A description/proof of that with respect to normal subgroup, set of cosets forms group
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of coset with respect to subgroup by element of group.
- 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 with respect to any normal subgroup, the set of the cosets forms a group with the canonical multiplication and inversion.
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: Description
For any group, \(G_1\), any normal subgroup, \(G_2 \subseteq G_1\), and the left or right coset map, \(\pi: G_1 \rightarrow G_1/G_2\), the set of the cosets, \(G_1/G_2\), forms a group with the canonical multiplication, \(\langle p_1 G_2, p_2 G_2 \rangle \mapsto p_1 p_2 G_2\) or \(\langle G_2 p_1, G_2 p_2 \rangle \mapsto G_2 p_1 p_2\) and the canonical inverse, \(p G_2 \mapsto p^{-1} G_2\) or \(G_2 p \mapsto G_2 p^{-1}\), respectively.
2: Proof
Let us prove it for the left coset map.
The multiplication is associative, because \((p_1 G_2 p_2 G_2) p_3 G_3 = p_1 p_2 G_2 p_3 G_2 = p_1 p_2 p_3 G_2 = p_1 G_2 p_2 p_3 G_2 = p_1 G_2 (p_2 G_2 p_3 G_2)\).
Let us prove that the inverse is well-defined, which is about when \(p_1 G_2 = p_2 G_2\), \({p_1}^{-1} G_2 = {p_2}^{-1} G_2\). 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, it is about \({p_1}^{-1} \in {p_2}^{-1} G_2\). Is there a \(q \in G_2\), \({p_1}^{-1} = {p_2}^{-1} q\)? While \(p_1 = p_2 q_1\) for a \(q_1 \in G_2\), \({p_1}^{-1} = {q_1}^{-1} {p_2}^{-1}\) and \({p_1}^{-1} = {q_1}^{-1} {p_2}^{-1} = {p_2}^{-1} q\)? \(p_2 {q_1}^{-1} {p_2}^{-1} = q\)? Yes, because \(G_2\) is a normal subgroup.
\(G_2\) is the identity element, because \(G_2 p_1 G_2 = e p_1 G_2 = p_1 G_2 = p_1 e G_2 = p_1 G_2 G_2\).
\({p_1}^{-1} G_2\) is the inverse of \(p_1 G_2\), because \({p_1}^{-1} G_2 p_1 G_2 = {p_1}^{-1} p_1 G_2 = e G_2 = G_2 = e G_2 = p_1 {p_1}^{-1} G_2 = p_1 G_2 {p_1}^{-1} G_2\).
Let us prove it for the right coset map.
Let us prove that the multiplication is well-defined, which is about when \(G_2 p_{11} = G_2 p_{12}\) and \(G_2 p_{21} = G_2 p_{22}\), \(G_2 p_{11} p_{21} = G_2 p_{12} p_{22}\). 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, it is about \(p_{11} p_{21} \in G_2 p_{12} p_{22} = G_2 p_{11} p_{22}\). While \(p_{21} = q_1 p_{22}\) for a \(q_1 \in G_2\), \(p_{11} p_{21} = p_{11} q_1 p_{22} \in G_2 p_{11} p_{22}\)? Is there a \(q \in G_2\) such that \(p_{11} q_1 p_{22} = q p_{11} p_{22}\)? \(p_{11} q_1 = q p_{11}\)? \(p_{11} q_1 {p_{11}}^{-1} = q\)? Yes, because \(G_2\) is a normal subgroup.
The multiplication is associative, because \((G_2 p_1 G_2 p_2) G_3 p_3 = G_2 p_1 p_2 G_2 p_3 = G_2 p_1 p_2 p_3 = G_2 p_1 G_2 p_2 p_3 = G_2 p_1 (G_2 p_2 G_2 p_3)\).
Let us prove that the inverse is well-defined, which is about when \(G_2 p_1 = G_2 p_2\), \(G_2 {p_1}^{-1} = G_2 {p_2}^{-1}\). 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, it is about \({p_1}^{-1} \in G_2 {p_2}^{-1}\). Is there a \(q \in G_2\), \({p_1}^{-1} = q {p_2}^{-1}\)? While \(p_1 = q_1 p_2\) for a \(q_1 \in G_2\), \({p_1}^{-1} = {p_2}^{-1} {q_1}^{-1}\) and \({p_1}^{-1} = {p_2}^{-1} {q_1}^{-1} = q {p_2}^{-1}\)? \({p_2}^{-1} {q_1}^{-1} {p_2} = q\)? Yes, because \(G_2\) is a normal subgroup.
\(G_2\) is the identity element, because \(G_2 G_2 p_1 = G_2 e p_1 = G_2 p_1 = G_2 p_1 e = G_2 p_1 G_2\).
\(G_2 {p_1}^{-1}\) is the inverse of \(G_2 p_1\), because \(G_2 {p_1}^{-1} G_2 p_1 = G_2 {p_1}^{-1} p_1 = G_2 e = G_2 = G_2 e = G_2 p_1 {p_1}^{-1} = G_2 p_1 G_2 {p_1}^{-1}\).
