definition of quotient group of group by normal subgroup
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: Note
Starting Context
- The reader knows a definition of normal subgroup of group.
Target Context
- The reader will have a definition of quotient group of group by normal 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:
\( G'\): \(\in \{\text{ the groups }\}\)
\( G\): \(\in \{\text{ the normal subgroups of } G'\}\)
\(*G' / G\): \(= \{[g'] \subseteq G' \vert g' \in G' \land \forall g'' \in G' (g'' \in g' G \iff g'' \in [g'])\}\) with the group operation
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Conditions:
\([g'] [g''] = [g' g'']\)
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2: Natural Language Description
For any group, \(G'\), and any normal subgroup, \(G\), of \(G'\), \(G' / G := \{[g'] \subseteq G' \vert g' \in G' \land \forall g'' \in G' (g'' \in g' G \iff g'' \in [g'])\}\) with the group operation, \([g'] [g''] = [g' g'']\)
3: Note
\(G' / G\) is indeed a set of equivalent classes: each \(g' \in G'\) belongs to \([g']\), because \(g' = g' 1 \in g' G\) while \(1 \in G\); if \(g' \in [g'']\) (which implies that \(g' = g'' g_1\) for a \(g_1 \in G\)), \([g''] = [g']\), because for each \(g''' \in [g'']\), \(g''' = g'' g_2\), where \(g_2 \in G\), \(= g' {g_1}^{-1} g_2 \in g' G\), and for each \(g''' \in [g']\), \(g''' = g' g_2\), where \(g_2 \in G\), \(= g'' g_1 g_2 \in g'' G\); so, each element of \(G'\) belongs to the unique class.
\([g'] [g''] = [g' g'']\) is indeed well-defined: for each \(g' g_1 \in [g']\) and \(g'' g_2 \in [g'']\), \(g' g_1 g'' g_2 = g' g'' g''^{-1} g_1 g'' g_2\), but \(g''^{-1} g_1 g'' \in G\) because \(G\) is a normal subgroup (in fact, for this, \(G\) is required to be a normal subgroup), so, \(g' g_1 g'' g_2 \in [g' g'']\) and so \([g' g_1 g'' g_2] = [g' g'']\), which mean that the definition does not depend on the representations of the classes.
\(G' / G\) is indeed a group: \([1]\) is the identity element, because \([1] [g'] = [1 g'] = [g'] = [g' 1] = [g'] [1]\); \([g'] [g''] = [g' g''] \in G' / G\); \([g'^{-1}] [g'] = [g'^{-1} g'] = [1] = [g' g'^{-1}] = [g'] [g'^{-1}]\); \(([g'] [g'']) [g'''] = [g' g''] [g'''] = [g' g'' g'''] = [g'] [g'' g'''] = [g'] ([g''] [g'''])\).
While this definition uses \(g' G\), using \(G g'\) instead does not make any difference, because \(g' G = G g'\): for each \(g'' \in g' G\), \(g'' = g' g_1\) for a \(g_1 \in G\), but \(= g' g_1 g'^{-1} g'\) and \(g' g_1 g'^{-1} \in G\) because \(G\) is a normal subgroup, which implies that \(g'' \in G g'\); for each \(g'' \in G g'\), \(g'' = g_1 g'\) for a \(g_1 \in G\), but \(= g' g'^{-1} g_1 g'\) and \(g'^{-1} g_1 g' \in G\) because \(G\) is a normal subgroup, which implies that \(g'' \in g' G\).
