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Let $G$ be a finite group and assume $G$ has a single $p$-Sylow subgroup. Let $q\neq p$ be prime for which $q$ divides the order of $G$. Show that there exists a subgroup of $G$ with order $pq$.

Following are some of my attempts at solving this:

if $P<G$ is a $p$-Sylow group then $P\vartriangleleft G$ from Sylow's theorems.

From Cauchy's theorem we have a $Q<G$ cyclic such that $\mid{Q}\mid = q$

Because $P$ is normal in $G$ I know that $PQ < G$, let's assume $\mid{PQ}\mid = p^{k}q$ for $k\geq 1$

My attempt mainly consisted from this point on taking a group of order $p$ from $PQ$, say $P'$ and showing that $P'Q<PQ$. I know that $\mid{P'Q}\mid = pq$ so that would conclude my proof (if it were the right approach, that is).

This approach didn't work for me because I can't show that if $h\in Q$ and $g \in P'$ then $gh = h'g' \in P'Q$. I do know that $gh = hg' \in QP$ because $P$ is also normal in $QP=PQ$.

Am I even in the right direction? I am not interested in a full proof but merely in a (good solid) hint.

Many thanks.

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Nice Question. I was deceived! –  user21436 Jan 21 '12 at 16:15

3 Answers 3

up vote 4 down vote accepted

You can't prove that there is a subgroup of order $pq$ in general, with just these hypotheses. For example, consider the case when $G$ is the alternating group $A_4$ of order $12,$ with $p=2,q = 3.$ Then $G$ has a single Sylow $2$-subgroup, but $G$ has no subgroup of order $6.$ Similar examples can be constructed for every choice of primes $p$ and $q$ such that $q$ does not divide $p-1.$

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+1 It's always good to go through a list of small groups to see if a claim fails, or to see if a proof in a specific case helps you prove it in general. –  Dylan Moreland Jan 21 '12 at 16:03
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I feel deceived. I spent a good amount of time trying to prove this. –  dankilman Jan 21 '12 at 16:14

You were pretty much on the right track. Except that you need to notice the following:

To Prove that $PQ$ is a subgroup of $G$:

Given $P\trianglelefteq G$ and $Q<G$, We claim that $PQ=\{pq|p\in P; q \in Q\}$ is a subgroup of $G$.

There are two ways of looking at this: One way of doing is to prove that $PQ$ is a subgroup if and only if $PQ=QP$.(Try Proving this and then show that this is true here!) Another way is to show through direct methods, which I'll show here:

  1. Note that identity of $G$ exists here in $PQ$.
  2. If $pq, p_1q_1 \in PQ$, $pq(p_1q_1)^{-1}=pqq_1^{-1}p^{-1}=p\cdot qq_1^{-1}p^{-1}(qq_1^{-1})^{-1}\cdot qq_1^{-1}$ Note that the term sandwiched between the dots is an element in $P$, because, $P$ is a normal subgroup in G. So, this completes the proof, thanks to the subgroup Test.

If $P$ and $Q$ are subgroups of $G$, then the subgroup $PQ$ is of cardinality, $$|PQ|=\dfrac{|P|\cdot |Q|}{|P \cap Q|}$$

Since $P$ and $Q$ are groups with co-prime orders, they intersect trivially.

(Does this not require a proof? It certainly does! Consider an element $x$ from the intersection $P \cap Q$, Since $P\cap Q$ is both a subgroup of $P$ and $Q$, by Lagrange's Theorem, $o(x)|p$ and $o(x)|q$ This forces that $o(x)=1$ which necessarily means that $x$ is the trivial element.)

So, $|P \cap Q|=1$ and hence $|PQ|=|P|\cdot |Q|=pq$.

This completes the proof!

I would like to point a few glaring errors in your way of proving.

My attempt mainly consisted from this point on taking a group of order $p$ from $PQ$, say $P'$ and showing that $P'Q<PQ$. I know that $∣P'Q∣=pq$ so that would conclude my proof (if it were the right approach, that is).

How will this prove that $k=1$. Are there not groups of order $p^2q$ with subgroups of order $pq$?

This approach didn't work for me because I can't show that if $h\in Q$ and $g\in P'$ then $gh=h'g'\in P'Q$. I do know that $gh=hg'\in QP$ because $P$ is also normal in $QP=PQ$.

Where are all these elements $(\cdot)'$ coming from? It's hard to comprehend that! From the sentence that, $gh=hg'\in QP$, I believe $g'=h^{-1}gh$.

The rest is fine!

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How can I know that $PQ$ is actually a group and not just a set of elements from $G$? My failed attempt that I described above consisted mainly of me not being able to show exactly this. –  dankilman Jan 21 '12 at 15:24
    
Oh I'll add that for you. Thank you for letting me know your difficulty exactly, @dankilman –  user21436 Jan 21 '12 at 15:33
    
Thanks for hanging on. let me clarify. First $\mid{PQ}\mid = p^{k}q$ for k≥1 and I make no assumption on $k$, in fact if $k = 1$ this is trivial. when I talk about $P'$ I am talking about a group of order $p$ which I know exists in $PQ$ from Cauchy's theorem (in fact I know that that exists an element of order $p$). same goes for $Q$. The rest which you mentioned as hard to comprehend resovled arround my attempt at showing that $P'Q < PQ$ –  dankilman Jan 21 '12 at 15:43
    
one more thing, $g' = hgh^{-1}$ because $P$ is normal in $PQ$ but i know that $g' \in P$. I can't be certain that it is in $P'$ –  dankilman Jan 21 '12 at 15:47
    
@dankilman I don't quite follow you, but is my proof clear? In fact, we can construct groups from the scratch with this kind of property with what is called the Semi Direct Products! –  user21436 Jan 21 '12 at 16:00

so first off, what makes you assume that the order of your q-sylow subgroup is of order q? Also you have that your sylow subgroups must have a non-trivial center. This would be an abelian p-subgroup of P, so it must have a subgroup of order p. The same goes for Q. Not sure if that leads to a proof, but it should at least be on the right track. You seem to be on the right track though, looking at product of subgroups. As for the last comment, thats definitely an important fact but does not constitute a proof, as you have not found subgroups of order p and q.

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I know that this $q$-sylow group is of order $q$ because $\mid{PQ}\mid = p^k\cdot q$. and I am looking at $PQ$ and not $G$. I know that I have subgroups of order $p$ and $q$ in $PQ$ because of Cauchy's theorem (which are even cyclic). My problem is at showing that $P'Q < PQ$ where $P'$ is a group of order $p$ and $Q$ is a group of order $q$. –  dankilman Jan 21 '12 at 15:36
    
sorry the comment was meant to go above –  dankilman Jan 21 '12 at 15:46

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