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How can I find all of the transitive groups of degree $4$ (i.e. the subgroups $H$ of $S_4$, such that for every $1 \leq i, j \leq 4$ there is $\sigma \in H$, such that $\sigma(i) = j$)? I know that one way of doing this is by brute force, but is there a more clever approach? Thanks in advance!

brick
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  • Well the possible orders of transitive subgroups are $4,8,12,24$. Only $S_4$ and $A_4$ have orders $24$ and $12$. Order $8$ would be a Sylow $2$-subgroup. For order $4$ $H$ is cyclic or a Klein $4$-group, so consider those cases separately. – Derek Holt Mar 27 '15 at 18:14
  • @DerekHolt is there a theoretical reason it cannot have order $6$? – Arthur Mar 27 '15 at 18:56
  • Yeah @DerekHolt because the number of elements in any orbit of an element of ${ 1,2,3,4}$ (which is $4$) has to divide the number of elements in the subgroup. – brick Mar 27 '15 at 19:03

2 Answers2

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Let $G$ be a transitive subgroup of $S_4$. Since the orbit of $1$ under the action of $G$ is $\{1, 2, 3, 4\}$, the order of $G$ must be divisible by $4$, and so must be equal to one of $4, 8, 12, 24$.

An order $4$ $G$ would be either cyclic (generated by a $4$-cycle, giving 3 subgroups) or Klein-Four. There are two $V_4$ subgroups of $D_8$, but only one of them is transitive in each $D_8$, and they're all equal to $\{1, (12)(34), (13)(24), (14)(23)\}$

The order $8$ subgroups are Sylow-$2$'s, so they're all conjugate to each other and isomorphic to $D_8$ (it's easy to find a subgroup isomorphic to $D_8$ by just looking at the symmetries of the vertices of a square). The number of them is either $1$ or $3$ by a Sylow Theorem, and it's $3$ because $D_8$ is not normal in $S_4$.

The order $12$ subgroup must be $A_4$, and the order $24$ subgroup is then $S_4$ itself.

mandella
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Nishant
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  • OK I understand about the cyclic group, $V_4 $, $A_4$ and $S_4$, but I don't know what is $D_8$ and 'Sylow-2'. Maybe there is some theory I'm missing can you explain about these $2$ groups. (or give me some link where I can read about them...) – brick Mar 27 '15 at 18:36
  • $D_8$ is the dihedral group of order $8$, which is the symmetry group of a square. A Sylow-$2$ subgroup of $G$ is a subgroup whose order is the largest power of $2$ dividing the order of $G$., in this case, $8$. Sylow's Theorem says that all such Sylow-$2$ subgroups are conjugate, and the number of Sylow-$2$ subgroups is equal to $1$ mod $2$ and also divides the index of each subgroup, in this case, $3$. This is why there are either $1$ or $3$, and since all of them are conjugate, one of them is normal iff it's the only one. – Nishant Mar 27 '15 at 19:02
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    The number is either 1 or 3 by Sylow's theorem, and it's 4 ... ????? – Derek Holt Mar 27 '15 at 20:28
  • Whoops, it's fixed. – Nishant Mar 27 '15 at 23:51
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We know subgroups of $S_4$ come in only a few different orders: $1,2,3,4,6,8,12,$ and $24$.

If we use the orbit stabilizer theorem, we have that $|H| = |\operatorname{Orb}_H(x)| \cdot |\operatorname{Stab}_H(x)|$; this limits us to only four possible subgroup orders, as $|\operatorname{Orb}_H(x)|$ can only be one thing.

Among those four orders, we can find $5$ non-isomorphic groups fulfilling the criteria. I don't know off-hand how many copies of such groups $S_4$ has offhand, but it shouldn't be too hard to pin that down.

Note that just because two subgroups of $S_4$ may be isomorphic, it doesn't mean that they all (do or don't) act transitively on $\{1,2,3,4\}$. For example, I can only think of one particular Klein four-group in $S_4$ that does, although there are several isomorphic subgroups in $S_4$.

pjs36
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