My question is how can I prove that a Group $G$ of order $|G|=p_{1} \cdots p_{k}$ has a normal sylow subgroup by counting elements ?

For example, let's assume no sylow subgroups are normal in G. Let $N_{p} $ be the number of $p$- sylow subgroups. Then $N_{p_{i}} >1$ for all $i$ as sylow subgroups are conjugate.

Since distinct sylow subgroups intersect trivially, the number of elements of prime order is $\sum_{i=1}^{k} N_{p_{i}}(p_{i}-1) > \sum_{i=1}^{k} (p_{i}-1)$.

Now according to https://arxiv.org/pdf/1104.3831.pdf this gives a contradiction as this sum is $>|G|$, but I fail to see why (however obvious it may be).

Could someone show how to finish the proof via this method if this is indeed correct?

  • $\begingroup$ Note that the assumption is that each $N_{p_i}$ is $>1$ which is stronger than your inequality. $\endgroup$ – Tobias Kildetoft Sep 26 '16 at 10:58
  • $\begingroup$ @TobiasKildetoft OK, well $N_{p_i} $ divides $|G|$ so we can get $ \sum_{i=1}^{k} N_{p_{i}}(p_{i}-1) \geq \sum_{i=1}^{k} p_{1}(p_{i}-1)$ if thats what you mean? $\endgroup$ – Isomorphism Sep 26 '16 at 11:06
  • $\begingroup$ Right, which should be more than $|G|$ at least in most cases. $\endgroup$ – Tobias Kildetoft Sep 26 '16 at 11:07
  • $\begingroup$ That arXiv paper does not look very convincing at the first glance. The references are a bit odd: Jungnickel's paper is fine, but the only other references are to basic textbooks! Not necessarily bad, but a bit strange IMO. Anyway, see Mikko Korhonen's post for another reference, links to the papers, and related discussion. $\endgroup$ – Jyrki Lahtonen Sep 27 '16 at 20:13
  • $\begingroup$ And see also this post by Pete L. Clark. If I got it right, this result is from the 40s. $\endgroup$ – Jyrki Lahtonen Sep 27 '16 at 20:21

I don't think the argument given works.

If $N_{p_i} > 1$ for all $i$, then $N_{p_i} \geq p_i + 1$ by Sylow's theorem. This gives a lower bound $$\sum_{i = 1}^k N_{p_i} (p_i - 1) \geq (p_i + 1)(p_i - 1) = \left( \sum_{i = 1}^k p_i^2 \right) - k$$ But in general this is not $\geq \prod_{i = 1}^k p_i$, for example for $|G| = 2 \cdot 3 \cdot 5 \cdot 7$. I think the argument works if $k = 2$ or $k = 3$, but fails for $k > 3$.

Atleast I believe you need more detailed information on the number of elements of prime order, or the number of Sylow $p$-subgroups etc. In its current form the argument given in https://arxiv.org/pdf/1104.3831.pdf does not work.

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  • $\begingroup$ Yes I agree, since we would require $ \sum_{i=1}^{k} (p_{i}-1) \geq p_{2} \cdots p_{k} $ $\endgroup$ – Isomorphism Sep 26 '16 at 11:42
  • $\begingroup$ Same problem I ran into when trying to answer your question .... $\endgroup$ – Nicky Hekster Sep 26 '16 at 11:43
  • $\begingroup$ @Isomorphism Indeed, I think I was too quick with my initial comments. I had thought the authors had thought it enough through that it should work out, but it does not seem to be the case. $\endgroup$ – Tobias Kildetoft Sep 26 '16 at 11:45

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