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Show that the subgroup $V=\{e, (12)(34), (13)(24), (14)(23)\}$ is normal in $S_4$. Make a multiplication table for the quotient group of $S_4$ and $V$

Proving that $V$ is a normal subgroup is not a problem, as each element is it's inverse, and conjugating gives something in the subgroup V. The problem I am having is doing the multiplication table.

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  • $\begingroup$ You must first choose a system of representatives of the cosets. $\endgroup$ – Bernard Mar 8 '15 at 19:35
  • $\begingroup$ Hint: the order of the quotient group is $24/4 = 6$. Do you know any symmetric groups of order 6? This might help find a transversal (system of representatives). $\endgroup$ – David Wheeler Mar 9 '15 at 5:43
  • $\begingroup$ I have an attempt at a solution up below in a couple of seconds if you want to take a look. $\endgroup$ – All About Groups Mar 10 '15 at 0:50
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The elements of the quotient group $S_4/V$ are the cosets of $V$ in $S_4$. So first I'd determine what elements of $S_4$ are in each of those six cosets and give them labels. For example $(1,2)$ is in the coset $B=\{(1,2),(3,4),(1,3,2,4),(1,4,2,3)\}.$ Remember that the cosets are disjoint.

From there start the multiplication table. To multiply the cosets pick a representative from each and multlitply in $S_4$. The choice of element in the coset doesn't matter as $V$ is normal so the quotient is well defined. For example to find $B*B$ pick a representative element from each, in this case, say, $(1,2)*(1,2)=e.$ Since $e$ is in $V$ we find that $B*B=V$.

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\begin{bmatrix} \ast& e & (12)(34) & (13)(24) & (14)(23) \\\hline\hline e & e & (12)(34) & (13)(24) & (14)(23) \\\hline (12) & (12) & (34) & (1423) & (1324) \\\hline (13) & (13) &(1432) & (24) & (1234) \\\hline (23) & (23) &(1243) &(1342) & (14) \\\hline (132) & (132) & (143) & (234) & (124) \\\hline (123) & (123) & (243) & (143) & (134)\end{bmatrix}

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