No group of order $pq^l$ with $1\le l\le k$ is simple.

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.

• 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. – Mindlack Jun 28 at 11:30
• Compare also with this post. – Dietrich Burde Jun 28 at 13:13
• @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. – Bach Jun 28 at 13:48
• @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). – Mindlack Jun 28 at 16:10
• @Mindlack I see. Thank you! – Bach Jun 29 at 10:16

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!