description/proof of that for group and subgroup, being in same coset is equivalence relation
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 equivalence relation on set.
Target Context
- The reader will have a description and a proof of the proposition that for any group and any subgroup, being in any same coset is an equivalence relation.
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:
\(G'\): \(\in \{\text{ the groups }\}\)
\(G\): \(\in \{\text{ the subgroups of } G'\}\)
\(\sim_l\): \(= \{(g'_1, g'_2) \in G' \times G' \vert \exists g' \in G' (g'_1, g'_2 \in g' G)\}\)
\(\sim_r\): \(= \{(g'_1, g'_2) \in G' \times G' \vert \exists g' \in G' (g'_1, g'_2 \in G g')\}\)
//
Statements:
\(\sim_l \in \{\text{ the equivalence relations }\}\)
\(\land\)
\(\sim_r \in \{\text{ the equivalence relations }\}\)
//
2: Note
\(G\) does not need to be any normal subgroup.
So, the quotient sets, \(G' / \sim_l\) and \(G' / \sim_r\), are well-defined for any subgroup, \(G\), although \(G' / \sim_l\) and \(G' / \sim_r\) are not any groups if \(G\) is not any normal subgroup: refer to the proposition that for any group and any subgroup, the quotient set by being in same coset is a group only if the subgroup is a normal subgroup.
3: Proof
Whole Strategy: Step 1: see that \(\sim_l\) satisfies the conditions to be an equivalence relation; Step 2: see that \(\sim_r\) satisfies the conditions to be an equivalence relation.
Step 1:
Let us see that \(\sim_l\) satisfies the conditions to be an equivalence relation.
1) \(\forall g' \in G' (g' \sim_l g')\): reflexivity: \(g', g' \in g' G\).
2) \(\forall g'_1, g'_2 \in G' (g'_1 \sim_l g'_2 \implies g'_2 \sim_l g'_1)\): symmetry: if \(g'_1 \sim_l g'_2\), there is a \(g' \in G'\) such that \(g'_1, g'_2 \in g' G\), then, \(g'_2, g'_1 \in g' G\), so, \(g'_2 \sim_l g'_1\).
3) \(\forall g'_1, g'_2, g'_3 \in G' ((g'_1 \sim_l g'_2 \land g'_2 \sim_l g'_3) \implies g'_1 \sim g'_3)\): transitivity: if \(g'_1 \sim_l g'_2 \land g'_2 \sim_l g'_3\), there are a \(g'_4 \in G'\) such that \(g'_1, g'_2 \in g'_4 G\) and a \(g'_5 \in G'\) such that \(g'_2, g'_3 \in g'_5 G\), but as \(g'_2 \in g'_4 G, g'_5 G\), \(g'_2 = g'_4 g_1 = g'_5 g_2\) for some \(g_1, g_2 \in G\), so, \(g'_5 = g'_4 g_1 {g_2}^{-1}\), and as \(g'_3 = g'_5 g_3\) for a \(g_3 \in G\), \(g'_3 = g'_4 g_1 {g_2}^{-1} g_3 \in g'_4 G\), so, \(g'_1, g'_3 \in g'_4 G\).
Step 2:
Let us see that \(\sim_r\) satisfies the conditions to be an equivalence relation.
1) \(\forall g' \in G' (g' \sim_r g')\): reflexivity: \(g', g' \in G g'\).
2) \(\forall g'_1, g'_2 \in G' (g'_1 \sim_r g'_2 \implies g'_2 \sim_r g'_1)\): symmetry: if \(g'_1 \sim_r g'_2\), there is a \(g' \in G'\) such that \(g'_1, g'_2 \in G g'\), then, \(g'_2, g'_1 \in G g'\), so, \(g'_2 \sim_r g'_1\).
3) \(\forall g'_1, g'_2, g'_3 \in G' ((g'_1 \sim_r g'_2 \land g'_2 \sim_r g'_3) \implies g'_1 \sim g'_3)\): transitivity: if \(g'_1 \sim_r g'_2 \land g'_2 \sim_r g'_3\), there are a \(g'_4 \in G'\) such that \(g'_1, g'_2 \in G g'_4\) and a \(g'_5 \in G'\) such that \(g'_2, g'_3 \in G g'_5\), but as \(g'_2 \in G g'_4, G g'_5\), \(g'_2 = g_1 g'_4 = g_2 g'_5\) for some \(g_1, g_2 \in G\), so, \(g'_5 = {g_2}^{-1} g_1 g'_4\), and as \(g'_3 = g_3 g'_5\) for a \(g_3 \in G\), \(g'_3 = g_3 {g_2}^{-1} g_1 g'_4 \in G g'_4\), so, \(g'_1, g'_3 \in G g'_4\).
