }\}\)
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2: Proof
Whole Strategy: Step 1: suppose that \(Ker (f) = \{1\}\), take any \(g, g' \in G_1\) such that \(g \neq g'\), suppose that \(f (g) = f (g')\), and find a contradiction; Step 2: suppose that \(f\) is injective, and see that \(Ker (f) = \{1\}\).
Step 1:
Let us suppose that \(Ker (f) = \{1\}\).
Let \(g, g' \in G_1\) be any such that \(g \neq g'\).
Let us suppose that \(f (g) = f (g')\) and find a contradiction.
\(1 = f (g) f (g')^{-1} = f (g) f (g'^{-1}) = f (g g'^{-1})\), which implies that \(g g'^{-1} = 1\), which implies that \(g = g'\), a contradiction.
Step 2:
Let us suppose that \(f\) is injective.
\(f (1) = 1\).
Let us suppose that there is a \(g \in G_1\) such that \(f (g) = 1\). As \(f (1) = 1\) and \(f\) is injective, there is no other \(g\) than \(1\), which means that \(Ker (f) = \{1\}\).
