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I am trying to prove first Sylow Theorem using the Lemma: if $G$ is a finite group such that has a Sylow $p$-subgroup and $H\subset G$, then $H$ has a Sylow $p$-subgroup.

The way I want to go about proving first Sylow Theorem using lemma is to notice that any finite group $G$ injects into $S_{|G|}$.

Given the Lemma, it suffices to show that $S_{|G|}$ has a Sylow $p$-subgroup. If $p^k$ is the highest power of $p$ such that $p^k| |G|$, then I can easily see that $\langle(123\cdots p^k)\rangle\subset S_{|G|}$ is a subgroup with order $p^k$. But this doesn't finish the proof as $|G|!$ may be divisible by $p^{k+1}$, which then means this subgroup is not a Sylow $p$-subgroup of $S_{|G|}$.

Would there be hints about what steps I should make in this way of proving the first Sylow Theorem?

kabenyuk
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mathlearner98
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  • Thanks for your comment, but I was more seeking a proof of the fact that every $S_n$ has a $p-$Sylow subgroup. Because if I can show this, then I get an alternative proof of first Sylow theorem other than shown in the post that you suggeste – mathlearner98 Mar 14 '22 at 21:24
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    You should like divide your steps into a list , like 1. , 2. , 3... So it becomes easier to understand but that is just my opinion. I've done some editing to make it clear to someone else that you require help on alternate method – Clemens Bartholdy Mar 14 '22 at 21:34
  • @Buraian that is the symmetric group $S_n$, the group of permutations on a set of $n$ elements, where in this case $n = |G|$ is the size of the group in question. – Rob Mar 14 '22 at 21:38
  • I think you've made a mistake or have some fault in reasoning. If $p^k $ divides $|G|$ then we can take it that $p$ is some number less than $G$. So, when we take $|G|!$ that product contains $G$ and all number less than it, meaning that $p$ mentioned before should come into the $p$ necessarily making it that $|G|!$ is at least divisible by $p^{k+1}$ – Clemens Bartholdy Mar 14 '22 at 21:51
  • What I meant by $\langle(123\cdots p^k)\rangle$ is the subgroup of $S_{|G|}$ generated by the element $(123\cdots p^k)\in S_{|G|}$. If $|G|$ is divisible by $p^k$, then we have $p^k\leq |G|$, so $(123\cdots p^k)$ is an element of $S_{|G|}$ – mathlearner98 Mar 14 '22 at 22:05
  • with order $p^k$. – mathlearner98 Mar 14 '22 at 22:05
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    Serre prefers to do this by embedding $H$ not into $S_n$ where $n = |H|$, but instead into ${\rm GL}_n(\mathbf F_p)$ by having $H$ act as permutations of the basis of $\mathbf F_p^n$ (index basis by elements of $H$), since it is much cleaner to describe a $p$-Sylow subgroup of ${\rm GL}_n(\mathbf F_p)$ then a $p$-Sylow subgroup of $S_n$. – KCd Mar 14 '22 at 22:37
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    I see that indeed resolves my question thanks. – mathlearner98 Mar 14 '22 at 23:11
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    If you want to know more about the Sylow subgroups of symmetric groups, then I recommed Ted's explanation. You may want to learn about wreath products, but the case $p=2$ is simpler, and the generalization to other primes is not too difficult either. For example, the permutations $(123)$ and $(147)(258)(369)$ generate a Sylow $3$-subgroup of $S_9$ etc. – Jyrki Lahtonen Mar 15 '22 at 04:43

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