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This question already has an answer here:

I've done most of the following problem, but I can't seem to get part (d).

Let $G$ be a group of order $pqr$ for primes $p > q > r$. By a counting argument one can see that there is either a normal Sylow $p$-subgroup or a normal Sylow $q$-subgroup (you may assume this). Prove:

(a) $G$ has a normal subgroup $H$ of order $pq$.

(b) Every subgroup of order $p$ or $q$ is contained in $H$.

(c) $G$ has a normal subgroup of order $p$.

(d) $G$ has a subgroup of order $qr$.

Proof: Let me first state the following lemma: if $N_0$ is a subgroup of a finite group $G_0$, and $[G_0 : N_0]$ is equal to the smallest prime which divides $|G_0|$, then $N_0$ is normal in $G_0$. The proof is not difficult. It can be established by a counting argument, or by looking at the natural homomorphism from $G_0$ to the symmetric group on the left cosets of $N_0$ in $G_0$.

For (a), let $N$ be a Sylow $p$-subgroup, and $K$ a Sylow $q$-subgroup. One of these subgroups is normal by assumption. So $H := NK$ is a subgroup of $G$ with order $pq$. Its index in $G$ is $r$, so it is normal in $G$ by the lemma.

(b) If $P$ is a subgroup of order $p$, it is a Sylow $p$-subgroup, so it equals $gNg^{-1}$ for some $g \in G$. So $g^{-1}Pg = N \subseteq H$, whence $P \subseteq gHg^{-1} = H$. Similarly for a subgroup of order $q$.

(c) Let $P$ be a subgroup of order $p$. Then it is a Sylow $p$-subgroup and it is contained in $H$. Since $|H| = pq$, and $[H : P] = q$, automatically $P$ is normal in $H$. If $P_1$ is another Sylow $p$-subgroup of $G$, then it is also contained in and normal in $H$, whence $P = P_1$.

(d) Let $Q$ be a Sylow $q$-subgroup, and $R$ a Sylow $r$-subgroup. If $Q$ is normal in $G$, we are done, since $QR$ is a subgroup of order $qr$. Otherwise, $n_q$ (the number of Sylow $q$ subgroups) is either $r, p,$ or $pr$. We cannot have $n_q = r$, since then $q$ divides $r-1 < q$. If $n_q = p$, then $|N_G(Q)| = qr$ and we are done. The last possibility is that $n_q = pr$. I don't know what to do here. I've tried making a counting argument involving the possibilities for $n_r$, but I haven't been able to make anything work.

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marked as duplicate by Claude Leibovici, Adam Hughes, Tobias Kildetoft, Pedro Tamaroff Jan 2 '15 at 20:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ It's not a duplicate, their $p$ and $r$ are in the opposite order. $\endgroup$ – D_S Jan 2 '15 at 6:45
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    $\begingroup$ You have $p > r > q$ in the title and $p > q > r$ in the body of the question. Which is it? $\endgroup$ – André 3000 Jan 2 '15 at 6:49
  • $\begingroup$ It's $p > q > r$, sorry. $\endgroup$ – D_S Jan 2 '15 at 14:27
  • $\begingroup$ By the Frattini argument (which is an easy application of Sylow's Theorem), the normalizer of a Sylow $q$-subgroup $Q$ has order $qr$ or $pqr$. In the first case you are done, and in the second case $Q$ is normal in $G$, so take the inverse image in $G$ of a Sylow $r$-subgroup of $G/Q$. $\endgroup$ – Derek Holt Jan 2 '15 at 15:12
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D follows easily from the following two theorems:

  1. Every group of square-free order is solvable.
  2. Hall Theorem: A solvable group admit all the possible Hall subgroups.
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  • $\begingroup$ Thanks for answering. The first theorem looks like an easy induction argument, but it seems like Hall's theorem isn't elementary. This problem showed up on a qualifying exam, so do you know if there is an ad hoc way to solve this problem? $\endgroup$ – D_S Jan 2 '15 at 14:33
  • $\begingroup$ @D_S I do not think there is an easy proof of either statement. For an elementary way to show what you need, see the duplicate question mentioned in the comments. $\endgroup$ – Tobias Kildetoft Jan 2 '15 at 15:05
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Ofir's answer is a high powered solution, and here is a more elementary approach as suggested by Derek.

Proof of (d): Let $Q$ be a Sylow $q$-subgroup of $G$. We first claim (this is the "Frattini argument")that $$G = N_G(Q)H$$ If $g \in G$, we know that $Q \subseteq H$ (therefore a Sylow $q$-subgroup of $H$), so $gQg^{-1} \subseteq gHg^{-1} = H$, since $H$ is normal in $G$. Since $gQg^{-1}$ is a Sylow $q$-subgroup of $H$, it is equal to $hQh^{-1}$ for some $h \in H$. Then $h^{-1}gQg^{-1}h$, or $h^{-1}g \in N_G(Q)$. So $g \in HN_G(Q) = N_G(Q)H$.

Therefore, $$|G| = |N_G(Q)H| = \frac{|N_G(Q)| \cdot |H|}{|N_G(Q) \cap H|} = \frac{|N_G(Q)| \cdot pq}{|N_G(Q) \cap H|}$$ so $n_q = [G :N_G(Q)] = \frac{pq}{|N_G(Q) \cap H|}$. The denominator is divisible by $q$, so $n_q$ is either $1$ or $p$.

If $n_q = 1$, then $Q$ is normal in $G$ and $QR$ is a subgroup of order $qr$, where $R$ is a Sylow $r$-subgroup. If $n_q = p$, then $N_G(Q)$ has order $qr$ and is the desired subgroup.

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