How big must the union of a group's Sylow p-subgroups be?

For various orders $n$ it's a common exercise to prove that a finite group $G$ of order $n$ can't be simple by using the Sylow theorems to show that there is some prime $p \mid n$ such that the number $n_p$ of Sylow $p$-subgroups equals $1$, so the unique Sylow $p$-subgroup is normal. One way these proofs can go is that you show that if $n_p$ isn't equal to $1$, then because $n_p \equiv 1 \bmod p$ it must be very large, so large that there isn't enough room in $G$ for all of its Sylow $p$-subgroups together plus the other Sylow subgroups.

I know how to run this argument if the exponent $a$ of $p$ in $n$ is $1$ and we can show that $n_p = \frac{n}{p}$; in this case the Sylow $p$-subgroups are cyclic, so intersect only in the identity, which means that $G$ has at least $\frac{n}{p}(p - 1)$ elements of order $p$, and hence only room for $\frac{n}{p}$ elements of other orders.

However, I don't know how to run this argument if $a \ge 2$; this came up when I was trying to answer this question and specifically trying to show that a group of order $|G| = 3 \cdot 5 \cdot 7^2 = 735$ can't be simple. The Sylow theorems give that if $n_7 \neq 1$ then $n_7 = 15$, so $G$ has the maximum possible number of Sylow $7$-subgroups. The specific question this gave rise to is:

Specific question: What is the sharpest lower bound on the size of the union of these $15$ Sylow $7$-subgroups?

I wanted to use the Bonferroni inequalities to address this question, using the fact that any two Sylow $7$-subgroups intersect in at most $7$ elements, but something very funny happened: if I apply Bonferroni to all $15$ subgroups I get a lower bound of

$$15 \cdot 49 - {15 \choose 2} \cdot 7 = 0.$$

The problem is that there are too many pairwise intersections between $15$ subgroups. If I instead apply Bonferroni with only $k$ of the $15$ subgroups I get a lower bound of

$$49k - 7 {k \choose 2}$$

which turns out to be maximized when $k = 8$, giving a lower bound of $210$. Is it possible to do better than this? I'm ignoring $7$ of the Sylows!

So the general question is:

General question: What is the sharpest lower bound on the size of the union of the Sylow $p$-subgroups of a finite group $G$ which can be written as a function of the size $p^a$ of such a subgroup and the number $n_p$ of such subgroups? What if $G$ is assumed to be simple?

When $a = 1$ the union has size exactly $(p - 1) n_p + 1$. In general any two Sylows intersect in at most $p^{a-1}$ elements, so Bonferroni with $k$ of the Sylows gives a lower bound of

$$k p^a - {k \choose 2} p^{a-1} = k p^{a-1} \left( p - \frac{k-1}{2} \right)$$

which is maximized when $k \approx p$ as above (or $k = n_p$, if $p$ is more than a little larger than $n_p$). But the smaller $p$ is compared to $n_p$ the less helpful of a bound this will be.

Solution 1:

To answer your specific question, the union of the $15$ Sylow $7$-subgroups must have cardinality $637$ (assuming that such a group existed).

By this question, if a group $G$ has fewer than $p^2$ Sylow $p$-subgroups then the Sylow $p$-subgroups have a common intersection of index $p$. In the case of this specific question, this gives $$7+15\cdot(49-7)=637$$ elements in total.

In general, the number of Sylow $p$-subgroups is of the form $kp+1$, and the cardinality of the union of the Sylow $p$-subgroups is of the form $p^n(t(p-1)+1)$. In the above example, $k=2$ and $t=2$.

There are some simple relationships between $k$ and $t$:

• $t=0\iff k=0$
• $t=1\iff k=1$
• $t\geq2\iff k\geq2$
• The case of pairwise trivial intersection gives the upper bound $t\leq\frac{(p^n-1)}{p^{n-1}(p-1)}k<\frac{p}{p-1}k$
• If $k<p$ then $t=k$ (this follows from the linked MO question)