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Let G be a finite non-abelian group having two distinct primes dividing its order. Is it always true that G has two non-commuting Sylow subgroups?

$\hskip300pt$ Thank you.

EDIT: I asked in a comment if the answer is also no if G is non-nilpotent. The answer turns out now to be affirmative.

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2 Answers

up vote 6 down vote accepted

The answer to the question as it stands, is no. A non-Abelian nilpotent group $G$ of order $p^{a}q^{b}$ where $p,q$ are distinct primes and $a,b$ are positive integers has only two Sylow subgroups (one for $p$ and one for $q$) say $P$ and $Q$ respectively. These commute with each other in the strongest sense that every element of $P$ commutes with every element of $Q$.

Perhaps a more natural condition for subgroups is whether they are permutable. Two subgroups $A$ and $B$ are said to be permutable if $AB = BA$. This does not necessarily mean that all elements of $A$ commute with all elements of $B$. In a finite nilpotent group, any two Sylow subgroups are permutable, and this condition characterizes nilpotent groups (I do not know who was the first to observe this). If $G$ is a finite group which is not nilpotent, $G$ has a Sylow $p$-subgroup $A$ which is not normal for some prime $p.$ Then $G$ must have another Sylow $p$-subgroup $B.$ The set $AB$ has cardinality a power of $p$ greater than $|A|$, so $AB$ is not a a subgroup of $G$ by Lagrange's theorem, so it is not the case that $AB = BA$, and $A$ and $B$ are not permutable.

Later edit: In response to an expansion of the original question, it is possible to go further. If $G$ is a finite group which is not nilpotent, then there are two primes $p$ and $q$ and an element $x \in G$ of order a power of $p$ and an element $y \in G$ of order a power of $q$ such that $x$ and $y$ do not commute. There may be a simpler way to see this, but one way to do it is to use Frobenius's normal $p$-complement theorem. Since $G$ is not nilpotent, there is a prime $p$ such that $G$ has no normal $p$-complement. By the mentioned theorem of Frobenius, there is a $p$-subgroup $P$ of $G$ (which need not be Sylow) such that $N_{G}(P)/C_{G}(P)$ is not a $p$-group. For some prime $q \neq p,$ there is then an element $y$ of $q$-power order in $N_{G}(P) \backslash C_{G}(P)$. There must be an element $x \in P$ such that $xy \neq yx,$ and we have the desired two elements.

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Thank you. Do you think the answer is still no if G is non-nilpotent? Now I see that I was actually looking at non-nilpotent groups when I came up with this question. –  Fred Aug 19 '11 at 7:57
    
@Fred: aswered in edit of original answer –  Geoff Robinson Aug 19 '11 at 8:06
    
Thank you very much. –  Fred Aug 19 '11 at 8:13
    
Nice!$\;\;\;\;$ –  Alex B. Aug 19 '11 at 8:15
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@Fred: I can't speak for Steve, but the way I see it is that there is a prime $q$ such that $N_G(P)$ does not contain a Sylow $q$-subgroup (since $P$ is not normal). Let $Q$ be a Sylow $q$-subgroup, and choose $x \in Q \backslash N_G(P).$ –  Geoff Robinson Aug 21 '11 at 9:21
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I don't know that you mean by "non-commuting Sylow subgroups", but whatever you mean, the answer is no. If you mean subgroups $\text{Syl}_p$ and $\text{Syl}_q$ such that there exist $g\in\text{Syl}_p$ and $h\in\text{Syl}_q$ that don't commute, then any direct product $\text{Syl}_p\times \text{Syl}_q$ of a non-commutative $p$-group by any $q$-group is a counterexample. If you mean subgroups as above such that $\text{Syl}_p\text{Syl}_q\neq \text{Syl}_q\text{Syl}_p$, then the same counterexample works. Finally, if instead of "non-commuting" you meant non-commutative, then any non-abelian semidirect product $C_p\rtimes C_q$ is a counterexample.

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I was playing with very small groups and thought the question could be true. Thanks for answering it. –  Fred Aug 19 '11 at 7:49
    
@Fred You didn't get as far as $D_8\times C_3$ then? :-) That's the smallest counterexample. –  Alex B. Aug 19 '11 at 7:53
    
No. :-( But thanks for letting me know what is the smallest example. I'll work it out. :-) –  Fred Aug 19 '11 at 8:00
    
I really wanted to accept both answers, but Geoff gave us a very nice and complete answer. Sorry. –  Fred Aug 19 '11 at 8:21
    
@Fred no need to be sorry. But next time, if you want to avoid this situation, ask a separate question as a separate question, not as a comment. Otherwise you can end up with five different answers to five different questions in the same thread, and then you would be in real trouble as to what to accept. –  Alex B. Aug 19 '11 at 8:31
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