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It seems that often in using counting arguments to show that a group of a given order cannot be simple, it is shown that the group must have at least $n_p(p^n-1)$ elements, where $n_p$ is the number of Sylow p-subgroups. It is explained that the reason this is the case is because distinct Sylow p-subgroups intersect only at the identity, which somehow follows from Lagrange's Theorem. I cannot see why this is true. Can anyone quicker than I tell me why, I know it's probably very obvious.

Note: This isn't a homework question, so if the answer is obvious I'd really just appreciate knowing why. Thanks!

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Yeah, this is definitely not true in general, and now that I look back at what I've been studying, this trick is only used in the case for a subgroup of order p. Thanks everybody! –  Jon Beardsley Mar 2 '11 at 23:28
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There are some groups where the Sylow p-subgroups intersect trivially even though they have order greater than p. These groups are very special and somewhat important. For p=2, Suzuki discovered exotic new simple groups by looking at such a configuration. –  Jack Schmidt Mar 3 '11 at 1:27

4 Answers 4

up vote 9 down vote accepted

That's because it is not true in general. Look at $2$-Sylows in $S_5$: they have nontrivial intersection.

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Suppose $P$ and $Q$ are Sylow p-subgroups of prime order p (so not just any power of p; as others remarked, then it is not true in general). Note that $P\cap Q$ is a subgroup of $P$ (and of $Q$). So by Lagrange, the order $|P\cap Q|$ divides p. As p is prime, it is 1 or p. But it cannot be p, as $P$ and $Q$ are distinct. So $|P\cap Q|=1$ and consequently the intersection is trivial.

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But the order of $P$ and $Q$ may be $p^n$ for some $n$, right? –  Jon Beardsley Mar 2 '11 at 23:22
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but if the p group has order $p^n, n\geq2$ this doesnt work and in fact the claim in the question isnt true. –  yoyo Mar 2 '11 at 23:23
    
Okay thankyou, that clears things up then! –  Jon Beardsley Mar 2 '11 at 23:25

In some situations, to prove that groups of order $n$ cannot be simple, you can use the counting argument if all Sylow subgroups have trivial intersection, and a different argument otherwise.

For example let $G$ be a simple group of order $n=144 = 16 \times 9$. The number $n_3$ of Sylow 3-subgroups is 1, 4 or 16. If $n_3 = 1$ then there is normal Sylow subgroup and if $n_3= 4$ then $G$ maps nontrivially to $S_4$, so we must have $n_3 = 16$.

If all pairs of Sylow 3-subgroups have trivial intersection, then they contain in total $16 \times 8$ non-identity elements, so the remaining 16 elements must form a unique and hence normal Sylow 2-subgroup of $G$.

Otherwise two Sylow 3-subgroups intersect in a subgroup $T$ of order 3. Then the normalizer $N_G(T)$ of $T$ in $G$ contains both of these Sylow 3-subgroups, so by Sylow's theorem it has at least 4 Sylow 3-subgroups, and hence has order at least 36, so $|G:N_G(T)| \le 4$ and $G$ cannot be simple.

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It simply isn't true, Sylow p-subgroups can very well intersect non-trivially, Plop gave an example thereof.

Well, it seems like you actually cannot say the following, see comments. I'm just leaving it here as a mistake one shouldn't make, so I won't mind if a moderator deletes it since it's not an actual answer.

[wrong]You could say that the number of elements of order $p$ is at least $n_p(p^n-p^{n-1})+p^{n-1}$, which is the case when all $p$-groups intersect maximally. [/wrong] Note that in this case however the intersection of all Sylow $p$-subgroups is a normal subgroup (even a characteristic subgroup, it is called $\mathbf O_p(G)$), so this cannot occur in a simple group.

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S5's Sylow 2-subgroups actually overlap a lot. The union of the Sylow 2-subgroups of S5 has 1+10+15+30=56 elements, but there are 15 Sylow 2-subgroups of order 8. 15*(8-4)+4=64 is actually larger than 56, the number of 2-elements. –  Jack Schmidt Mar 3 '11 at 1:19
    
This being a comment to my answer, I'm presuming you mean the non-trivial intersection of those sylow 2-subgroups is in contradiction with myself saying this could indeed be the case? –  Myself Mar 3 '11 at 1:22
    
Hehe, I'm not sure I can parse that. I just mean in the first sentence of the second paragraph: a group need not have as many elements of order (a power of) p as you said, because Sylow p-subgroups can overlap like crazy. Your number is 64, but the real number is only 56. –  Jack Schmidt Mar 3 '11 at 1:29
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Indeed, theorem 1.38 there states that if $D = S\cap T$ is the minimal such intersection of two Sylow p-subgroups, then then $\mathbf O_p(G)$ is the unique largest subgroup of $D$ that is normal in both $S$ and $T$. And there are other fascinating theorems in that wonderful book about the world of Sylow-intersections that turns out to be even more interesting than I believed. –  Myself Mar 3 '11 at 1:55
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I found a family of finite groups with q³ Sylow 2-subgroups but only 3q²+1 elements of order a power of 2, so it looks like the ratio of p-elements to Sylow p-subgroups can be arbitrarily small ((p²−1)/q + 1/q³). –  Jack Schmidt Mar 3 '11 at 19:17

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