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I was trying to prove that if $p<q$ are two prime numbers, $n$ is a positive integer and $G$ is a group of order $p^nq$, then $G$ is not simple.

What I know so far is that, if I assume $G$ simple, $n_p = q \equiv 1 \pmod p$, $n_q= p^k \equiv 1 \pmod q$, for some positive $k$ (by Sylow's theorems), and that there is an element $g$ in a $p$-Sylow whose coniugacy class $g^G$ has size $q$ (by taking a nonidentical element in the center of a $p$-Sylow). Now I'm stuck. How can I go on? Is there anything particular about coniugacy classes of prime size?

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Let $G$ act on the Sylow $p$-subgroups (of which there are $q$) via conjugation. Because we assume $G$ is simple, this action embeds $G$ into $S_q$. In $S_q$, the elements of order $q$ are self-centralizing. Thus there are $p^n$ elements of order $q$ in $G$. A counting argument shows there is only one Sylow $p$-subgroup in $G$, which is a contradiction.


EDIT: Sorry, the argument above doesn't work. Here is one that does, from Isaacs's Finite Group Theory:

Let $P_n$, $1\le n\le q$ be the Sylow $p$-subgroups. Pick $K=P_i\cap P_j$ to be the intersection of two of these Sylow subgroups, such that $|K|$ is maximal. (Note a simple counting argument shows $|K|>1$.)

Now the normalizer $N_G(K)$ is such that $N_G(K)\cap P_i > K$ and $N_G(K)\cap P_j > K$, because normalizers grow in $p$-groups. By the maximality of $|K|$, $N_G(K)$ is not contained in a $p$-group. So there is a Sylow $q$-subgroup $Q$ contained in $N_G(K)$. Since $P_iQ=G$, we can write any element in $G$ as $hm$, with $h\in P_i$ and $m\in Q$. Then $P_i^{hm}=P_i^m\supset K^m=K$, since $Q\in N_G(K)$. So $K$ is contained in all Sylow $p$-subgroups, and is thus normal.

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  • $\begingroup$ How do you go from the elements of order $q$ are self-centralizing to there are $p^n$ elements of order $q$ in $G$? $\endgroup$ – Human Oct 27 '16 at 7:09
  • $\begingroup$ @DerekHolt. But aren't the cycles of the same order conjugates?? $\endgroup$ – Human Oct 27 '16 at 9:14
  • $\begingroup$ Yes you are right, some of them are. So I don't think this argument works. I will delete my earlier comments. $\endgroup$ – Derek Holt Oct 27 '16 at 9:22

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