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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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    See another [thread](http://math.stackexchange.com/q/155785/11619) for a description of the Sylow subgroups of symmetric groups. In some sense this question is a duplicate of that one, but that assumes that you are familiar with a construction called wreath product. May be we should close this question as an abstract duplicate? OTOH small specific cases can be described explicitly without the language of wreath products, and some users may appreciate that?2012-06-10
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    But isn't $2^{19}$ the highest power of two that is a factor of $22!$?2012-06-10
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    $2^1\mid 2,6,10,14,18,22$; $2^2\mid 4,12,20$; $2^3\mid 8$; $2^4\mid 16$. $$6\cdot1+3\cdot2+1\cdot3+1\cdot4=19,$$ so $2^{19}\mid 22!$2012-06-13
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    OK agreed :) $2^{19}$ is the highest power dividing $22!$. thanks heaps :)2012-06-20

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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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    +1 An interesting way of looking at it! Somewhat specific to the case $p=2$ (AFAICT), but a perfect fit for this thread!2012-06-10
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    @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 $S_{p^k}$.2012-06-11
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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 $22!$.