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$|G| = pqr$ with $p$, $q$ and $r$ distinct primes. Show G is not simple. I know this might have been asked and answered before. I just wanted someone to tell me if my argument is OK:

Let $|G| = pqr$, and assume $p < q < r$. We have at least one Sylow-$q$ subgroup (call it $Q$) and at least another Sylow-$r$ subgroup (call it $R$). Consider the subgroup $K = $ $<Q, R>$. Now $K$ is not the whole group because there is no element of order $p$ in there. Further, $|K|$ must equal $qr$ because nothing smaller is possible by Lagrange's theorem.

Therefore [G : K] = p, which is the smallest prime dividing G. Hence by a standard theorem (for instance, see corollary 4.5, p. 44, Isaacs' Algebra), K must be normal. Hence $G$ is not simple.

Thanks!

WH

Wulfgang
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  • Why, or in fact: how would you prove $;K;$ contains no element of order $;p;$ ?? – DonAntonio Oct 28 '13 at 05:04
  • My original reasoning was that $K$ is subgroup generated by $Q$ and $R$, which are Sylow-p subgroups of distinct primes $q$ and $r$. If an element of order $p$ distinct from $q$ and $r$ was present in $K$, $p$ would have to divide the order of $K$, which could not include a prime factor, $p$, because of the way it was constructed: $K = <Q, R>$. – Wulfgang Oct 29 '13 at 01:00
  • That's wrong, @Wulfang. For example, $;S_5=\langle (12),,,(12345)\rangle;$ , so according to your argument $;S_5;$ cannot have elements of order $;3;$...yet it does. – DonAntonio Oct 29 '13 at 04:44
  • If $|G| = pqr$, then is not $G \approx Z_p \times Z_q \times Z_r$? How could $<Z_q, Z_r>$ generate the whole group? – Wulfgang Oct 29 '13 at 14:04
  • Well, I just gave you an example of a group of order $;5!=2^3\cdot 3\cdot 5;$ generated by a cyclic group of order two and one of order five. What else do you want?! And no: if $;|G|=pqr;$ then it is not necessarily true that $;G\cong C_p\times C_q\times C_r;$ , as this would imply yhe group is abelian (even cyclic, if the three primes are distinct) ... – DonAntonio Oct 29 '13 at 14:07

2 Answers2

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There is nothing stopping $K$ from being equal to $G$ - ie. $K$ might have elements of order $p$. However, I cannot think of a counterexample of the top of my head. Here is an alternate solution though.

You can do this by counting elements of orders $p, q$ and $r$ : Let $n_s$ denote the number of $s$-Sylow subgroups, for $s\in \{p,q,r\}$. Since any two distinct $s-$Sylow subgroups intersect trivially (why?), it follows that

The number of elements in $G$ of order $s$ is $n_s(s-1)$

Now assume $n_s \neq 1$ for all $s\in \{p,q,r\}$. Then

  1. $n_r \equiv 1\pmod{r}, n_r\mid pq \Rightarrow n_r = pq$

  2. $n_q \equiv 1\pmod{q}, n_q \mid pr \Rightarrow n_q \in\{ r,pr\} \Rightarrow n_q \geq r$

  3. $n_p \equiv 1\pmod{q}, n_p\mid qr \Rightarrow n_p \geq q$

Hence, $$ |G| \geq n_r(r-1) + n_q(q-1) + n_p(p-1) + 1 $$ $$ \Rightarrow pqr \geq pq(r-1) + r(q-1) + q(p-1) + 1 $$ Solving this gives you a contradiction, and so $G$ cannot be simple (in fact, one of these Sylow subgroups must be normal)

Prahlad Vaidyanathan
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  • Many thanks for suggesting an alternate solution. However, since Q and R are Sylow p-subgroups of distinct primes q and r. How could a subgroup <Q, R> thus formed have an element of order p (distinct from q and r) inside <Q, R>? I just want to see if I could still salvage my argument. Thank you again. – Wulfgang Oct 28 '13 at 03:34
  • @Wulfgang : In general, what you say is not true (for instance, $A_5$ is generated by 3-cycles), but in this case it is (because the r-Sylow subgroup turns out to be normal. However, proving that is much harder. – Prahlad Vaidyanathan Oct 28 '13 at 03:39
  • OK, but what about the distinct Sylow-p subgroup in G (call that P). Now P cannot be contained in K = <Q, R>. So, K cannot be the whole group. So, should not [G : K] still be p? – Wulfgang Oct 28 '13 at 03:43
  • Well, if you can prove that $P$ cannot be contained in $K$, you are good to go. But what I mean to say is that you cannot just say that is true because of an order argument - it is more subtle than that. – Prahlad Vaidyanathan Oct 28 '13 at 03:53
  • Thank you again, and for the warning. As a beginning graduate student, I am beginning to see that most "obvious" things (esp. in Algebra) are wrong when details are sought. – Wulfgang Oct 28 '13 at 04:04
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Can you use Feit-Thompson? An odd order simple group must have prime order. So, in this case, the smallest prime must be 2. But if a group has order equal to twice an odd number, it must have a normal subgroup of index two.

Bill Kleinhans
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  • The asker shows a rather elementary approach to try to answer his own question. It seems obvious that (s)he's attending an introductory course in group theory. To use the F-T theorem is like trying to use a bazooka to get rid of a pesky mosquito. – DonAntonio Oct 28 '13 at 05:03
  • Thanks to everyone for helping me. Yes, I am pretty sure I could not quote the F-T theorem in a homework problem without proving it first! – Wulfgang Oct 29 '13 at 00:56
  • I am indepentantly studying algebra from Dummit and Foote. They present the F-T theorem quite early. They imply that the result is so simple and useful, but the proof so difficult, that it would be a shame to limit its use to the very limited audiences who could reproduce it. 255 pages of hard math! How many people could actually reproduce this? – Bill Kleinhans Oct 29 '13 at 04:14
  • My understanding is that the F-T theorem (although quoted by Dummit and Foote) is actually a very deep result--and that the 255 pages are well justified. Thompson won the Fields Medal for this effort. It is simple to state, and the meaning can be understood by anyone with undergrad algebra. That is why Dummit and Foote quote the result. – Wulfgang Oct 31 '13 at 16:16