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I'm Very confused as to what my lecturer means in the final few lines of a proof of one of sylows theorems means. The theorem in question is the one that says the number of sylow P-subgroups is 1+kp.

Here's the part of the proof I'm confused about :

Let S be the set of all subsets of order $p^\alpha$ let $A\in S$ and let G act on it by right multiplication

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It follows then that the number of P-sylow subgroups is the number of orbits of S of size m . let $n_p$ be the number of p sylow subgroups then

$$ \begin{aligned} &\ |S|=\sum_{\text{orbits of size m}}|O|+\sum_{\text{orbits size >m}}|O|\\ \Rightarrow &\ |S|=n_pm+mpr\\ \Rightarrow &\ \binom{mp^\alpha}{p^\alpha}/m=n_p+pr \\ \Rightarrow &\ \binom{mp^\alpha}{p^\alpha}/m\equiv n_p\pmod p \end{aligned}$$

But we know that the cyclic group of order $mp^{\alpha}$ has exactly one subgroup of order $p^{\alpha}$ Thus $ \binom{mp^\alpha}{p^\alpha}/m\equiv1\pmod p$ and so $n_p\equiv1\pmod p$

I understodd the rest of the proof just fine , but this has really stumped me. for starters I'm not familiar with the notation he seems to be using to describe the size of S( the brackets term). secondly I don't understand how he made the jump $\binom{mp^\alpha}{p^\alpha}/m\equiv1\pmod p$ and so $n_p\equiv1\pmod p$

Could anyone plese help me to understand this ?

Note: the dots represent part of the proof I left out if needs be I can add it for context but I thought it seemed superfluous given that this little part is the source of all my confusion.

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  • $\begingroup$ Not sure I can help (yet), but I'm curious how $m$ is defined. $\endgroup$ – Chris Custer Dec 23 '18 at 3:51
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    $\begingroup$ When you say "bracket term" do you mean $\binom{mp^a}{p^s}$? That's a binomial coefficient. $\endgroup$ – Lord Shark the Unknown Dec 23 '18 at 5:01
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    $\begingroup$ @ChrisCuster No doubt the order of $G$ is $mp^a$ where $p\nmid m$. $\endgroup$ – Lord Shark the Unknown Dec 23 '18 at 5:01
  • $\begingroup$ Yeah. I figured something like that. I'm still a little stuck on the first part. @LordSharktheUnknown $\endgroup$ – Chris Custer Dec 23 '18 at 5:07
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    $\begingroup$ The point seems to be that any group can be used, since nothing but the order was used to get there. Clearly $n_p=1$ for the cyclic group. Maybe this clears it up. $\endgroup$ – Chris Custer Dec 23 '18 at 16:43

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