Prove Group of square-free order has normal Sylow Subgroup by counting elements

Question

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?

Note that the assumption is that each $N_{p_i}$ is $>1$ which is stronger than your inequality. — Tobias Kildetoft, Sep 26 '16 at 10:58
@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? — Isomorphism, Sep 26 '16 at 11:06
Right, which should be more than $|G|$ at least in most cases. — Tobias Kildetoft, Sep 26 '16 at 11:07
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. — Jyrki Lahtonen, Sep 27 '16 at 20:13
And see also this post by Pete L. Clark. If I got it right, this result is from the 40s. — Jyrki Lahtonen, Sep 27 '16 at 20:21

score 3 · Answer 1 · answered Sep 26 '16 at 11:28

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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.

answered Sep 26 '16 at 11:28

spin

12,267

Yes I agree, since we would require $ \sum_{i=1}^{k} (p_{i}-1) \geq p_{2} \cdots p_{k} $ – Isomorphism Sep 26 '16 at 11:42
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Same problem I ran into when trying to answer your question .... – Nicky Hekster Sep 26 '16 at 11:43
@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. – Tobias Kildetoft Sep 26 '16 at 11:45

Prove Group of square-free order has normal Sylow Subgroup by counting elements

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