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If I have a group $G$ with order $pq$ with $p,q$ primes, $p\neq q$. Then I want to try to apply Sylow's third theorem.

I want to argue that there is only 2 proper subgroups of $G$.

Since $p$ divides $|G|$ then let $n_p$ be the number of subgroups of $G$ with order $p$. Then by sylow's third theorem we have $$n_p =_p 1 \quad $$ $$n_p | \frac{pq}{p}=q$$

Since $q$ is prime, so we can only have $n_p = 1$ or $n_p=q$. But $n_p=q$ doesn't satisfy the first, and so $n_p=1$. A similar argument can be said on $n_q$

So there is 1 subgroup of order $p$, and 1 subgroup of order $q$. Does that work?

AspiringMat
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  • Your conjecture is false. E.g. let $G = S_3$, so $|G| = 6 = 2\cdot 3$. This group has three subgroups of order $2$, and one of order $3$. –  Dec 13 '18 at 03:58
  • related https://math.stackexchange.com/questions/1502186/groups-of-order-pq-are-cyclic/2975741#2975741 – qwr Dec 13 '18 at 04:00
  • I see. Thank you for clarifying – AspiringMat Dec 13 '18 at 04:02
  • I guess my proof holds only for the smaller prime of $p.q$. The other one might have other solutions. – AspiringMat Dec 13 '18 at 04:03
  • There are two possible groups of order $pq$. One is cyclic, in which case your conjecture is correct. The other can only occur when $p|q-1$ (assuming $p\lt q$). in which case there is the possibility that the group is nonabelian, there is a unique $q$-subgroup, but there are more $p$-subgroups, as in the case of $S_3$. – Arturo Magidin Dec 13 '18 at 05:53
  • "But $n_p=q$ doesn't satisfy the first" That statement is false. For example if $p=2$ then every prime $q>2$ satisfies $q=_p 1$. Other example is $p=5$ and $q=11$ and so on. – freakish Dec 13 '18 at 10:27

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