description/proof of the 1st isomorphism theorem for groups
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of %structure kind name% homomorphism.
- The reader knows a definition of %category name% isomorphism.
- The reader knows a definition of quotient group of group by normal subgroup.
- The reader admits the proposition that for any group homomorphism, the kernel of the homomorphism is a normal subgroup of the domain.
- The reader admits the proposition that for any group homomorphism, the range of the homomorphism is a subgroup of the codomain.
- The reader admits the proposition that any bijective group homomorphism is a 'groups - homomorphisms' isomorphism.
Target Context
- The reader will have a description and a proof of the proposition that for any group homomorphism, the quotient group of the domain by the kernel of the homomorphism is 'groups - homomorphisms' isomorphic to the range of the homomorphism: the 1st isomorphism theorem for groups.
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:
\(G_1\): \(\in \{\text{ the groups }\}\)
\(G_2\): \(\in \{\text{ the groups }\}\)
\(f\): \(: G_1 \to G_2\), \(\in \{\text{ the group homomorphisms }\}\)
\(Ker (f)\): \(= \text{ the kernel of } f\)
\(Ran (f)\): \(= \text{ the range of } f\)
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Statements:
\(Ker (f) \in \{\text{ the normal subgroups of } G_1\}\)
\(\land\)
\(Ran (f) \in \{\text{ the subgroups of } G_2\}\)
\(\land\)
\(G_1 / Ker (f) \cong_{groups} Ran (f)\) by \(f': G_1 / Ker (f) \to Ran (f), [g] \mapsto f (g)\), where \(\cong_{groups}\) denotes being 'groups - homomorphisms' isomorphic
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2: Proof
Whole Strategy: Step 1: see that \(Ker (f)\) is a normal subgroup of \(G_1\); Step 2: see that \(Ran (f)\) is a subgroup of \(G_2\); Step 3: see that \(f'\) is well-defined; Step 4: see that \(f'\) is bijective; Step 5: see that \(f'\) is a group homomorphism; Step 6: conclude the proposition.
Step 1:
\(Ker (f)\) is a normal subgroup of \(G_1\), by the proposition that for any group homomorphism, the kernel of the homomorphism is a normal subgroup of the domain.
So, \(G_1 / Ker (f)\) is well-defined.
Step 2:
\(Ran (f)\) is a subgroup of \(G_2\), by the proposition that for any group homomorphism, the range of the homomorphism is a subgroup of the codomain.
Step 3:
\(f'\) is well-defined, because for any \(g' \in G_1\) such that \([g'] = [g]\), \(g' = g g''\) where \(g'' \in Ker (f)\), \(f (g') = f (g) f (g'') = f (g) 1 = f (g)\).
Step 4:
\(f'\) is injective, because for any \([g_1] \neq [g_2]\), \(f (g_1 {g_2}^{-1}) = f (g_1) {f (g_2)}^{-1}\), but \(g_1 {g_2}^{-1} \notin Ker (f)\), so, \(f (g_1) {f (g_2)}^{-1} \neq 1\), \(f (g_1) \neq f (g_2)\).
\(f'\) is surjective, obviously.
So, \(f'\) is bijective.
Step 5:
\(f'\) is a group homomorphism, because \(f' ([g_1] [g_2]) = f' ([g_1 g_2]) = f (g_1 g_2) = f (g_1) f (g_2) = f' ([g_1]) f' ([g_2])\); \(f' ([1]) = f (1) = 1\); \(f' ([g]^{-1}) = f' ([g^{-1}]) = f (g^{-1}) = {f (g)}^{-1} = {f' ([g])}^{-1}\).
Step 6:
So, \(f'\) is a 'groups - homomorphisms' isomorphism, by the proposition that any bijective group homomorphism is a 'groups - homomorphisms' isomorphism.
