description/proof of that for group, subgroup, and left or right cosets of subgroup quotient set, composition of preimage after classification map of subset is subgroup multiplied by subset from left or right
Topics
About: group
The table of contents of this article
Starting Context
- 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 admits the proposition that with respect to any subgroup, the coset by any element of the group equals a coset if and only if the element is a member of the latter coset, whether they are left cosets or right cosets.
Target Context
- The reader will have a description and a proof of the proposition that for any group, any subgroup, and the left or right cosets of the subgroup quotient set, the composition of the preimage after the classification map of any subset is the subgroup multiplied by the subset from left or right.
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:
\(G'\): \(\in \{\text{ the groups }\}\)
\(G\): \(\in \{\text{ the subgroups of } G'\}\)
\(G' / \sim_{G, l}\): \(= \text{ the left cosets of } G \text{ quotient set of } G'\)
\(G' / \sim_{G, r}\): \(= \text{ the right cosets of } G \text{ quotient set of } G'\)
\(f_l\): \(: G' \to G' / \sim_{G, l}\), \(= \text{ the classification map }\)
\(f_r\): \(: G' \to G' / \sim_{G, r}\), \(= \text{ the classification map }\)
\(S'\): \(\subseteq G'\)
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Statements:
\({f_l}^{- 1} \circ f_l (S') = S' G\)
\(\land\)
\({f_r}^{- 1} \circ f_r (S') = G S'\)
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2: Note
Each of \(G' / \sim_{G, l}\) and \(G' / \sim_{G, r}\) is not any group nor any topological space in general but just a set: unless \(G\) is a normal subgroup, \(G' / \sim_{G, l}\) will not be any group, by 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, and as \(G'\) has no topology, the quotient set cannot be given the quotient topology.
But anyway, each of \(G' / \sim_{G, l}\) and \(G' / \sim_{G, r}\) is valid, by the proposition that for any group and any subgroup, being in any same coset is an equivalence relation.
3: Proof
Whole Strategy: Step 1: see that for each \(g' \in {f_l}^{- 1} \circ f_l (S')\), \(g' \in S' G\), and for each \(g' \in S' G\), \(g' \in {f_l}^{- 1} \circ f_l (S')\); Step 2: see that for each \(g' \in {f_r}^{- 1} \circ f_r (S')\), \(g' \in G S'\), and for each \(g' \in G S'\), \(g' \in {f_r}^{- 1} \circ f_r (S')\).
Step 1:
Let \(g' \in {f_l}^{- 1} \circ f_l (S')\) be any.
\(f_l (g') \in f_l (S')\).
\(f_l (g') = g' G \in f_l (S')\).
So, \(g' G = f_l (s') = s' G\) for an \(s' \in S'\).
\(g' \in s' G\), by the proposition that with respect to any subgroup, the coset by any element of the group equals a coset if and only if the element is a member of the latter coset, whether they are left cosets or right cosets.
So \(g' \in S' G\).
So, \({f_l}^{- 1} \circ f_l (S') \subseteq S' G\).
Let \(g' \in S' G\) be any.
\(g' = s' g\) for an \(s' \in S'\) and a \(g \in G\).
\(f_l (g') = s' g G = s' G\), because \(s' g \in s' G\), and the proposition that with respect to any subgroup, the coset by any element of the group equals a coset if and only if the element is a member of the latter coset, whether they are left cosets or right cosets applies, \(\in f_l (S')\).
So, \(g' \in {f_l}^{- 1} \circ f_l (S')\).
So, \(S' G \subseteq {f_l}^{- 1} \circ f_l (S')\).
So, \({f_l}^{- 1} \circ f_l (S') = S' G\).
Step 2:
Let \(g' \in {f_r}^{- 1} \circ f_r (S')\) be any.
\(f_r (g') \in f_r (S')\).
\(f_r (g') = G g' \in f_r (S')\).
So, \(G g' = f_r (s') = G s'\) for an \(s' \in S'\).
\(g' \in G s'\), by the proposition that with respect to any subgroup, the coset by any element of the group equals a coset if and only if the element is a member of the latter coset, whether they are left cosets or right cosets.
So \(g' \in G S'\).
So, \({f_r}^{- 1} \circ f_r (S') \subseteq G S'\).
Let \(g' \in G S'\) be any.
\(g' = g s'\) for an \(s' \in S'\) and a \(g \in G\).
\(f_r (g') = G g s' = G s'\), because \(g s' \in G s'\), and the proposition that with respect to any subgroup, the coset by any element of the group equals a coset if and only if the element is a member of the latter coset, whether they are left cosets or right cosets applies, \(\in f_r (S')\).
So, \(g' \in {f_r}^{- 1} \circ f_r (S')\).
So, \(G S' \subseteq {f_r}^{- 1} \circ f_r (S')\).
So, \({f_r}^{- 1} \circ f_r (S') = G S'\).
