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Let $p$ and $q$ be distinct primes. Let $q\in\mathbb Z/p\mathbb Z$ denote the class of $q$ modulo $p$ and let $k$ denote the order of $\bar q$ as an element of $(\mathbb Z/p\mathbb Z)^*$. Prove that no group of order $pq^l$ with $1\le l\le k$ is simple.


Side Note:

I know that this is a direct corollary of Burnside theorem, but I am not supposed use that.


My attempt:

Suppose that $G$ is a group of order $pq^l$. Let $n_p$ and $n_q$ denote the number of Sylow $p$-subgroups and Sylow $q$-subgroup respectively. Then we have $n_p|q^l$, $n_p\equiv 1\pmod p$ and $n_q|p, n_q\equiv 1\pmod q$. If either $n_p$ or $n_q$ is congruent to $1$ then we are done by Sylow theorems, otherwise we have $n_q=p$ and $n_p=q^h$, $q^h\equiv 1\pmod p,h\in\mathbb Z_{\ge 1}$.

Now suppose $P_1,P_2$ are distinct Sylow $p$-subgroups of $G$, then $$1=|P_1\cap P_2|=\frac{|P_1||P_2|}{|P_1P_2|}\ge\frac{p^2}{pq^l}=\frac{p}{q^l}.$$ Similarly, suppose $Q_1,Q_2$ are distinct Sylow $q$-subgroups of $G$, then we have $$ q^{l-1}\ge|Q_1\cap Q_2|=\frac{|Q_1||Q_2|}{|Q_1Q_2|}\ge\frac{q^{2l}}{pq^l}=\frac{q^l}{p} .$$
Therefore, we have $q\le p\le q^l$. Let $Q$ be a subgroup of order $q^{l-1}$, then consider the set $$ H:=P_1 Q .$$ Note that $$|H|=\frac{|P_1||Q|}{|P_q\cap Q|}=pq^{l-1}$$ since $q$ is the smallest prime dividing $|G|$, it suffices to show that $H$ is actually a subgroup of $G$. Then I am stuck... In addition, I am confused about all the conditions regarding $k$. Can someone give me a hint? Thank you.

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    $\begingroup$ Recall that $n_p | q^l$, thus $n_p=q^t$ for some $0 \leq t \leq l$. Thus, as you wrote, $q^t=1$ mod $p$... By the definition of $k$, either $t=0$ or $t \geq k$, ie $t=l=k$ or $n_p=1$. Assume $n_p,n_q > 1$. Then there are $q^l$ $p$-Sylows (since they do meet only at identity, this makes $(p-1)q^l$ elements of order $p$) and more than $3/2 q^l-1$ distinct non-identity elements of some $q$-Sylow, which makes too many elements for the group. $\endgroup$ – Mindlack Jun 28 at 11:30
  • $\begingroup$ Compare also with this post. $\endgroup$ – Dietrich Burde Jun 28 at 13:13
  • $\begingroup$ @Mindlack Thank you! By the way, can you explain how do you get $3/2q^l-1$? Besides, I have also posted an answer below with perhaps a slightly different approach deducing the number of non-identity elements exceeds the group order. $\endgroup$ – Bach Jun 28 at 13:48
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    $\begingroup$ @Bach: Let $A,B$ be two distinct $q-Sylow$, then $A \cap B$ is a strict subgroup of $A$, thus has cardinality at most $|A|/2=q^l/2$ (actually $q^{l-1}$ is a better bound, but it is not necessary). Therefore $|A\cup B| \geq 2q^l-q^l/2=3/2q^l$ ($(2q-1)q^{l-1}$ is a better lower bound again, but it is not important). $\endgroup$ – Mindlack Jun 28 at 16:10
  • $\begingroup$ @Mindlack I see. Thank you! $\endgroup$ – Bach Jun 29 at 10:16
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As @Mindlack suggested in the comment, $q^h\equiv 1\pmod p$ implies $k|h$, but $h\le l\le k$, we have $h=l=k$. In this case, we have at least $N=n_p(p-1)+n_q(q^k-q^{k-1})$ distinct non-identity elements. However, \begin{align} N&=n_p(p-1)+n_q(q^k-q^{k-1})\\ &=q^k(p-1)+p(q^k-q^{k-1})\\ &\ge q^k(p-1)+q(q^k-q^{k-1})\\ \end{align} since we have already proved $q\le p$.

Note that \begin{align} q^k(p-1)+q(q^k-q^{k-1})&=q^kp-2q^k+q^{k+1}\\ &=pq^k+q^k(q-2)\\ &\ge pq^k \end{align} since $q\ge 2$. Thus we have $N\ge pq^k$, but the identity has not been counted yet. Contradiction!

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