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Find a Sylow 2-subgroup of $S_5 $ and calculate how many sylow 2-subgroups of $S_5$ there are.

I have the answers but I'm having trouble understanding them.

By prime decomposition I can see it will have order $2^3$=$8$, but then it states

$|C_{S_4}((12)(34))|= \frac{S_4}{|C_{S_4}((12)(34))|} =\frac{4!}{|(12)(34),(13)(24),(14)(23)|} = \frac{24}{3}=8$

How do I know from finding the order 8 to knowing it is the cyclic group of two disjoint 2-cycles? If anyone could explain step-by-step what is happening, I'd greatly appreciate it.

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Let's start by finding one subgroup of order $8$. Since this will have the correct order, it will be a Sylow $2$-subgroup. Note that $S_5$ is the group of permutations of $\{1,2,3,4,5\}$. If we view $\{1,2,3,4\}$ as the vertices of a square, and consider the dihedral group of that square, then we have identified a subgroup of order $8$.

Since all Sylow subgroups of a given order are conjugate, this means that all Sylow $2$-subgroups of $S_5$ are isomorphic to $D_8$, the dihedral group of a square. The others are found by choosing different subsets of $\{1,2,3,4,5\}$ for the vertices of the square. Also note that reordering the vertices as, for example, $1,2,4,3$ versus $1,2,3,4$ gives us a different subgroup.

There are $\displaystyle {5 \choose 4} = 5$ ways to choose $4$ vertices from $5$ elements. For each choice of $4$ vertices, there are $6$ ways to order them (assuming we view cyclic shifts such as $1,2,3,4$ and $2,3,4,1$ as the same ordering), resulting in $3$ distinct dihedral groups (because "opposite" pairs of orderings such as $1,2,3,4$ and $1,4,3,2$ result in the same group; just flip the square upside-down to go from one ordering to the other). Thus there are $5 \cdot 3 = 15$ distinct subgroups of order $8$.

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  • $\begingroup$ Thank you for explaining it so clearly, that's very helpful. $\endgroup$ – user377174 Dec 2 '16 at 21:21
  • $\begingroup$ @smallscot My pleasure, I'm glad it helped. This approach won't work in general (the geometric interpretation of an arbitrary Sylow subgroup of $S_n$ may not be obvious), but in this case it does, and it gives nice insight into what is going on. $\endgroup$ – Bungo Dec 2 '16 at 21:29

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