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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.2012-11-08
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    @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.2012-11-08
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    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.2012-11-08
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    Hint: Take $Q$ be maximal among the intersections of $p$-Sylow subgroups. How many $p$-Sylow subgroups contain $Q$, and what do you know about their intersections? [Apply Sylow's theorem to $N_G(Q)$.]2012-11-09
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    @jug: that might be a little complicated since Q need not be normal in the Sylows that contain it (A6 wr S2, p=2 for example).2012-11-09
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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.2012-11-09
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    @Derek: thanks. I hadn't been able to prove they were in distinct Sylows yet (busy with classes), but had not found any counterexample. The obvious proof that they were in fact Sylows is flawed, but perhaps there is another obvious proof :-) Ah, and Miller and M.K. give such an easy proof. Marvelous.2012-11-09
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    @DerekHolt: Yes, this was the idea.2012-11-09
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    @JackSchmidt: They are all in Sylows. If they are not in distinct Sylows, the intersection is not maximal.2012-11-09
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    @Jug: thanks! That is very simple and very useful. Every p-group containing a maximal sylow intersection is contained in a unique sylow, by definition of maximal. I was only considering subgroups that were intersections themselves, which was a bad perspective.2012-11-09
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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).2012-11-10

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