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Suppose $G$ is a finite group with Sylow $p$-subgroups of order $p^n$. I am trying to show that if there are $\geq p+1$ Sylow $p$-subgroups, then the union of all these subgroups has order $\geq p^{n+1}$.

In the case where there are exactly $p+1$ Sylow $p$-subgroups, we can find a homomorphism $G \rightarrow S_{p+1}$, which has kernel $K$ of size $p^{n-1}$ since $(p+1)!$ is divisible by $p$ but not by $p^2$. Then if $P$ is a Sylow $p$-subgroup, the subgroup $PK$ has order power of $p$, hence $|PK| \leq p^n$ and by the order formula for $PK$ we find that $|P \cap K| \geq p^{n-1}$. Therefore $K \leq P$. Since $K$ is contained in every Sylow $p$-subgroup, the intersection of all Sylow $p$-subgroups has order $p^{n-1}$. Thus there are $(p+1)(p^n - p^{n-1}) + p^{n-1} = p^{n+1}$ elements in the union of all Sylow $p$-subgroups.

Now I'm stuck. How can I handle the case where there are $> p+1$ Sylow $p$-subgroups?

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The last equality in your second paragraph is pretty special. Be careful that if P/K is not cyclic of order p, a similar formula need not hold. –  Jack Schmidt Nov 8 '12 at 19:07
    
@JackSchmidt: Sorry, what do you mean by "special"? My idea there was that in each sylow subgroup there are $p^n - p^{n-1}$ elements not contained in any other sylow subgroup, and the rest of the elements are in every sylow subgroup. –  spin Nov 8 '12 at 19:14
    
yes and such a calculation requires that Sylow p-subgroups overlap in this very special way. In general there can be more Sylow subgroups than there are elements inside the union of the Sylow subgroups. There need not be ANY elements not contained in any other Sylow subgroup. –  Jack Schmidt Nov 8 '12 at 19:38
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@Jack: But $Q$ is contained in more than one Sylow subgroup, and it is properly contained in the normalizer of each of these, so $N_G(Q)$ has more than one Sylow subgroup, and hence at least $p+1$, and each of these is contained in a distinct Sylow subgroup of $G$. I think this does work. –  Derek Holt Nov 9 '12 at 16:59
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The idea of the "maximal Sylow intersection" proof is nice, you can also use it to give a short proof of the fact that groups of order $p^nq$, ($p$, $q$ primes) are solvable (it's also due to Miller I think, and can be found in the book I refer to in my answer). –  Mikko Korhonen Nov 10 '12 at 15:34
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1 Answer

There is a proof of this fact in Theory and Applications of Finite Groups by G. A. Miller, H .F. Blichfeldt and L. E. Dickson, in the chapter about Frobenius' theorem (pages 79-80). They actually prove a stronger result as well: when there are $> p+1$ Sylow $p$-subgroups, then the number of elements in the union of Sylow $p$-subgroups is $> p^{n+1}$. But here's a proof of just the fact asked about, using the ideas in Miller/Blichfeldt/Dickson (See here and here for the relevant pages from the book).

Let $P_1$ and $P_2$ be distinct Sylow $p$-subgroups such that their intersection $D = P_1 \cap P_2$ has largest possible order, say $|D| = p^k$. Let $g \in N_{P_1}(D)$ such that $g \not\in D$. The existence of such $g$ is guaranteed since $p$-groups satisfy the normalizer condition.

Then $g$ has order $\geq p$ and $g \not\in P_2$, so there are at least $p$ distinct conjugates of the form $g^i P_2 g^{-i}$. Each of these conjugates contains $D$ since $g^i D g^{-i} = D$, thus by the maximality of $D$ the intersection of these conjugates is equal to $D$.

When we include $P_1$ among these conjugates, we get a family of at least $p+1$ distinct Sylow subgroups which have the common intersection $D$. Thus there are $$(p+1)(p^n - p^k) + p^k = p^{n+1} + p^n - p^{k+1} \geq p^{n+1}$$ elements among this family.

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