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Suppose that $p$ is an odd prime. Does every Sylow $p$-subgroup of a finite group contain an element that is not contained in any other Sylow $p$-subgroup? Or does there exist a group $G$ with Sylow $p$-subgroups $P, P_1, \ldots, P_s$ such that $P$ is contained in $P_1 \cup \ldots \cup P_s$?

One immediate observation here is that if a Sylow $p$-subgroup contains an element that is not contained in any other Sylow $p$-subgroup, then the same is true for every other Sylow $p$-subgroup since they are conjugate. Hence we only need to check the statement for one Sylow $p$-subgroup.

I've tried different approaches to this problem but I don't think I have found out anything useful so far. Some special cases where the statement is true is when there are $\leq p + 1$ Sylow $p$-subgroups, or when the Sylow $p$-subgroups are cyclic.

The reason I am assuming that $p$ is odd because there are counterexamples when $p = 2$. One example is given by $\operatorname{PSL}(2,11)$, where every element of a Sylow $2$-subgroup is contained in at least two Sylow $2$-subgroups. Plenty of more examples can be found with GAP, the smallest example seems to be of order $108$. I have checked all groups of order $\leq 1000$ except for those of orders $576$ and $864$. All the examples I've found so far are given by $2$-sylow subgroups.

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2 Answers 2

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A counterexample is ${\rm PSL}_3(7)$ with $p=3$ - I don't know whether that is the smallest. You can check that, if $T \in {\rm Syl}_3(G)$, then $T$ is elementary abelian of order 9, and all of its elements of order 3 are conjugate in $N_G(T)$, so we just need to check that some element $t$ of order 3 is contained in more that one Sylow 3-subgroup. In fact $C_G(t) \cong A_4 \times C_3$ has 4 Sylow 3-subgroups.

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Derek Holt's example is nice because it is an important and well-studied group. However, one actually had to think about its Sylow 3-structure since it had enough elements for each Sylow to get its own (even two of its own!): it has 26068 Sylow 3-subgroups and 52136 elements of order 3. The group itself has 1876896 elements.

A smaller but more extreme example is a semidirect product of some diagonal matrices that has more Sylows 3-subgroups than elements of order 3 (even though its Sylow 3-subgroup is $C_3 \times C_3$, so consists only of elements of order dividing 3).

The example group has order $3^2 \cdot q^4$ and exponent $3q$, exactly $q^4$ Sylow $3$-subgroups, but only $(3^2-1)\cdot q^3+1$ elements of order a power of $3$. Taking $q > 8$ equivalent to 1 mod 3, for instance $q=13$, one gets an example with more Sylows than elements, so there definitely cannot be an element per Sylow. For instance, when $q=13$, we get order 257049, exactly 28561 Sylow 3-subgroups, but only 17576 elements of order 3.

In GAP this group can be constructed as PcGroupCode(1806487088154759822895022317673299049648051327,9*13^4)

Similar examples exist for all primes $p$ (and ranks $r\geq 2$, so that $P\cong C_p^r$; they have order $p^r q^{(p^r-1)/(p-1)}$, exponent $pq$, exactly $q^{(p^r-1)/(p-1)}$ Sylow $p$-subgroups, but only $(p^r-1)q^{p^{(r-1)}}$ elements of order $p$), and a few of their properties are described in https://math.stackexchange.com/a/459878/583.

PcGroupCode(29377269791674521176286309280230372124250662632997733647904724186158072442663109244307500580188835723522039725999999,5^2*31^(5+1)) is an example for $p=5$; it has order $5^2 \cdot 31^6$ and exponent $5 \cdot 31$; it has $31^6$ Sylow 5-subgroups but only $(5^2-1) 31^5$ elements of order 5.

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    $\begingroup$ Two great answers! $\endgroup$ Aug 5, 2013 at 2:13

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