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Describe the structure of the Sylow $2$-subgroups of the symmetric group of degree $22$.

The only thing I've managed to deduce about the structure of $P\in \operatorname{Syl}_p(G)$ is that $|P| = 2^{12}$.

Help please :)

edit: I obviously can't count. Silly, I (for some reason) only counted $8$ as one $2$ and $16$ as one $2$. But as far as I can tell now, $2^{17}$ is the highest power of $2$ dividing $22$.

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    $\begingroup$ But isn't $2^{19}$ the highest power of two that is a factor of $22!$? $\endgroup$ Jun 10, 2012 at 9:14
  • $\begingroup$ $2^1\mid 2,6,10,14,18,22$; $2^2\mid 4,12,20$; $2^3\mid 8$; $2^4\mid 16$. $$6\cdot1+3\cdot2+1\cdot3+1\cdot4=19,$$ so $2^{19}\mid 22!$ $\endgroup$ Jun 13, 2012 at 7:02
  • $\begingroup$ OK agreed :) $2^{19}$ is the highest power dividing $22!$. thanks heaps :) $\endgroup$
    – Chong NZ
    Jun 20, 2012 at 0:50

2 Answers 2

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First let's consider the case of Sylow 2-subgroups for symmetric groups of degree $2^k$. I claim that the Sylow 2-subgroups of $S_{2^k}$ are isomorphic to the automorphism group $G_k$ of $T_k$, the complete, rooted binary tree of depth $k$ (having $2^k$ leaves).

Since an automorphism of $T_k$ is determined uniquely by automorphisms of the two subtrees of the root, together with a decision of whether to exchange the subtrees of the root, we have $|G_k| = 2 |G_{k-1}|^2$. Since $|G_0|=1$, by induction, $|G_k|$ is always a power of 2. Furthermore, $G_k$ maps injectively into $S_{2^k}$ (by sending an automorphism to the permutation it induces on the leaves of $T_k$), so $S_{2^k}$ has a 2-subgroup with the order of $G_k$. Finally, it is easy to check that the order of the Sylow 2-subgroup of $S_{2^k}$ satisfies the same recurrence relation that is satisfied for $|G_k|$. This proves the claim.

For the general case of $S_n$, write $n$ as a sum of distinct powers of 2, and take the direct product of the Sylow 2-subgroups for each. For example, the Sylow 2-subgroups of $S_{14}$ are isomorphic to the direct product of the Sylow 2-subgroups of $S_8$, $S_4$, and $S_2$.

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    $\begingroup$ +1 An interesting way of looking at it! Somewhat specific to the case $p=2$ (AFAICT), but a perfect fit for this thread! $\endgroup$ Jun 10, 2012 at 7:05
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    $\begingroup$ @Jyrki The generalization to arbitrary primes $p$ is: The Sylow $p$-subgroups of $S_{p^k}$ are isomorphic to the automorphisms of the complete, rooted $p$-ary tree of depth $k$ which preserve the cyclic order of the children of each node. We have $|G_k| = p |G_{k-1}|^p$, and the same relation for the Sylow $p$-subgroups of $S_{p^k}$. $\endgroup$
    – Ted
    Jun 11, 2012 at 4:26
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I very much endorse Ted's answer. Just in case you appreciate a concrete list of generators and/or want a very elementary approach, I will give you such a list while walking you through this exercise. The size of the Sylow 2-subgroup of $S_n$ (as a function of $n$) grows whenever $n$ is even. As you observed (by stuyding the highest power of two that divides $n!$) something special happens, whenever we reach a power of two.

Start out with the permutation $g_1=(12)$ that generates the Sylow 2-subgroup $P_1$ of $S_2$. That will also do for $S_3$, but let's spend some time with $S_4$, and see how we can build its Sylow 2-subgroup, call it $P_2$, out of $P_1$. Consider the permutation $g_2=(13)(24)$. Notice that it interchanges the "lower half" ($=\{1,2\}$) with the "upper half" ($=\{3,4\}$) of the set $\{1,2,3,4\}$ elementwise. We see that the permutation $g_1'=g_2g_1g_2^{-1}=(34)$ alone generates a copy of $P_1$, call it $P_1'$, that acts on the upper half, i.e. a Sylow 2-subgroup of $\mathrm{Sym}(\{3,4\})$. The two groups $P_1$ and $P_1'$ act on disjoint subsets (=the two halves), so they commute with each other, and we can form their direct product $P_1\times P_1'$ inside $S_4$. As $g_2$ has order two, repeating the conjugation gives back $g_1=g_2g_1'g_2^{-1}$. This means that $g_2$ is in the normalizer of the group $P_1\times P_1'$. It follows that the subset $$ P_2=(P_1\times P_1')\cup g_2(P_1\times P_1') $$ is closed under multiplication and, hence a subgroup of $S_4$ of the desired size 8. From the above relations it follows that $P_2$ is generated by $g_1$ and $g_2$. We also recover the fact that $P_2$ is the dihedral group: for example $g_1g_2=(1324)$ is a 4-cycle. Do observe that you need to number the vertices of a square in a not the most obvious way to realize $P_2$ as the dihedral group. In what follows we forget about symmetries of geometric objects.

Next up is $S_6$. This is easy, because in addition to $P_2$ we only need to add yet another copy of $P_1$, namely the one generated by $g_1''=(56)$, call this group $P_1''$. The groups $P_2$ and $P_1''$ move different elements of $\{1,2,3,4,5,6\}$, so they commute inside $S_6$, and their direct product $P_2\times P_1''$ is a group of order 16. Hence it must be a Sylow 2-subgroup of $S_6$.

The story really begins with $S_8$. Let us introduce a new permutation $g_3=(15)(26)(37)(48)$. As earlier, we see that this order two permutation intechanges the lower half ($=\{1,2,3,4\}$) and the upper half ($=\{5,6,7,8\}$) of the set $\{1,2,3,4,5,6,7,8\}$ elementwise. Therefore the permutations $g_3g_1g_3^{-1}=(56)$ and $g_3g_2g_3^{-1}=(57)(68)$ generate a copy of $P_2$ acting on the upper half, call it $P_2'$. As in the case of $S_4$ we see that $g_3$ normalizes the direct product $P_2\times P_2'$, and the set $$ P_3=(P_2\times P_2')\cup g_3(P_2\times P_2') $$ is then a group of order $2\cdot8^2=2^7$. As $2^7$ is the highest power of two dividing $8!$, $P_3$ must be a Sylow 2-subgroup of $S_8$. Furthermore, the permutations $g_1,g_2,g_3$ already generate all of $P_3$.

Moving on we skip all the way to $S_{16}$, and introduce yet another permutation of order two $$ g_4=(19)(2A)(3B)(4C)(5D)(6E)(7F)(8G). $$ To avoid any confusion I used single character substitutes for the integers in the range $[10,16]$, so $A=10$, $B=11$, $\ldots, G=16$, and $g_4$ interchanges the lower half (1 to 8) with the upper half (9 to 16). It's the same story again. The group $P_3'=g_4P_3g_4^{-1}$ is a copy of $P_3$ acting on the upper half, i.e. a Sylow 2-subgroup of $\mathrm{Sym}(\{9,10,\ldots,16\})$. Also the set $$ P_4=(P_3\times P_3')\cup g_4(P_3\times P_3') $$ is easily seen to be a subgroup of $S_{16}$. It is generated by $g_1,g_2,g_3,g_4$ and its order is $2\cdot(2^{7})^2=2^{15}$ is just right for it to be a Sylow 2-subgroup of $S_{16}$.

In the end we get a Sylow 2-subgroup of $S_{22}$ as a direct product of three subgroups: a copy of $P_4$ acting on the subset $\{1,2,\ldots,16\}$, a copy of $P_2$ acting on the subset $\{17,18,19,20\}$ and finally a copy of $P_1$ acting on the subset $\{21,22\}$. This group is generated by $g_1,g_2,g_3,g_4$, $(17;18)$, $(17;19)(18;20)$ and $(21;22)$. Its order is $2^{15}\cdot2^3\cdot2=2^{19}$, which is also the highest power of two dividing $22!$.

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  • $\begingroup$ "This means that $g_2$ is in the normalizer of the group $P_1×P′_1$. It follows that the subset $P_2=(P_1×P′_1)∪g_2(P_1×P′_1)$ is closed under multiplication." I don't understand why the second line follows from the first. Can you please explain? $\endgroup$
    – HerrWarum
    Jan 14, 2020 at 15:26
  • $\begingroup$ @HerrWarum If $x_1$ and $x_2$ are elements of $P_1\times P_1'$, then $x_1g_2x_2=g_2((g_2^{-1}x_1g_2)(x_2)$ with the latter factor in $P_1\times P_1'$ due to $g_2$ being in the normalizer. Similarly $g_2x_1g_2x_2\in P_1\times P_1'$. Product of other types are more straightforward. $\endgroup$ Jan 14, 2020 at 17:56

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