description/proof of that for group and subgroup, quotient set by being in same coset is group only if subgroup is normal 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 left or right coset of subgroup by element of group.
- The reader knows a definition of quotient set. The reader knows a definition of normal subgroup of group.
The reader knows a definition of quotient group of group by normal subgroup.
The reader admits the proposition that for any group and any subgroup, being in any same coset is an equivalence relation.
The reader admits the proposition that with respect to any normal subgroup, the set of the cosets forms a group with the canonical multiplication and inversion.
The reader admits the proposition that for any group and its any subgroup, the subgroup is a normal subgroup if its conjugate subgroup by each element of the group is contained in it.
Target Context
- The reader will have a description and a proof of the proposition that for any group and any subgroup, the quotient set by being in same coset is a group with respect to the canonical multiplication and inversion only if the subgroup is a 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:
\(G'\): \(\in \{\text{ the groups }\}\)
\(G\): \(\in \{\text{ the subgroups of } G'\}\)
\(\sim_l\): \(= \{(g'_1, g'_2) \in G' \times G' \vert \exists g' \in G' (g'_1, g'_2 \in g' G)\}\), \(\in \{\text{ the equivalence relations }\}\)
\(\sim_r\): \(= \{(g'_1, g'_2) \in G' \times G' \vert \exists g' \in G' (g'_1, g'_2 \in G g')\}\), \(\in \{\text{ the equivalence relations }\}\)
\(G' / \sim_l\): \(= \text{ the quotient set }\)
\(G' / \sim_r\): \(= \text{ the quotient set }\)
//
Statements:
(
\(G' / \sim_l \text{ with the canonical multiplication and inversion } \in \{\text{ the groups }\}\)
\(\implies\)
\(G \in \{\text{ the normal subgroups }\}\)
)
\(\land\)
(
\(G' / \sim_r \text{ with the canonical multiplication and inversion } \in \{\text{ the groups }\}\)
\(\implies\)
\(G \in \{\text{ the normal subgroups }\}\)
)
//
2: Note
\(G' / \sim_l\) and \(G' / \sim_r\) are guaranteed to be well-defined as quotient sets, by the proposition that for any group and any subgroup, being in any same coset is an equivalence relation.
If \(G\) is a normal subgroup, \(G' / \sim_l = G' / \sim_r\) is a quotient group, by the proposition that with respect to any normal subgroup, the set of the cosets forms a group with the canonical multiplication and inversion; this proposition is claiming that \(G\) needs to be a normal subgroup in order for \(G' / \sim_l\) or \(G' / \sim_r\) to be a group with respect to the canonical multiplication and inversion.
In fact, if \(G\) is not any normal subgroup, the canonical multiplication and inversion are not well-defined.
3: Proof
Whole Strategy: Step 1: for \(G' / \sim_l\), suppose that the canonical multiplication is well-defined, let \(g \in G\) and \(g' \in G'\) be any, and see that \(g' g {g'}^{-1} \in G\); Step 2: for \(G' / \sim_r\), suppose that the canonical multiplication is well-defined, let \(g \in G\) and \(g' \in G'\) be any, and see that \(g' g {g'}^{-1} \in G\).
Step 1:
Let us suppose that for \(G' / \sim_l\), the canonical multiplication is well-defined.
"the canonical multiplication" means that \([g'_1] [g'_2] = [g'_1 g'_2]\).
Let \(g \in G\) and \(g' \in G'\) be any.
Let \(g'_1 \in G'\) be any.
Let \(g_1 \in G\) be any.
\([g'_1 g] = [g'_1]\), because \(g'_1, g'_1 g \in g'_1 G\).
\([{g'}^{-1} g_1] = [{g'}^{-1}]\), because \({g'}^{-1}, {g'}^{-1} g_1 \in {g'}^{-1} G\).
\([g'_1 g] [{g'}^{-1} g_1] = [g'_1] [{g'}^{-1}]\), which implies that \([g'_1 g {g'}^{-1} g_1] = [g'_1 {g'}^{-1}]\).
That means that \(g'_1 g {g'}^{-1} g_1 \in g'_1 {g'}^{-1} G\), which means that \(g'_1 g {g'}^{-1} g_1 = g'_1 {g'}^{-1} g_2\) for a \(g_2 \in G\).
So, \(g {g'}^{-1} g_1 = {g'}^{-1} g_2\), so, \(g' g {g'}^{-1} = g_2 {g_1}^{-1} \in G\).
That means that for each \(g' \in G'\), \(g' G {g'}^{-1} \subseteq G\).
By the proposition that for any group and its any subgroup, the subgroup is a normal subgroup if its conjugate subgroup by each element of the group is contained in it, \(G\) is a normal subgroup.
Step 2:
Let us suppose that for \(G' / \sim_r\), the canonical multiplication is well-defined.
"the canonical multiplication" means that \([g'_1] [g'_2] = [g'_1 g'_2]\).
The logic is parallel to that for \(G' / \sim_l\).
Let \(g \in G\) and \(g' \in G'\) be any.
Let \(g'_1 \in G'\) be any.
Let \(g_1 \in G\) be any.
\([g g'_1] = [g'_1]\), because \(g'_1, g g'_1 \in G g'_1\).
\([g_1 g'] = [g']\), because \(g', g_1 g' \in G g'\).
\([g_1 g'] [g g'_1] = [g'] [g'_1]\), which implies that \([g_1 g' g g'_1] = [g' g'_1]\).
That means that \(g_1 g' g g'_1 \in G g' g'_1\), which means that \(g_1 g' g g'_1 = g_2 g' g'_1\) for a \(g_2 \in G\).
So, \(g_1 g' g = g_2 g'\), so, \(g' g {g'}^{-1} = {g_1}^{-1} g_2 \in G\).
That means that for each \(g' \in G'\), \(g' G {g'}^{-1} \subseteq G\).
By the proposition that for any group and its any subgroup, the subgroup is a normal subgroup if its conjugate subgroup by each element of the group is contained in it, \(G\) is a normal subgroup.
