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Given the symmetric group $S_4$ and a subgroup $H\subset S_4$ I want to construct a Young subgroup $Y\subset S_4$ such that $Y$ is minimal, meaning that there is no other young subgroup $Y'$ such that $H\subset Y'\subset Y$. My understanding of the problem is in two ways:

1) Consider a partition of the set $\{1,2,3,4\}$. A Young subgroup is the direct product of the symmetric groups on the components of the partition. So the Young subgroup $Y$ must contain all combinations of all permutations on all subsets forming the partition. So we need only to complete the list of permutations that are missing in $H$ to obtain $Y$. For example, consider $H=\{1,(12)(34)\}$ then this subgroup "corresponds" to the partition $\{1,2\}\cup\{3,4\}$, and to obtain $Y$ we just need to add to $H$ the permutations $(12)$ and $(34)$.

2) The second way to do this is to write every permutation in $H$ as a product of disjoint cycles then to write every cycle as a product of transpositions. Then $Y$ will be the subgroup generated by all these transpositions. For example if $H=\{1,(134),(143)\}$ then we write $H=\{1,(13)(14),(14)(13)\}$ and then $Y$ is just the subgroup generated by $(13)$ and $(14)$ which is isomorphic to $S_3$.

Are my thoughts correct? and is there another way to solve the problem?

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1 Answer 1

It might be easier to think in terms of the orbits of the actions of $H$ and $Y$ on the set $\{1,2,3,4\}$, as the $Y$-orbits must contain the $H$-orbits (since $H$ is a subgroup of $Y$).

The $Y$-orbits are simply the partition of $\{1,2,3,4\}$ which defines $Y$. This obviously agrees with your answer to the first example. For the second example you wrote there that $Y$ would be the group generated by $(13)$ and $(14)$, namely the subgroup of $S_4$ that fixes 2. This agrees with the observation that the orbits under $H$ are $\{1,3,4\}$ and $\{2\}$, and this set of orbits is the partition that defines $Y$.

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