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.
