Distinct Sylow $p$-subgroups intersect only at the identity, which somehow follows from Lagrange's Theorem. Why?

Question

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!

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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.