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I proved that $G$ isn't simple but I am having troubles with how to prove it that the normal subgroup has order $p$,$q$, or $r$.

p Groups
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  • How are you intending to show $G$ is not simple if not by showing there exist such normal subgroups? – Pedro Oct 05 '16 at 22:58

3 Answers3

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Let $n_s$ be the number of $s$-Sylow subgroups of $G$ for $s \in \{p,q,r\}$. By Sylow's Theorem, for each $s$ we have that $n_s$ divides $|G| = pqr$ and $n_s \equiv 1 \pmod{s}$. In particular, $s$ does not divide $n_s$. One way of proceeding is to show that there is at least one $s$-Sylow subgroup which is normal, that is, $n_s \geq 1$ for some $s \in \{p,q,r\}$.

Suppose not. Then, $n_s \geq s + 1$ for $s \in \{p,q,r\}$. Since $n_p \in \{q,r,qr\}$ and $q < r$ by assumption, we must have $n_p \geq q$. Moreover, since $n_q \in \{p,r,pr\}$ and $p<q$, $r\leq qr$ we get $n_q \geq r$. Finally, since $n_r \in \{p,q,pq\}$ and $p < q < r$, we obtain $n_r = pq$. Each $s$-Sylow subgroup has prime order and is thus cyclic. Hence, each of these subgroups has $s-1$ elements of order $s$ and so $G$ contains at least

  • $q(p - 1)$ elements of order $p$,
  • $r(q - 1)$ elements of order $q$,
  • $pq(r - 1)$ elements of order $r$.

Can you derive a contradiction from here?

Hubble
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Once you know that $G$ cannot be simple, it must have a normal subgroup. By Lagrange, it has order $p,q,r$, or $pq,pr$, or $qr$. In the first case we are done. In the second, use the fact that there are no such simple groups of order $pq$ where $p$ and $q$ are distinct primes.

pancini
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  • hey elliot I don't understand why is groups of order pq normal in our original group can you explain more ? I thought I understood it but I am second guessing myself. –  Oct 06 '16 at 16:50
  • Case $1$ there is a normal subgroup of order $p,q$, or $r$. Case $2$ there is a (normal) subgroup of order $pq, pr,$ or $qr$. In the second case, none of those groups could be simple, so, for example, a group of order $pq$ must have a nontrivial subgroup, and it must be order $p$ or $q$ by Lagrange. – pancini Oct 06 '16 at 19:27
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A group of square-free order is solvable, so you can consider minimal normal subgroup that is elementary abelian and so has order p,q or r.