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Find the group G generated under matrix multiplication by the matrices
A=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, B=\begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}.
Determine its proper subgroups, and verify for each of them that its cosets exhaust G.

Step-by-step

Before we can draw up a group multiplication table to search for subgroups, we must determine the multiple products of A and B with themselves and with each other:
{A}^{2}=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}=I.
Since B = iA, it follows that {B}^{2} = −I, that AB = iI = BA, and that {B}^{3} = −B. In brief, A is of order 2, B is of order 4, and A and B commute. The eight distinct elements of the group are therefore: I,A,B,{B}^{2},{B}^{3}, AB,A{B}^{2} and A{B}^{3}.
The group multiplication table is
By inspection, the subgroups and their cosets are as follows:
{I,A} : {I,A}, {B,AB}, {{B}^{2},A{B}^{2}}, {{B}^{3},A{B}^{3}};
{I,{B}^{2}} : {I,{B}^{2}}, {A,A{B}^{2}}, {B,{B}^{3}}, {AB,A{B}^{3}};
{I,A{B}^{2}} : {I,A{B}^{2}}, {A,{B}^{2}}, {B,A{B}^{3}}, {{B}^{3},AB};
{I,B,{B}^{2},{B}^{3}} : {I,B,{B}^{2},{B}^{3}}, {A,AB,A{B}^{2},A{B}^{3}};
{I,AB,{B}^{2},A{B}^{3}} : {I,AB,{B}^{2},A{B}^{3}}, {A,B,A{B}^{2},{B}^{3}}.
As expected, in each case the cosets exhaust the group, with each element in one and only one coset.

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