Answer on Question #44359 – Math - Abstract Algebra
Problem.
Let G = S 4 G = S4 G = S 4 H = A 4 H = A4 H = A 4 K = f 1 K = f1 K = f 1
i) Check that H = K = h ( 1 2 3 ) H i H = K = h(1\ 2\ 3)Hi H = K = h ( 1 2 3 ) H i
ii) Check that K K K H H H h 2 H , h 62 K h \ 2\ H, h \ 62\ K h 2 H , h 62 K h K = K h hK = Kh h K = K h
iii) Check whether ( 1 2 3 4 ) H (1\ 2\ 3\ 4)\mathrm{H} ( 1 2 3 4 ) H ( 1 3 4 2 ) H (1\ 3\ 4\ 2)\mathrm{H} ( 1 3 4 2 ) H S 4 = H S4 = H S 4 = H
Remark.
The statement isn't correctly formatted. I suppose that the correct statement is
"Let G = S 4 G = S_4 G = S 4 H = A 4 H = A_4 H = A 4 K = { 1 ; ( 1 2 ) ( 3 4 ) ; ( 1 3 ) ( 2 4 ) ; ( 1 4 ) ( 2 3 ) } K = \{1; (1\ 2)(3\ 4); (1\ 3)(2\ 4); (1\ 4)(2\ 3)\} K = { 1 ; ( 1 2 ) ( 3 4 ) ; ( 1 3 ) ( 2 4 ) ; ( 1 4 ) ( 2 3 )}
i) Check that H / K = ⟨ ( 1 2 3 ) H ⟩ H / K = \langle (1\ 2\ 3)H\rangle H / K = ⟨( 1 2 3 ) H ⟩
ii) Check that K K K H H H h ∈ H , h ∉ K h \in H, h \notin K h ∈ H , h ∈ / K h K = K h hK = Kh h K = K h
iii) Check whether ( 1 2 3 4 ) H (1\ 2\ 3\ 4)\mathrm{H} ( 1 2 3 4 ) H ( 1 3 4 2 ) H (1\ 3\ 4\ 2)\mathrm{H} ( 1 3 4 2 ) H S 4 / H S_4 / H S 4 / H
Solution.
H = { 1 ; ( 1 2 ) ( 3 4 ) ; ( 1 3 ) ( 2 4 ) ; ( 1 4 ) ( 2 3 ) ; ( 1 2 3 ) ; ( 1 2 4 ) ; ( 1 3 2 ) ; ( 1 3 4 ) ; ( 1 4 2 ) ; ( 1 4 3 ) ; ( 2 3 4 ) ; ( 2 4 3 ) } . H = \left\{ \begin{array}{c} 1; (1\ 2)(3\ 4); (1\ 3)(2\ 4); (1\ 4)(2\ 3); (1\ 2\ 3); (1\ 2\ 4); (1\ 3\ 2); (1\ 3\ 4); (1\ 4\ 2); \\ (1\ 4\ 3); (2\ 3\ 4); (2\ 4\ 3) \end{array} \right\}. H = { 1 ; ( 1 2 ) ( 3 4 ) ; ( 1 3 ) ( 2 4 ) ; ( 1 4 ) ( 2 3 ) ; ( 1 2 3 ) ; ( 1 2 4 ) ; ( 1 3 2 ) ; ( 1 3 4 ) ; ( 1 4 2 ) ; ( 1 4 3 ) ; ( 2 3 4 ) ; ( 2 4 3 ) } .
i) ( 1 2 3 ) H = { ( 1 2 3 ) ; ( 1 2 3 ) ( 1 2 ) ( 3 4 ) ; ( 1 2 3 ) ( 1 3 ) ( 2 4 ) ; ( 1 2 3 ) ( 1 4 ) ( 2 3 ) ; ( 1 2 3 ) ( 1 2 3 ) ; ( 1 2 3 ) ( 1 2 4 ) ; ( 1 2 3 ) ( 1 3 2 ) ; ( 1 2 3 ) ( 1 3 4 ) ; ( 1 2 3 ) ( 1 4 2 ) ; ( 1 2 3 ) ( 1 4 3 ) ; ( 1 2 3 ) ( 2 3 4 ) ; ( 1 2 3 ) ( 2 4 3 ) } = (1\ 2\ 3)H = \left\{ \begin{array}{c} (1\ 2\ 3); (1\ 2\ 3)(1\ 2)(3\ 4); (1\ 2\ 3)(1\ 3)(2\ 4); (1\ 2\ 3)(1\ 4)(2\ 3); \\ (1\ 2\ 3)(1\ 2\ 3); (1\ 2\ 3)(1\ 2\ 4); (1\ 2\ 3)(1\ 3\ 2); (1\ 2\ 3)(1\ 3\ 4); (1\ 2\ 3)(1\ 4\ 2); \\ (1\ 2\ 3)(1\ 4\ 3); (1\ 2\ 3)(2\ 3\ 4); (1\ 2\ 3)(2\ 4\ 3) \end{array} \right\} = ( 1 2 3 ) H = ⎩ ⎨ ⎧ ( 1 2 3 ) ; ( 1 2 3 ) ( 1 2 ) ( 3 4 ) ; ( 1 2 3 ) ( 1 3 ) ( 2 4 ) ; ( 1 2 3 ) ( 1 4 ) ( 2 3 ) ; ( 1 2 3 ) ( 1 2 3 ) ; ( 1 2 3 ) ( 1 2 4 ) ; ( 1 2 3 ) ( 1 3 2 ) ; ( 1 2 3 ) ( 1 3 4 ) ; ( 1 2 3 ) ( 1 4 2 ) ; ( 1 2 3 ) ( 1 4 3 ) ; ( 1 2 3 ) ( 2 3 4 ) ; ( 1 2 3 ) ( 2 4 3 ) ⎭ ⎬ ⎫ =
\{(1\ 2\ 3); (1\ 3\ 4); (2\ 4\ 3); (1\ 4\ 2); (1\ 3\ 2); \dots \}
There are no class in $H / K$ that has 5 different elements (it has at most 4 elements).
ii) We will check that for all $h \in H, h \notin K$, check that $hK = Kh$:
- $(1\ 2\ 3)H = \{(1\ 2\ 3); (1\ 2\ 3)(1\ 2)(3\ 4); (1\ 2\ 3)(1\ 3)(2\ 4); (1\ 2\ 3)(1\ 4)(2\ 3)\} =$ { ( 1 2 3 ) ; ( 1 3 4 ) ; ( 2 4 3 ) ; ( 1 4 2 ) } and \{(1\ 2\ 3); (1\ 3\ 4); (2\ 4\ 3); (1\ 4\ 2)\} \text{ and } {( 1 2 3 ) ; ( 1 3 4 ) ; ( 2 4 3 ) ; ( 1 4 2 )} and H ( 1 2 3 ) = { ( 1 2 3 ) ; ( 1 2 ) ( 3 4 ) ( 1 2 3 ) ; ( 1 3 ) ( 2 4 ) ( 1 2 3 ) ; ( 1 4 ) ( 2 3 ) ( 1 2 3 ) } = H(1\ 2\ 3) = \{(1\ 2\ 3); (1\ 2)(3\ 4)(1\ 2\ 3); (1\ 3)(2\ 4)(1\ 2\ 3); (1\ 4)(2\ 3)(1\ 2\ 3)\} = H ( 1 2 3 ) = {( 1 2 3 ) ; ( 1 2 ) ( 3 4 ) ( 1 2 3 ) ; ( 1 3 ) ( 2 4 ) ( 1 2 3 ) ; ( 1 4 ) ( 2 3 ) ( 1 2 3 )} = { ( 1 2 3 ) ; ( 2 4 3 ) ; ( 1 4 2 ) ; ( 1 3 4 ) } . \{(1\ 2\ 3); (2\ 4\ 3); (1\ 4\ 2); (1\ 3\ 4)\}. {( 1 2 3 ) ; ( 2 4 3 ) ; ( 1 4 2 ) ; ( 1 3 4 )} .
Hence ( 1 2 3 ) H = H ( 1 2 3 ) (1\ 2\ 3)H = H(1\ 2\ 3) ( 1 2 3 ) H = H ( 1 2 3 )
- ( 1 2 4 ) H = { ( 1 2 4 ) ; ( 1 2 4 ) ( 1 2 ) ( 3 4 ) ; ( 1 2 4 ) ( 1 3 ) ( 2 4 ) ; ( 1 2 4 ) ( 1 4 ) ( 2 3 ) } = (1\ 2\ 4)H = \{(1\ 2\ 4); (1\ 2\ 4)(1\ 2)(3\ 4); (1\ 2\ 4)(1\ 3)(2\ 4); (1\ 2\ 4)(1\ 4)(2\ 3)\} = ( 1 2 4 ) H = {( 1 2 4 ) ; ( 1 2 4 ) ( 1 2 ) ( 3 4 ) ; ( 1 2 4 ) ( 1 3 ) ( 2 4 ) ; ( 1 2 4 ) ( 1 4 ) ( 2 3 )} =
\{(1\ 2\ 4); (1\ 4\ 3); (1\ 3\ 2); (2\ 3\ 4)\} \text{ and }
H(1\ 2\ 4) = \{(1\ 2\ 4); (1\ 2)(3\ 4)(1\ 2\ 4); (1\ 3)(2\ 4)(1\ 2\ 4); (1\ 4)(2\ 3)(1\ 2\ 4)\} =
\{(1\ 2\ 4); (2\ 3\ 4); (1\ 4\ 3); (1\ 3\ 2)\}.
Hence $(1\ 2\ 4)H = H(1\ 2\ 4)$.
- $(1\ 3\ 2)H = \{(1\ 3\ 2); (1\ 3\ 2)(1\ 2)(3\ 4); (1\ 3\ 2)(1\ 3)(2\ 4); (1\ 3\ 2)(1\ 4)(2\ 3)\} =$ { ( 1 3 2 ) ; ( 2 3 4 ) ; ( 1 2 4 ) ; ( 1 4 3 ) } and \{(1\ 3\ 2); (2\ 3\ 4); (1\ 2\ 4); (1\ 4\ 3)\} \text{ and } {( 1 3 2 ) ; ( 2 3 4 ) ; ( 1 2 4 ) ; ( 1 4 3 )} and H ( 1 3 2 ) = { ( 1 3 2 ) ; ( 1 2 ) ( 3 4 ) ( 1 3 2 ) ; ( 1 3 ) ( 2 4 ) ( 1 3 2 ) ; ( 1 4 ) ( 2 3 ) ( 1 3 2 ) } = H(1\ 3\ 2) = \{(1\ 3\ 2); (1\ 2)(3\ 4)(1\ 3\ 2); (1\ 3)(2\ 4)(1\ 3\ 2); (1\ 4)(2\ 3)(1\ 3\ 2)\} = H ( 1 3 2 ) = {( 1 3 2 ) ; ( 1 2 ) ( 3 4 ) ( 1 3 2 ) ; ( 1 3 ) ( 2 4 ) ( 1 3 2 ) ; ( 1 4 ) ( 2 3 ) ( 1 3 2 )} = { ( 1 3 2 ) ; ( 1 4 3 ) ; ( 2 3 4 ) ; ( 1 2 4 ) } . \{(1\ 3\ 2); (1\ 4\ 3); (2\ 3\ 4); (1\ 2\ 4)\}. {( 1 3 2 ) ; ( 1 4 3 ) ; ( 2 3 4 ) ; ( 1 2 4 )} .
Hence ( 1 3 2 ) H = H ( 1 3 2 ) (1\ 3\ 2)H = H(1\ 3\ 2) ( 1 3 2 ) H = H ( 1 3 2 )
- ( 1 3 4 ) H = { ( 1 3 4 ) ; ( 1 3 4 ) ( 1 2 ) ( 3 4 ) ; ( 1 3 4 ) ( 1 3 ) ( 2 4 ) ; ( 1 3 4 ) ( 1 4 ) ( 2 3 ) } = (1\ 3\ 4)H = \{(1\ 3\ 4); (1\ 3\ 4)(1\ 2)(3\ 4); (1\ 3\ 4)(1\ 3)(2\ 4); (1\ 3\ 4)(1\ 4)(2\ 3)\} = ( 1 3 4 ) H = {( 1 3 4 ) ; ( 1 3 4 ) ( 1 2 ) ( 3 4 ) ; ( 1 3 4 ) ( 1 3 ) ( 2 4 ) ; ( 1 3 4 ) ( 1 4 ) ( 2 3 )} =
\{(1\ 3\ 4); (1\ 2\ 3); (1\ 4\ 2); (2\ 4\ 3)\} \text{ and }
H(1\ 3\ 4) = \{(1\ 3\ 4); (1\ 2)(3\ 4)(1\ 3\ 4); (1\ 3)(2\ 4)(1\ 3\ 4); (1\ 4)(2\ 3)(1\ 3\ 4)\} =
\{(1\ 3\ 4); (1\ 4\ 2); (2\ 4\ 3); (1\ 2\ 3)\}.
Hence $(1\ 3\ 4)H = H(1\ 3\ 4)$.
- $(1\ 4\ 2)H = \{(1\ 4\ 2); (1\ 4\ 2)(1\ 2)(3\ 4); (1\ 4\ 2)(1\ 3)(2\ 4); (1\ 4\ 2)(1\ 4)(2\ 3)\} =$ { ( 1 4 2 ) ; ( 2 4 3 ) ; ( 1 3 4 ) ; ( 1 2 3 ) } and \{(1\ 4\ 2); (2\ 4\ 3); (1\ 3\ 4); (1\ 2\ 3)\} \text{ and } {( 1 4 2 ) ; ( 2 4 3 ) ; ( 1 3 4 ) ; ( 1 2 3 )} and H ( 1 4 2 ) = { ( 1 4 2 ) ; ( 1 2 ) ( 3 4 ) ( 1 4 2 ) ; ( 1 3 ) ( 2 4 ) ( 1 4 2 ) ; ( 1 4 ) ( 2 3 ) ( 1 4 2 ) } = H(1\ 4\ 2) = \{(1\ 4\ 2); (1\ 2)(3\ 4)(1\ 4\ 2); (1\ 3)(2\ 4)(1\ 4\ 2); (1\ 4)(2\ 3)(1\ 4\ 2)\} = H ( 1 4 2 ) = {( 1 4 2 ) ; ( 1 2 ) ( 3 4 ) ( 1 4 2 ) ; ( 1 3 ) ( 2 4 ) ( 1 4 2 ) ; ( 1 4 ) ( 2 3 ) ( 1 4 2 )} = { ( 1 4 2 ) ; ( 1 3 4 ) ; ( 1 2 3 ) ; ( 2 4 3 ) } . \{(1\ 4\ 2); (1\ 3\ 4); (1\ 2\ 3); (2\ 4\ 3)\}. {( 1 4 2 ) ; ( 1 3 4 ) ; ( 1 2 3 ) ; ( 2 4 3 )} .
Hence ( 1 4 2 ) H = H ( 1 4 2 ) (1\ 4\ 2)H = H(1\ 4\ 2) ( 1 4 2 ) H = H ( 1 4 2 )
- ( 1 4 3 ) H = { ( 1 4 3 ) ; ( 1 4 3 ) ( 1 2 ) ( 3 4 ) ; ( 1 4 3 ) ( 1 3 ) ( 2 4 ) ; ( 1 4 3 ) ( 1 4 ) ( 2 3 ) } = (1\ 4\ 3)H = \{(1\ 4\ 3); (1\ 4\ 3)(1\ 2)(3\ 4); (1\ 4\ 3)(1\ 3)(2\ 4); (1\ 4\ 3)(1\ 4)(2\ 3)\} = ( 1 4 3 ) H = {( 1 4 3 ) ; ( 1 4 3 ) ( 1 2 ) ( 3 4 ) ; ( 1 4 3 ) ( 1 3 ) ( 2 4 ) ; ( 1 4 3 ) ( 1 4 ) ( 2 3 )} =
{ ( 1 4 3 ) ; ( 1 2 4 ) ; ( 2 3 4 ) ; ( 1 3 2 ) } and \{(1\ 4\ 3); (1\ 2\ 4); (2\ 3\ 4); (1\ 3\ 2)\} \text{ and } {( 1 4 3 ) ; ( 1 2 4 ) ; ( 2 3 4 ) ; ( 1 3 2 )} and
H(143) = \{(143); (12)(34)(143); (13)(24)(143); (14)(23)(143)\} = \{(143); (132); (124); (234)\}.
Hence $(143)H = H(143)$.
- $(234)H = \{(234); (234)(12)(34); (234)(13)(24); (234)(14)(23)\} = \{(234); (132); (143); (124)\}$ and
H(234) = \{(234); (12)(34)(234); (13)(24)(234); (14)(23)(234)\} = \{(234); (124); (132); (143)\}.
Hence $(234)H = H(234)$.
- $(243)H = \{(243); (243)(12)(34); (243)(13)(24); (243)(14)(23)\} = \{(243); (142); (123); (134)\}$ and
H(243) = \{(243); (12)(34)(243); (13)(24)(243); (14)(23)(243)\} = \{(243); (123); (134); (142)\}.
$$
Hence ( 243 ) H = H ( 243 ) (243)H = H(243) ( 243 ) H = H ( 243 )
Therefore K K K H H H
iii) ( 1234 ) H ( 1342 ) H = ( 1234 ) ( 1342 ) H = ( 143 ) H ≠ H (1234)H(1342)H = (1234)(1342)H = (143)H \neq H ( 1234 ) H ( 1342 ) H = ( 1234 ) ( 1342 ) H = ( 143 ) H = H ( 1234 ) H (1234)H ( 1234 ) H ( 1342 ) H (1342)H ( 1342 ) H
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