Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
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
up vote 10 down vote accepted

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

share|cite|improve this answer

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.

share|cite|improve this answer
But the order of $P$ and $Q$ may be $p^n$ for some $n$, right? – Jon Beardsley Mar 2 '11 at 23:22
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.

share|cite|improve this answer

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.

share|cite|improve this answer
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
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.